Today's concept : Finding HCF n LCM of typical values.
#1 : to find the HCF, LCM quickly.
i know most of us wud know this....if v have few nos., 20,40,50,80,180...to find their LCMs, HCF...there's a slightly quick method...
express them in prime nos.
20 = 2^2 x 5
40 = 2^3 x 5
50 = 5^2 x 2
80 = 2^4 x 5
180 =3^2 x 2^2 x 5
now to HCF, see highest power of all prime nos. that are common to all nos.
2 - 2
3- 0
5 - 1
hence hcf is 2^2 x 5 = 20
to find lcm...see highest power of all prime nos across all nos.
2 - 4
3 - 2
5 - 1
hence, lcm = 2^4 x 3^2 x 5 = 1620.
#2 To find HCF and LCM of the form-
2222....30 times.
3333....70 times.
to solve such questions...
for HCF..
take hcf of no. thats being repeated...i.e. hcf of 2 & 3. i.e. 1
take hcf of no. of time these nos. are being repeated...i.e. hcf of 30 n 70...thats 10.
so the hcf is 111...written 10 times.
For LCM...
take lcm of no. thats being repeated...i.e. lcm of 2 & 3. i.e. 6
take lcm of no. of time these nos. are being repeated...i.e. lcm of 30 n 70...thats 210.
so the hcf is 666...written 210 times.
#3...to find hcf and lcm of following form...
2^300 - 1, 8^250 - 1.
the idea is..a^n - b^n is always divisible by a-b. so v need to find highest a-b that will divide a^n - b^n and smallest term that'll be divisible by a^n - b^n.
express them in a common base.
2^300 - 1 and 2^750 -1.
to find hcf...
take hcf of powers i.e. hcf of 300 and 750...i.e. 150
so the hcf is 2^150 - 1.
to find lcm....
take lcm of powers i.e. lcm of 300 and 750...i.e. 1500
so the hcf is 2^1500 - 1.
Questions :
find hcf and lcm of:
1.2222...250 times and 8888...300 times
2. 333....120 times and 1111...400 times
3. 111...700 times and 9999...200 times.
4. HCF of 33333...200 times. and 777777.....300 times
5. 32^250 -1 & 16 ^ 100 - 1.
6. 81^100 -1 & 243 ^ 200 - 1.
7. 343^150-1 & 2401^100 - 1.
8. 125^200 - 1 & 625^120 - 1.
9. 169^320 - 1 & 32^160 - 1.
for the following questns...
mark 1. - stmt 1 is sufficient.
2- stmt 2 is suff.
3-both are reqd to solve the questn.
4-either is suff.
5-both insufficient.
10. what is the hcf of 5 nos., a,b,c,d,e?
stmt 1 - a=72,b=4,c=6
stmt 2 - d= 8, e = 27.
easy set...hope many wud get all correct...
Friday, July 3, 2009
finding squares and close multiplications quickly...
Today's concept....finding squares and close multiplications quickly...
In a hurry guys....so few small concepts that'll help u save sum time...n avoid cramming....which i've always maintained is the best ill-preparation for cat.
#1 finding squares.
step 1. think of a base which is a multiple of 10 or 100 (whichever nearer to the no. whose square is to be determined.)
suppose 34^2.
a gud base wud be 30.
express 34 as 30 + 4 (i.e. base + 4)...i'll call this base + deviation
step 2.
ldigit(s) before 0 in base x (no. + deviation) | deviation^2 where | is an imaginary line separating 2 parts of calculation.
i.e. 3x(34 + 4) | 4^2
i.e 114 | 16.
now the important thing....no. of digits on right side of the imaginary line shud be exactly same as the no. of zeros in our base..dont forget this!!!!
since our base 30 has only one 0 at the end, v can have only one digit on right side of the imaginary line....the less powerful digit...thats 6.
and thus, one has to be carried to the left side....so 114 will become 115.
so the expression becomes...115 | 6
hence, the answer is 1156.
look at few more examples for practice...
28^2
base 20
2x(28 + 8 ) | 8x8
= 72 | 64
= 784
106^2
106 + 6 | 6x6
base 100
=112 | 36
=11236 . since 100 has 2 zeroes, there shud be 2 digits on right side of the line....
103^2
base 100
103 + 3 | 3x3
=106 | 9
= 10609. ensure the no. of digits are exactly same as the no. of zeroes in base....so expressed 9 as 09....a 2 digit no.
312^2
base 300
3x(312 + 12) | 12^2
3x324 | 144
972 | 144
= 97344....1 of 144 carried.
smart ways of using this method....
98^2. now if v take base 100, v can avoid multiplication by 9.
so base 100.
98 - 2 | -2^2
= 96 | 04
= 9604....funny ha?
197^2
base 200
2x(197 - 3) | -3^2
=388 | 09
= 38809.
#2 : finding product of nos when they lie nearby.
all digits before 0 in base x (first no. + deviation of second no.) | deviation of first no. x deviation of second no.
rest, everything is same...
examples...
27 x 22 =
base 20
2x(27 + 2) | 7x2
= 58 | 14
= 594
103 x 108
base 100
= 103 + 8 | 3x8
= 111 | 24
=11124
97 x 102
base 200
97 + 2 | -3 x 2
=99 | -6
= 9900 - 6
= 9894
197 x 199
base 200
= 2x(197 - 1) | -3 x -1
= 392 | 03
= 39203
37 x 31
base 30
37 + 1) x 3 | 7x1
= 114 | 7
= 1147.
32 x 24
base 30
32 - 8 | 2 x -8
= 24 | -16
= 240 - 16
= 226.
In a hurry guys....so few small concepts that'll help u save sum time...n avoid cramming....which i've always maintained is the best ill-preparation for cat.
#1 finding squares.
step 1. think of a base which is a multiple of 10 or 100 (whichever nearer to the no. whose square is to be determined.)
suppose 34^2.
a gud base wud be 30.
express 34 as 30 + 4 (i.e. base + 4)...i'll call this base + deviation
step 2.
ldigit(s) before 0 in base x (no. + deviation) | deviation^2 where | is an imaginary line separating 2 parts of calculation.
i.e. 3x(34 + 4) | 4^2
i.e 114 | 16.
now the important thing....no. of digits on right side of the imaginary line shud be exactly same as the no. of zeros in our base..dont forget this!!!!
since our base 30 has only one 0 at the end, v can have only one digit on right side of the imaginary line....the less powerful digit...thats 6.
and thus, one has to be carried to the left side....so 114 will become 115.
so the expression becomes...115 | 6
hence, the answer is 1156.
look at few more examples for practice...
28^2
base 20
2x(28 + 8 ) | 8x8
= 72 | 64
= 784
106^2
106 + 6 | 6x6
base 100
=112 | 36
=11236 . since 100 has 2 zeroes, there shud be 2 digits on right side of the line....
103^2
base 100
103 + 3 | 3x3
=106 | 9
= 10609. ensure the no. of digits are exactly same as the no. of zeroes in base....so expressed 9 as 09....a 2 digit no.
312^2
base 300
3x(312 + 12) | 12^2
3x324 | 144
972 | 144
= 97344....1 of 144 carried.
smart ways of using this method....
98^2. now if v take base 100, v can avoid multiplication by 9.
so base 100.
98 - 2 | -2^2
= 96 | 04
= 9604....funny ha?
197^2
base 200
2x(197 - 3) | -3^2
=388 | 09
= 38809.
#2 : finding product of nos when they lie nearby.
all digits before 0 in base x (first no. + deviation of second no.) | deviation of first no. x deviation of second no.
rest, everything is same...
examples...
27 x 22 =
base 20
2x(27 + 2) | 7x2
= 58 | 14
= 594
103 x 108
base 100
= 103 + 8 | 3x8
= 111 | 24
=11124
97 x 102
base 200
97 + 2 | -3 x 2
=99 | -6
= 9900 - 6
= 9894
197 x 199
base 200
= 2x(197 - 1) | -3 x -1
= 392 | 03
= 39203
37 x 31
base 30
37 + 1) x 3 | 7x1
= 114 | 7
= 1147.
32 x 24
base 30
32 - 8 | 2 x -8
= 24 | -16
= 240 - 16
= 226.
to solve tricky questions based on no. of factors of a number.
Today's concept...to solve tricky questions based on no. of factors of a number.
Despite the interest shown in the concept yesterday, not many cud solve the question efficiently...so, best concept for the day can be to discuss yesterday's problems...have a quick, practical solution to them...and practice a lot of similar, different questions on same concept. here we go...
Find the smallest no. that has exactly.....
1. 16 factors
2. 12 factors
3. 60 factors
questions are lil tricky...but if u get the concept...they become child's play...see how...
16 factors...that mean product of (powers +1) of all the prime nos = 16.
now 16 can be achieved in following ways...
16
8x2
4x4
2x4x2
2x2x2x2.
by sheer common sense, v can say the highest power shud go to the smallest prime no. i.e. 2....and as v proceed, smaller powers shud be given to higher prime nos.
Concept:
powers shud reduce and the corresponding prime nos. shud increase.
so...
16...the no. is 2^15
8x2...the no. is 2^7 x 3
4x4...the no. is 2^3 x 3^3.
2x4x2...the no. is 2^3 x 3 x 5
2x2x2x2.....the no. is 2x3x5x7.
first three nos can be easily discarded as they are too big...just calculate last two nos, they are 120 and 210...120 is smaller and hence the answer.
important: please avoid craming...in few cases last way might give best answer..in other cases, 2nd last one...its always advicable to form patterns n check the closer ones.
2. 12 factors
12 = 12 or 4x3 or 2x2x3. easily answer wud be 2^3 x 3^2 or 2^2 x 3 x 5
the corresponding values are 72 n 60. hence, the answer is 60.
similarly, Q3 also.
4. how many factors of 27000 are perfect cubes?
i realy wonder y nobody cud get this right...u just need to form combinations and check which combinations give u cubes...they are 27000,27,1000, etc. am not discussing this question...i hope when i give a similar question today...i get few correct answers.
This question is still open for answers/discussion (so are others...but if u can answer this...with an xplanation, it'll be gr8 )
how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...
5. 3 terms
6. 4 terms
7. 5 terms
8. 6 terms
concept:
to have first n last term as the given terms...the difference b/w the two terms shud be divisible by the common difference. so u basically have to find how many such common differences exist...for every common difference...u'll get a new AP.
for practice, lets take a small interval [1,15]
the difference is 15-1=14. now 14 is divisible by 1,2,7,14...four different integers... so v can have four different APs...if v take a common difference other than these four values...the last term wont lie in that AP.
e.g if v take the common diff = 4, the AP wud be 1,5,9,13,17...see 15 didnt lie...
so, 4 APs are possible.
now if a conditions is attached...there shud be atleast three terms...it means that AP with 2 terms shud be neglected...
1,15 is an AP with jus 2 terms...so it shud be beglected...remember, an AP with 2 terms always lies in any interval.
so the answer wud be 4-1 = 3
if the condition is atleast four terms...then the AP with 2 terms as well as the AP with 3 terms shud be neglected.
we know that an AP with 2 terms is bound to exist...lets c if an AP with 3 terms also exists.
an AP with 3 terms will look like... 1, x, 15
see...there are 2 intervals... x-1 and 15-x. hence for a 3 term AP to exist, the difference shud be divisible by 2.
since 14 is divisible by 2, we further reduce the answer by 1...so APs with atleast 4 terms are 3-1 =2
now, if the conditions is...atleast 5 terms...v need to check if AP with 4 terms exists...
such an AP wud luk like...
1, x, y, 15
see, there are 3 intervals...since 14 is not divisible by 3, such an AP does not exist. so the answer remains 2.
similarly, for atleast 6 terms, v check if 14%4 = 0...since no, the answer is again 2
for 7 terms 14%5 is not 0, the answer is 2
for atleast 8 terms...14%6 is not 0 so the answer is again 2.
for 9 terms, 14%7 = 0. hence answer becomes 2-1 =1
keep on proceeding like this...the soln wont be so bulky...its been done like this for ease of understanding...for ease of calculation...see how to proceed...
how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...
5. 3 terms
6. 4 terms
7. 5 terms
8. 6 terms
3535 - 1235 = 2300.
2300 = 23x2^2x5^2 . hence, no. of factors = 3x3x2 = 18. (check yesterday's concept if missed)
how many APs...18.
how many with atleast 3 terms?
since 2300%1 = 0, 18-1 = 17
how many with atleast 4 terms?
2300%2 = 0, hence 17-1 = 16
how many with atleast 5 terms?
2300%3 =/ 0, answer remains 16. where =/ means not equal to
how many with atleast 6 terms?
2300%4 = 0, 16-1 = 15
how many with atleast 7 terms?
2300%5 = 0, 15-1 = 14
how many with atleast 8 terms?
2300%6 =/ 0 , answer remains 14
how many with atleast 9 terms?
2300%7 =/ 0, answer remains 14
and so on...
to check for atleast n terms, v need divisibility till n-2...i'll appreciate if u dont cram...but understand it...i've neva learnt it...its jus an observation...
so, the answers to above 3 questions wud be 17,16,16,15.
finally...
How many values of a are possible if x^2 + ax + 2400 has...
9. integral roots
10. roots which are natural nos.
this is an actually tricky problem...sad that nobody came up with this...see...
concept:
to solve ax2 + bx + c =0,
we break it as ax2 + mx + nx + c = 0, such that m*n = a*c.
here, a*c = 1*2400 = 2400. so v need to find in how many ways can 2400 be expressed as product of 2 nos. every such pair of nos. will give a new value of the coefficient of x.
2400 = 2^5 x 5^2 x 3
no. of factors = 6x3x2 = 36.
but these factors have to exist in pairs...e.g when v use one factor 2 (to express 2400 as 2x1200) the other factor...1200 is automatically used...so total pairs possible are ...
36/2 = 18.
but relax...this is not it....again...equal no. of negative pairs exist...i.e. 2x1200 corresponds to -2 x -1200. although the product is same as the reqd product i.e 2400...the sum is different...its 1202 n -1202...n v need to find different values of sum...hence the answer wud
18x2 = 36 again...
9. integral roots
10. roots which are natural nos.
9. for integral roots...the answer wud be 36
10. values wud be 1/2 the total possible values as negative roots aint allowed...36/2 = 18 is the answer
Kudos to rockeezee...who got 5 correct answers...so too to junoonmba,vani for their active participation!!! wud love to see all of u gett'n of the following correct today...
Today's questions...
smallest no. that has exactly.
1. 20 factors
2. 36 gactors
3. 30 factors
how many APs are possible such that the first term is 2454 , last term is 4254 and there are atleast...
4. 5 terms
5. 7 terms
6. 10 terms.
how many of the factors of 640000 have
7. perfect square roots.
8. perfect fourth roots
9. perfect cube roots
how many different values can 'A' take if x2 + Ax + 2500 has ...
10. integral roots
11. negative roots.
12. non-negative integral roots.
13. find the no. of factors 15! has (here comes the season of fresh concepts...which'll derive from older ones...he he)
14. find the no. of factors 18! has
Minimum value that A can take if x2 + Ax + 900 has...
15. Integral roots.
16. Negative integral roots.
17. Positive integral roots.
Despite the interest shown in the concept yesterday, not many cud solve the question efficiently...so, best concept for the day can be to discuss yesterday's problems...have a quick, practical solution to them...and practice a lot of similar, different questions on same concept. here we go...
Find the smallest no. that has exactly.....
1. 16 factors
2. 12 factors
3. 60 factors
questions are lil tricky...but if u get the concept...they become child's play...see how...
16 factors...that mean product of (powers +1) of all the prime nos = 16.
now 16 can be achieved in following ways...
16
8x2
4x4
2x4x2
2x2x2x2.
by sheer common sense, v can say the highest power shud go to the smallest prime no. i.e. 2....and as v proceed, smaller powers shud be given to higher prime nos.
Concept:
powers shud reduce and the corresponding prime nos. shud increase.
so...
16...the no. is 2^15
8x2...the no. is 2^7 x 3
4x4...the no. is 2^3 x 3^3.
2x4x2...the no. is 2^3 x 3 x 5
2x2x2x2.....the no. is 2x3x5x7.
first three nos can be easily discarded as they are too big...just calculate last two nos, they are 120 and 210...120 is smaller and hence the answer.
important: please avoid craming...in few cases last way might give best answer..in other cases, 2nd last one...its always advicable to form patterns n check the closer ones.
2. 12 factors
12 = 12 or 4x3 or 2x2x3. easily answer wud be 2^3 x 3^2 or 2^2 x 3 x 5
the corresponding values are 72 n 60. hence, the answer is 60.
similarly, Q3 also.
4. how many factors of 27000 are perfect cubes?
i realy wonder y nobody cud get this right...u just need to form combinations and check which combinations give u cubes...they are 27000,27,1000, etc. am not discussing this question...i hope when i give a similar question today...i get few correct answers.
This question is still open for answers/discussion (so are others...but if u can answer this...with an xplanation, it'll be gr8 )
how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...
5. 3 terms
6. 4 terms
7. 5 terms
8. 6 terms
concept:
to have first n last term as the given terms...the difference b/w the two terms shud be divisible by the common difference. so u basically have to find how many such common differences exist...for every common difference...u'll get a new AP.
for practice, lets take a small interval [1,15]
the difference is 15-1=14. now 14 is divisible by 1,2,7,14...four different integers... so v can have four different APs...if v take a common difference other than these four values...the last term wont lie in that AP.
e.g if v take the common diff = 4, the AP wud be 1,5,9,13,17...see 15 didnt lie...
so, 4 APs are possible.
now if a conditions is attached...there shud be atleast three terms...it means that AP with 2 terms shud be neglected...
1,15 is an AP with jus 2 terms...so it shud be beglected...remember, an AP with 2 terms always lies in any interval.
so the answer wud be 4-1 = 3
if the condition is atleast four terms...then the AP with 2 terms as well as the AP with 3 terms shud be neglected.
we know that an AP with 2 terms is bound to exist...lets c if an AP with 3 terms also exists.
an AP with 3 terms will look like... 1, x, 15
see...there are 2 intervals... x-1 and 15-x. hence for a 3 term AP to exist, the difference shud be divisible by 2.
since 14 is divisible by 2, we further reduce the answer by 1...so APs with atleast 4 terms are 3-1 =2
now, if the conditions is...atleast 5 terms...v need to check if AP with 4 terms exists...
such an AP wud luk like...
1, x, y, 15
see, there are 3 intervals...since 14 is not divisible by 3, such an AP does not exist. so the answer remains 2.
similarly, for atleast 6 terms, v check if 14%4 = 0...since no, the answer is again 2
for 7 terms 14%5 is not 0, the answer is 2
for atleast 8 terms...14%6 is not 0 so the answer is again 2.
for 9 terms, 14%7 = 0. hence answer becomes 2-1 =1
keep on proceeding like this...the soln wont be so bulky...its been done like this for ease of understanding...for ease of calculation...see how to proceed...
how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...
5. 3 terms
6. 4 terms
7. 5 terms
8. 6 terms
3535 - 1235 = 2300.
2300 = 23x2^2x5^2 . hence, no. of factors = 3x3x2 = 18. (check yesterday's concept if missed)
how many APs...18.
how many with atleast 3 terms?
since 2300%1 = 0, 18-1 = 17
how many with atleast 4 terms?
2300%2 = 0, hence 17-1 = 16
how many with atleast 5 terms?
2300%3 =/ 0, answer remains 16. where =/ means not equal to
how many with atleast 6 terms?
2300%4 = 0, 16-1 = 15
how many with atleast 7 terms?
2300%5 = 0, 15-1 = 14
how many with atleast 8 terms?
2300%6 =/ 0 , answer remains 14
how many with atleast 9 terms?
2300%7 =/ 0, answer remains 14
and so on...
to check for atleast n terms, v need divisibility till n-2...i'll appreciate if u dont cram...but understand it...i've neva learnt it...its jus an observation...
so, the answers to above 3 questions wud be 17,16,16,15.
finally...
How many values of a are possible if x^2 + ax + 2400 has...
9. integral roots
10. roots which are natural nos.
this is an actually tricky problem...sad that nobody came up with this...see...
concept:
to solve ax2 + bx + c =0,
we break it as ax2 + mx + nx + c = 0, such that m*n = a*c.
here, a*c = 1*2400 = 2400. so v need to find in how many ways can 2400 be expressed as product of 2 nos. every such pair of nos. will give a new value of the coefficient of x.
2400 = 2^5 x 5^2 x 3
no. of factors = 6x3x2 = 36.
but these factors have to exist in pairs...e.g when v use one factor 2 (to express 2400 as 2x1200) the other factor...1200 is automatically used...so total pairs possible are ...
36/2 = 18.
but relax...this is not it....again...equal no. of negative pairs exist...i.e. 2x1200 corresponds to -2 x -1200. although the product is same as the reqd product i.e 2400...the sum is different...its 1202 n -1202...n v need to find different values of sum...hence the answer wud
18x2 = 36 again...
9. integral roots
10. roots which are natural nos.
9. for integral roots...the answer wud be 36
10. values wud be 1/2 the total possible values as negative roots aint allowed...36/2 = 18 is the answer
Kudos to rockeezee...who got 5 correct answers...so too to junoonmba,vani for their active participation!!! wud love to see all of u gett'n of the following correct today...
Today's questions...
smallest no. that has exactly.
1. 20 factors
2. 36 gactors
3. 30 factors
how many APs are possible such that the first term is 2454 , last term is 4254 and there are atleast...
4. 5 terms
5. 7 terms
6. 10 terms.
how many of the factors of 640000 have
7. perfect square roots.
8. perfect fourth roots
9. perfect cube roots
how many different values can 'A' take if x2 + Ax + 2500 has ...
10. integral roots
11. negative roots.
12. non-negative integral roots.
13. find the no. of factors 15! has (here comes the season of fresh concepts...which'll derive from older ones...he he)
14. find the no. of factors 18! has
Minimum value that A can take if x2 + Ax + 900 has...
15. Integral roots.
16. Negative integral roots.
17. Positive integral roots.
finding factors of a no. n sorting relevant questions.
Today's concept...finding factors of a no. n sorting relevant questions.
concepts...
#1. to find total no. of factors of a number.
=>first of all...express it in terms of prime numbers
e.g 1260^4.
express it as (2x2x3x3x5x7)^4 = 2^8 * 3^8 * 5^4*7^4.
=>now add 1 to the powers of every prime no. n multiply them all...u get the total no. of factors of 1260^4.
i.e (8+1)*(8+1)*(4+1)*(4+1) = 2025 = total no. of factors of 1260^4.
#2 to find no. of odd factors of a number.
leave power of 2 and multiply powers of all other prime nos after adding one to them.
i.e. (8 + 1)*(4+1)*(4+1) = 225 = no. of odd factors of 1260^4
#3 to find no. of even factors of a number.
take the difference of total factors and odd factors.
i.e. 2025 - 225 = 1800. = no. of even factors of 1260^4
#4 to find in how many ways can a given no. be represented as product of two relatively prime factors.
in this case, the power of prime no. becomes irrelevant as all the powers shud lie with the same factor else the two factors wont be relatively prime.
i.e in the above case 1260 = (2x2x3x3x5x7)^4 = 2^8 * 3^8 * 5^4*7^4.
now powers 8,8,4,4 have no importance...whats important is how many prime nos. are there...they are four...viz 2,3,5,7. hence 4.
now v have to form nos. using these 4 primr nos. its a like a question askin u...if u have 4 sweets...in how many ways can u eat them? the answer is 4C0..when u eat none... + 4C1 when eat any one...+4C2 + ...4C4...when u eat all.
similarly here, answer wud be 4C0 + 4C1 + 4C2 ...4C4 = 2^4 = 16.
but the factors have to exist in pairs...hence 2^4/2 = 8 factors are possible.
on a general note...a no. formed of n prime nos has 2^(n-1) pair of co-prime factors
Answers to yesterday's questions
1. 32^23 + 17^23 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
2. 32^23 - 17^23 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
3. 32^232 + 17^232 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
4. 32^232 - 17^232 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
5. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by....
a. 40 b. 20 c. 80 d. all of these. e. none of these.
what is the remainder when 42527152653425416242624272427215287 is divided by :
6. 16----7
7. 32----23
8. 64----55
kudos to irevani, gk who got all correct...so too to deep, rockeezee...who got jus 1 wrong....bhaiyon...if a no. is divisible by 80...its obviously divisible by 40 n 20.
questions for today...
on the very simple fundas above, very tricky questions can be framed.
Find the smallest no. that has exactly.....
1. 16 factors
2. 12 factors
3. 60 factors
4. how many factors of 27000 are perfect cubes?
how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...
5. 3 terms
6. 4 terms
7. 5 terms
8. 6 terms
How many values of a are possible if x^2 + ax + 2400 has...
9. integral roots
10. roots which are natural nos.
concepts...
#1. to find total no. of factors of a number.
=>first of all...express it in terms of prime numbers
e.g 1260^4.
express it as (2x2x3x3x5x7)^4 = 2^8 * 3^8 * 5^4*7^4.
=>now add 1 to the powers of every prime no. n multiply them all...u get the total no. of factors of 1260^4.
i.e (8+1)*(8+1)*(4+1)*(4+1) = 2025 = total no. of factors of 1260^4.
#2 to find no. of odd factors of a number.
leave power of 2 and multiply powers of all other prime nos after adding one to them.
i.e. (8 + 1)*(4+1)*(4+1) = 225 = no. of odd factors of 1260^4
#3 to find no. of even factors of a number.
take the difference of total factors and odd factors.
i.e. 2025 - 225 = 1800. = no. of even factors of 1260^4
#4 to find in how many ways can a given no. be represented as product of two relatively prime factors.
in this case, the power of prime no. becomes irrelevant as all the powers shud lie with the same factor else the two factors wont be relatively prime.
i.e in the above case 1260 = (2x2x3x3x5x7)^4 = 2^8 * 3^8 * 5^4*7^4.
now powers 8,8,4,4 have no importance...whats important is how many prime nos. are there...they are four...viz 2,3,5,7. hence 4.
now v have to form nos. using these 4 primr nos. its a like a question askin u...if u have 4 sweets...in how many ways can u eat them? the answer is 4C0..when u eat none... + 4C1 when eat any one...+4C2 + ...4C4...when u eat all.
similarly here, answer wud be 4C0 + 4C1 + 4C2 ...4C4 = 2^4 = 16.
but the factors have to exist in pairs...hence 2^4/2 = 8 factors are possible.
on a general note...a no. formed of n prime nos has 2^(n-1) pair of co-prime factors
Answers to yesterday's questions
1. 32^23 + 17^23 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
2. 32^23 - 17^23 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
3. 32^232 + 17^232 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
4. 32^232 - 17^232 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
5. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by....
a. 40 b. 20 c. 80 d. all of these. e. none of these.
what is the remainder when 42527152653425416242624272427215287 is divided by :
6. 16----7
7. 32----23
8. 64----55
kudos to irevani, gk who got all correct...so too to deep, rockeezee...who got jus 1 wrong....bhaiyon...if a no. is divisible by 80...its obviously divisible by 40 n 20.
questions for today...
on the very simple fundas above, very tricky questions can be framed.
Find the smallest no. that has exactly.....
1. 16 factors
2. 12 factors
3. 60 factors
4. how many factors of 27000 are perfect cubes?
how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...
5. 3 terms
6. 4 terms
7. 5 terms
8. 6 terms
How many values of a are possible if x^2 + ax + 2400 has...
9. integral roots
10. roots which are natural nos.
few tricky questions on divisibility
Today's concept...few tricky questions on divisibility
concepts...
a^n - b^n is always divisible by a-b
a^n - b^n is divisible by a+b when n is even.
a^n + b ^n is divisible by a+b when n is odd.
a^n + b^n + c^n +... is divisible by a+b+c+.... when n is odd.
when last n digits of a no. are divided by 2^n, the remainder is same as the remainder when the entire no. is divided by 2^n.
Answers to yesterday's questions.
1. smallest no. that must be added to 1000 so that the resultant no. leaves remainders 2,3,4,5 with 5,6,7,11 respectivelt----667
2. smallest no. that leaves remainders 1,2,5,6, when divided successively by 2,3,4,23.----such a no. doesnt exist as 5 rem with aint possible.
3. smallest no. that leaves remainders 4 everytime when successively divided by 7,5,10,13 respectively.----872
4. a no. leaves remainders 2,5,3,7 when successively divided by 3,7,6,9. what is the remainder when this no. is divided by 126?----80
5.find the largest no. that leaves same remainder when it divides 2345,7645,9845,6595 and 10095.----50
6.a no. when divided 88 leaves remainder 3. what is the remainder when its divided by 11? ----3
7.a no. when divided by 391 leaves rem. of 49. find the remainder when its divided by 39...options 29,10,none of these,cannot be determined.---cannot be detrmnd
kudos to rockeezee who got most of them right!
questions for today...
1. 32^23 + 17^23 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
2. 32^23 - 17^23 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
3. 32^232 + 17^232 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
4. 32^232 - 17^232 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
5. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by....
a. 40 b. 20 c. 80 d. all of these. e. none of these.
what is the remainder when 42527152653425416242624272427215287 is divided by :
6. 16
7. 32
8. 64
concepts...
a^n - b^n is always divisible by a-b
a^n - b^n is divisible by a+b when n is even.
a^n + b ^n is divisible by a+b when n is odd.
a^n + b^n + c^n +... is divisible by a+b+c+.... when n is odd.
when last n digits of a no. are divided by 2^n, the remainder is same as the remainder when the entire no. is divided by 2^n.
Answers to yesterday's questions.
1. smallest no. that must be added to 1000 so that the resultant no. leaves remainders 2,3,4,5 with 5,6,7,11 respectivelt----667
2. smallest no. that leaves remainders 1,2,5,6, when divided successively by 2,3,4,23.----such a no. doesnt exist as 5 rem with aint possible.
3. smallest no. that leaves remainders 4 everytime when successively divided by 7,5,10,13 respectively.----872
4. a no. leaves remainders 2,5,3,7 when successively divided by 3,7,6,9. what is the remainder when this no. is divided by 126?----80
5.find the largest no. that leaves same remainder when it divides 2345,7645,9845,6595 and 10095.----50
6.a no. when divided 88 leaves remainder 3. what is the remainder when its divided by 11? ----3
7.a no. when divided by 391 leaves rem. of 49. find the remainder when its divided by 39...options 29,10,none of these,cannot be determined.---cannot be detrmnd
kudos to rockeezee who got most of them right!
questions for today...
1. 32^23 + 17^23 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
2. 32^23 - 17^23 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
3. 32^232 + 17^232 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
4. 32^232 - 17^232 is definetly divisible by....
a. 49 b. 15 c. 49 & 15 d. none of these.
5. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by....
a. 40 b. 20 c. 80 d. all of these. e. none of these.
what is the remainder when 42527152653425416242624272427215287 is divided by :
6. 16
7. 32
8. 64
concept regarding numbers giving specific remainders with specific divisors.
Today's concept...an extension of concept regarding numbers giving specific remainders with specific divisors.
Type#5
Smallest no. that must be subtracted from 1000 so that the resultant no. leaves remainders 1,3,4,8 with divisors 2,6,5,13 respectively.
from yesterday's approach, v can find out...the smallest no. satisfying all 4 conditions is 99. now to 99 if v add lcm of 2,6,5,13 i.e. 390, the remainders will remain unchanged.
so v need 99 + 390k such that the resultant value is just below 1000.
easily, for k=2, we get one such value....99+390x2 = 879.
hence, ans is 1000-879=121
Type#6
smallest no. that leaves remainders 3,2,4 when successively divided by 5,6,7 respectively.
for such questions...start approaching from the rear end...
we want 4 remainder with 7...the smallest such no. is 4 itself.
now this 4 must have come after a no. was divided by 6
so the no. must have been 4x6+2(remainder with 6) = 26
now, 26 was the quotient when sum no. was divided by 5
so the no. must have been 26x5 + 3(remainder with 5) = 133
so, the answer is 133.
Type#7
a no. leaves remainder 3 when divided by 5 and remainder 8 when successively divided by 11. what is the remainder when this no. is divided by 55?
look at this question carefully...55 is lcm of earlier divisors 11,5... in such a case...the remainder with lcm as divisor wud be constant.
an easy approach for this problem...start from the rear end...take a small no. that leaves rem. 8 with 11...lets take 8.
this 8 is quotient when the main no. is divided by 5...it also leaves remainder 3.
hence, the main no. is
8x5 + 3 = 43
43%55 = 43 answer.
Type#8
Find the largest no. that leaves same remainder when it divides 3398 and 6578.
the concept is very simple...to leave same remainder...difference between two dividents must be divisible by the divisor.
i.e. 6578-3398 = 3180 shud be divisible by the divisor to leave same remainders.
largest no. that divides 3180 is 3180 itself.
hence, the answer is 3180.
Type#9
Find the largest no. that leaves same remainder when it divides 16009,9009,7509 and 14009.
the approach is same... take difference of the nos in ascending or descending order....
i.e. 16009-14009=2000,
14009-9009=5000
9009-7509=1500.
now to leave same remainder, each of the interval shud be divisible by the divisor.
hence, take hcf of 2000,5000,1500. i.e. 500
so, the answer is 500.
Type#10
If a no. is divided by 15, it leaves a remainder 7, if thrice the no. is divided by 5, then what is the remainder?
options...1,2,3,4,0
such questions are difficult to frame as one has to find a pattern b/w divisors n remainders...i know these questions are easy n v all can crack it easily...but the reason y am putting it here is bcoz i have a very short...practical approach for solving this question..
choose a no. that leaves 7 remainder with 15....lets take 7 only.
thrice 7 = 21
21%5 =1 (edited after vani's post)
since, the no. shud give same result for all values that give 7 rem. with 15, its better to take sum value n solve it instead of takin an algebraic approach...
hence, answer is 1.
Answers for yesterday's questions...
Q1. 2 when divided by 3,5,6 or 9 (other than 2)----92
Q2. 2,5,7 when divided by 7,10 and 12 respectively----415
Q3. 1,2,3,4 with 3,4,5,7 respectively.----298
Q4. 6 with 7,8,9,10 and 3 with 11.----20166
Q5. 3 with 6, 0 with 11, 3 with 5, 7 with 8----1023
Q6. 2 with 5, 7 with 8, 3 with 4, 5 with 7&11----2007
Q7. 1 with 11, 4 with 5, 9 with 10, 7 with 9.----529
kudos to vani for the active participation n gett'n most answers correct.
Today's questions...
1. smallest no. that must be added to 1000 so that the resultant no. leaves remainders 2,3,4,5 with 5,6,7,11 respectivelt
2. smallest no. that leaves remainders 1,2,5,6, when divided successively by 2,3,4,23.
3. smallest no. that leaves remainders 4 everytime when successively divided by 7,5,10,13 respectively.
4. a no. leaves remainders 2,5,3,7 when successively divided by 3,7,6,9. what is the remainder when this no. is divided by 126?
5.find the largest no. that leaves same remainder when it divides 2345,7645,9845,6595 and 10095.
6.a no. when divided 88 leaves remainder 3. what is the remainder when its divided by 11?
7.a no. when divided by 391 leaves rem. of 49. find the remainder when its divided by 39...options 29,10,none of these,cannot be determined.
Type#5
Smallest no. that must be subtracted from 1000 so that the resultant no. leaves remainders 1,3,4,8 with divisors 2,6,5,13 respectively.
from yesterday's approach, v can find out...the smallest no. satisfying all 4 conditions is 99. now to 99 if v add lcm of 2,6,5,13 i.e. 390, the remainders will remain unchanged.
so v need 99 + 390k such that the resultant value is just below 1000.
easily, for k=2, we get one such value....99+390x2 = 879.
hence, ans is 1000-879=121
Type#6
smallest no. that leaves remainders 3,2,4 when successively divided by 5,6,7 respectively.
for such questions...start approaching from the rear end...
we want 4 remainder with 7...the smallest such no. is 4 itself.
now this 4 must have come after a no. was divided by 6
so the no. must have been 4x6+2(remainder with 6) = 26
now, 26 was the quotient when sum no. was divided by 5
so the no. must have been 26x5 + 3(remainder with 5) = 133
so, the answer is 133.
Type#7
a no. leaves remainder 3 when divided by 5 and remainder 8 when successively divided by 11. what is the remainder when this no. is divided by 55?
look at this question carefully...55 is lcm of earlier divisors 11,5... in such a case...the remainder with lcm as divisor wud be constant.
an easy approach for this problem...start from the rear end...take a small no. that leaves rem. 8 with 11...lets take 8.
this 8 is quotient when the main no. is divided by 5...it also leaves remainder 3.
hence, the main no. is
8x5 + 3 = 43
43%55 = 43 answer.
Type#8
Find the largest no. that leaves same remainder when it divides 3398 and 6578.
the concept is very simple...to leave same remainder...difference between two dividents must be divisible by the divisor.
i.e. 6578-3398 = 3180 shud be divisible by the divisor to leave same remainders.
largest no. that divides 3180 is 3180 itself.
hence, the answer is 3180.
Type#9
Find the largest no. that leaves same remainder when it divides 16009,9009,7509 and 14009.
the approach is same... take difference of the nos in ascending or descending order....
i.e. 16009-14009=2000,
14009-9009=5000
9009-7509=1500.
now to leave same remainder, each of the interval shud be divisible by the divisor.
hence, take hcf of 2000,5000,1500. i.e. 500
so, the answer is 500.
Type#10
If a no. is divided by 15, it leaves a remainder 7, if thrice the no. is divided by 5, then what is the remainder?
options...1,2,3,4,0
such questions are difficult to frame as one has to find a pattern b/w divisors n remainders...i know these questions are easy n v all can crack it easily...but the reason y am putting it here is bcoz i have a very short...practical approach for solving this question..
choose a no. that leaves 7 remainder with 15....lets take 7 only.
thrice 7 = 21
21%5 =1 (edited after vani's post)
since, the no. shud give same result for all values that give 7 rem. with 15, its better to take sum value n solve it instead of takin an algebraic approach...
hence, answer is 1.
Answers for yesterday's questions...
Q1. 2 when divided by 3,5,6 or 9 (other than 2)----92
Q2. 2,5,7 when divided by 7,10 and 12 respectively----415
Q3. 1,2,3,4 with 3,4,5,7 respectively.----298
Q4. 6 with 7,8,9,10 and 3 with 11.----20166
Q5. 3 with 6, 0 with 11, 3 with 5, 7 with 8----1023
Q6. 2 with 5, 7 with 8, 3 with 4, 5 with 7&11----2007
Q7. 1 with 11, 4 with 5, 9 with 10, 7 with 9.----529
kudos to vani for the active participation n gett'n most answers correct.
Today's questions...
1. smallest no. that must be added to 1000 so that the resultant no. leaves remainders 2,3,4,5 with 5,6,7,11 respectivelt
2. smallest no. that leaves remainders 1,2,5,6, when divided successively by 2,3,4,23.
3. smallest no. that leaves remainders 4 everytime when successively divided by 7,5,10,13 respectively.
4. a no. leaves remainders 2,5,3,7 when successively divided by 3,7,6,9. what is the remainder when this no. is divided by 126?
5.find the largest no. that leaves same remainder when it divides 2345,7645,9845,6595 and 10095.
6.a no. when divided 88 leaves remainder 3. what is the remainder when its divided by 11?
7.a no. when divided by 391 leaves rem. of 49. find the remainder when its divided by 39...options 29,10,none of these,cannot be determined.
Some speed maths stuff
The approach would be to get the denominator to either a 100 or a 1000 because that is what percentages is all bout.
Simply focus on the fact that how do the given denominators reach 100/1000. I have left some blank because they are very obvious.
The following are the stations between 100 and 1000
100
111 - Reduce 10%
125 - Multiply by 8
133 - Reduce 1/4
150 - Reduce 1/3
166 - Multiply by 6
182 - Add 10%
200
222 - Reduce 10%
250
273 - Add 10%
300
333 - Reduce 10%
364
400
455 - Add 10%
500
555 - Reduce 10%
600
666 - Add Half
700
750 - Add 1/3
800
833 - Add 20%
875 - Add 1/7
900
910 - Add 9%
950 - Add 5%
1000
So Task (1) you have to mug up the values of these stations. It is very important that these values are memorized because this will help you in
knowing which number to reach from any given number. e.g. if the denominator is 887 you know you have to reach for 900 or 875 and so on.
Task (2) Practice!
Below is an approach to tackle (three digit /three digit) with consummate ease. We shall attempt to understand it with examples.
Example 1:
What is 145/182 ------------------79.5
Steps
1. add 10% to numerator and denominator...
2. it becomes 159/200 ...............which is 79.5
( the answer from the calculator is 79.6)
Example 2:
123/178....???
Step1:- which station is closest to it?.............................200?...or some say 150. Either is good. (identification took 2 seconds)
Step 2:- what do I have to do to go from 178 to 200/150..................add 22/subtract 28... (another 2 seconds.)
Step 3:- so if I add 22................i am actually adding slightly less than 13% of 178 to itself.
This part is tricky..here is how I got it:
This is how u need to think---
Our number is 178. Thus, 10% of it is 17.8
22 definitely more than 10% .
if 10% = 17.8 then 5% = 8.xx
And 2.5% = 4.xxx (don't even bother to calculate xxx)
which means its around 12.5........or 13 or 12..........
Add same to numerator...........10% of 123 = 12.3
1% = 1.23 so 2% = 2.46
So 12% = 15?
So it becomes 138/200 = 69%?
(calculator answer is 69.1).
All you require to calculate is what is 10% 1%...and approx stuff...any damn calculation works in less than 8 seconds.:satisfie:
Some thing like (2456*4567 - 2134*3214)/2134*3214 will taken 10 seconds maximum...
it works coz I had that time 1.5 years ago..and I aint kidding..
Lets take one more.
Example 3: 532/745?
Tough???
This's how it can be approached:
# Nearest station.........750.................so add 5..............about 1% or less..add 1% damnit
# Numerator now is 537.............(added 1%)
# Fraction is 537/750 add 1/3 each...........it is 71 something.
Remember here don't even attempt to do...537/3. Because denominator is a 1000 and not a 100. So one digit is redundant. So all I do is........53/3 approx..18...plus..53 = 71%..(will do if answer are spaced...)
answers not spaced? Then 537 +537/3 = 537+179 = 716 which makes it 71.6 (calculator is 71.5)
Dont worry if the last few statements were difficult to digest. Try solving a few questions and you'll get the crux of it.
Remember in CAT we don't find answers... we choose them!
Happy Computing
Simply focus on the fact that how do the given denominators reach 100/1000. I have left some blank because they are very obvious.
The following are the stations between 100 and 1000
100
111 - Reduce 10%
125 - Multiply by 8
133 - Reduce 1/4
150 - Reduce 1/3
166 - Multiply by 6
182 - Add 10%
200
222 - Reduce 10%
250
273 - Add 10%
300
333 - Reduce 10%
364
400
455 - Add 10%
500
555 - Reduce 10%
600
666 - Add Half
700
750 - Add 1/3
800
833 - Add 20%
875 - Add 1/7
900
910 - Add 9%
950 - Add 5%
1000
So Task (1) you have to mug up the values of these stations. It is very important that these values are memorized because this will help you in
knowing which number to reach from any given number. e.g. if the denominator is 887 you know you have to reach for 900 or 875 and so on.
Task (2) Practice!
Below is an approach to tackle (three digit /three digit) with consummate ease. We shall attempt to understand it with examples.
Example 1:
What is 145/182 ------------------79.5
Steps
1. add 10% to numerator and denominator...
2. it becomes 159/200 ...............which is 79.5
( the answer from the calculator is 79.6)
Example 2:
123/178....???
Step1:- which station is closest to it?.............................200?...or some say 150. Either is good. (identification took 2 seconds)
Step 2:- what do I have to do to go from 178 to 200/150..................add 22/subtract 28... (another 2 seconds.)
Step 3:- so if I add 22................i am actually adding slightly less than 13% of 178 to itself.
This part is tricky..
here is how I got it:This is how u need to think---
Our number is 178. Thus, 10% of it is 17.8
22 definitely more than 10% .
if 10% = 17.8 then 5% = 8.xx
And 2.5% = 4.xxx (don't even bother to calculate xxx)
which means its around 12.5........or 13 or 12..........
Add same to numerator...........10% of 123 = 12.3
1% = 1.23 so 2% = 2.46
So 12% = 15?
So it becomes 138/200 = 69%?
(calculator answer is 69.1).
All you require to calculate is what is 10% 1%...and approx stuff...any damn calculation works in less than 8 seconds.:satisfie:
Some thing like (2456*4567 - 2134*3214)/2134*3214 will taken 10 seconds maximum...
it works coz I had that time 1.5 years ago..and I aint kidding..
Lets take one more.
Example 3: 532/745?
Tough???
This's how it can be approached:
# Nearest station.........750.................so add 5..............about 1% or less..add 1% damnit
# Numerator now is 537.............(added 1%)
# Fraction is 537/750 add 1/3 each...........it is 71 something.
Remember here don't even attempt to do...537/3. Because denominator is a 1000 and not a 100. So one digit is redundant. So all I do is........53/3 approx..18...plus..53 = 71%..(will do if answer are spaced...)
answers not spaced? Then 537 +537/3 = 537+179 = 716 which makes it 71.6 (calculator is 71.5)
Dont worry if the last few statements were difficult to digest. Try solving a few questions and you'll get the crux of it.
Remember in CAT we don't find answers... we choose them!
Happy Computing
Finding out smallest no. which leaves specific remainders with specific divisors.
Today's concept: Finding out smallest no. which leaves specific remainders with specific divisors.
Type # 1.
find smallest no. other than k, that leaves remainder k when divided by w,x,y...
to solve such questions, take lcm of w,x,y...and add k to it.
e.g. find Smallest no. other than 4, that leaves remainder 4 when divided by 6,7,8 or 9...
take lcm of 6,7,8,9 and add 4
i.e. 504 + 4 = 508
Type # 2
find smallest no. that leaves remainder 3,5,7 when divided by 4,6,8 respectively.
unlike last case, this time the remainder is not constant. but if u see carefully, difference b/w divisor n remainder is constant. i.e. 4-3=6-5=8-7=1
in such questions, take lcm of divisors n subtract the common difference from it
here, the answer wud be lcm of 4,6,8 i.e 24 - 1 = 23
Type # 3
Smallest no. that leaves remainder 3,4,5 whn divided by 5,6,7 respectively and leaves remainder 1 with 11,
we have just seen a way to tackle the first 3 conditions...the no. wud be lcm of 5,6,7 - 2 = 208
now we have one more condition...remainder 1 with 11.
concept => to a no. if v add lcm of divisors...the corresponding remainders dont change.
i.e to 208, if v keep adding 210 ... the first 3 conditions will continue being fulfilled.
so, let 208 + 210k be the no. that will satisfy the 4th condition...viz (208 + 210k)% 11 = 1
208%11 = 10
210k%11 = k
therefore, 10 + k shud leave remainder 1 when divided by 11.
hence, k = 2. and the no. is 208 + 210 x 2 = 628
e.g. find the smallest no. that leaves remainder 2 when divided by 3,4 or 5 and is divisible by 7
for first 3 conditions....no. is 120 + 2 = 122
hence, 122 + 120k is the required no. which reduces to 3 + 2k when divided by 7...now 3+2k shud be a multiple of 7...easily, k=2 and the required no. is 122 + 120 x 2 = 362
Type # 4
What if there is no relation between divisors n remainders?
e.g. find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.
we can c...there's no relation among these divisor-remainder sets...neither is the remainder constant...nor is the difference b/w divisor n remainder a constant.
in such cases...take 1 case n target another case...
e.g. i take the case 7 with 13...and target 6 with 11.
which is the smallest no. that leaves 7 with 13? 7 itself...right? so all nos of the form 7 + 13k will give 7 rem with 13.
now am targeting 6 with 11...so i divide 7 + 13k by 11...i get remainder 7 + 2k...now 7 + 2k = 6,17,28,39,50...so that the remainder with 11 is 6.
a no. that gives integral value of k is 17 i.e. 7 + 2k = 17. hence, k =5 and the no. that satisfies these two conditions is 7 + 13 x 5 = 72
now that 2 conditions are fulfilled, lets target a third condition...say 4 with 7.
to 72, if v add lcm of 11, 13 i.e 143, 2 conditions awready satisfied wud continue being satisfied...
hence the no. is of the form 72 + 143 k.
72 + 143k % 7 = 2 + 3k
now 2 + 3k shud be = 4,11,18,25,32... to satisfy the condition of 4 rem with 7..
a no. that gives integral soln is 11..i.e. 2 + 3k = 11, k = 3.
hence, the no. that satisfies all 3 conditions is 72 + 143 x 3 = 501.
now if v see carefully...4th condition...remainder 1 with 501 has already been satisfied...so the no. v have been looking for is 501.
For ease of calculation, start from biggest divisor n gradually move to smaller ones...u'll always see that last 1-2 conditions will be satisfied automatically.
there are theorems for solving above questions...viz chinese theorem etc...but i solve such questions by the way i've suggested...i find this approach very practical as the flow of nos. is very much visible...n i believe i can tackle any twist in the question devised by cat makers thru this method...there are lotsa other questions based on this concept which i'll soon post but the basic concept remains the same...
few points to be noted
*you can always re-check ur answer
**at times, u can use options to solve such questions.
***dont let concepts go away believing such questions can be easily dealt with thru options...the question may not always be find the smallest no. which...... at times it may be ..find the sum of integers of smallest no. which leaves remainders...blah blah...
****there may be a case when they put an option which satisfies all the conditions but is not the smallest poss value...n put another option...our favorite...none of these!!! lets not undermine genius of cat makers!!
questions for practice...
find the smallest no. that leaves remainder (s)
Q1. 2 when divided by 3,5,6 or 9 (other than 2)
Q2. 2,5,7 when divided by 7,10 and 12 respectively
Q3. 1,2,3,4 with 3,4,5,7 respectively.
Q4. 6 with 7,8,9,10 and 3 with 11.
Q5. 3 with 6, 0 with 11, 3 with 5, 7 with 8
Q6. 2 with 5, 7 with 8, 3 with 4, 5 with 7&11
Q7. 1 with 11, 4 with 5, 9 with 10, 7 with 9.
Type # 1.
find smallest no. other than k, that leaves remainder k when divided by w,x,y...
to solve such questions, take lcm of w,x,y...and add k to it.
e.g. find Smallest no. other than 4, that leaves remainder 4 when divided by 6,7,8 or 9...
take lcm of 6,7,8,9 and add 4
i.e. 504 + 4 = 508
Type # 2
find smallest no. that leaves remainder 3,5,7 when divided by 4,6,8 respectively.
unlike last case, this time the remainder is not constant. but if u see carefully, difference b/w divisor n remainder is constant. i.e. 4-3=6-5=8-7=1
in such questions, take lcm of divisors n subtract the common difference from it
here, the answer wud be lcm of 4,6,8 i.e 24 - 1 = 23
Type # 3
Smallest no. that leaves remainder 3,4,5 whn divided by 5,6,7 respectively and leaves remainder 1 with 11,
we have just seen a way to tackle the first 3 conditions...the no. wud be lcm of 5,6,7 - 2 = 208
now we have one more condition...remainder 1 with 11.
concept => to a no. if v add lcm of divisors...the corresponding remainders dont change.
i.e to 208, if v keep adding 210 ... the first 3 conditions will continue being fulfilled.
so, let 208 + 210k be the no. that will satisfy the 4th condition...viz (208 + 210k)% 11 = 1
208%11 = 10
210k%11 = k
therefore, 10 + k shud leave remainder 1 when divided by 11.
hence, k = 2. and the no. is 208 + 210 x 2 = 628
e.g. find the smallest no. that leaves remainder 2 when divided by 3,4 or 5 and is divisible by 7
for first 3 conditions....no. is 120 + 2 = 122
hence, 122 + 120k is the required no. which reduces to 3 + 2k when divided by 7...now 3+2k shud be a multiple of 7...easily, k=2 and the required no. is 122 + 120 x 2 = 362
Type # 4
What if there is no relation between divisors n remainders?
e.g. find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.
we can c...there's no relation among these divisor-remainder sets...neither is the remainder constant...nor is the difference b/w divisor n remainder a constant.
in such cases...take 1 case n target another case...
e.g. i take the case 7 with 13...and target 6 with 11.
which is the smallest no. that leaves 7 with 13? 7 itself...right? so all nos of the form 7 + 13k will give 7 rem with 13.
now am targeting 6 with 11...so i divide 7 + 13k by 11...i get remainder 7 + 2k...now 7 + 2k = 6,17,28,39,50...so that the remainder with 11 is 6.
a no. that gives integral value of k is 17 i.e. 7 + 2k = 17. hence, k =5 and the no. that satisfies these two conditions is 7 + 13 x 5 = 72
now that 2 conditions are fulfilled, lets target a third condition...say 4 with 7.
to 72, if v add lcm of 11, 13 i.e 143, 2 conditions awready satisfied wud continue being satisfied...
hence the no. is of the form 72 + 143 k.
72 + 143k % 7 = 2 + 3k
now 2 + 3k shud be = 4,11,18,25,32... to satisfy the condition of 4 rem with 7..
a no. that gives integral soln is 11..i.e. 2 + 3k = 11, k = 3.
hence, the no. that satisfies all 3 conditions is 72 + 143 x 3 = 501.
now if v see carefully...4th condition...remainder 1 with 501 has already been satisfied...so the no. v have been looking for is 501.
For ease of calculation, start from biggest divisor n gradually move to smaller ones...u'll always see that last 1-2 conditions will be satisfied automatically.
there are theorems for solving above questions...viz chinese theorem etc...but i solve such questions by the way i've suggested...i find this approach very practical as the flow of nos. is very much visible...n i believe i can tackle any twist in the question devised by cat makers thru this method...there are lotsa other questions based on this concept which i'll soon post but the basic concept remains the same...
few points to be noted
*you can always re-check ur answer
**at times, u can use options to solve such questions.
***dont let concepts go away believing such questions can be easily dealt with thru options...the question may not always be find the smallest no. which...... at times it may be ..find the sum of integers of smallest no. which leaves remainders...blah blah...
****there may be a case when they put an option which satisfies all the conditions but is not the smallest poss value...n put another option...our favorite...none of these!!! lets not undermine genius of cat makers!!
questions for practice...
find the smallest no. that leaves remainder (s)
Q1. 2 when divided by 3,5,6 or 9 (other than 2)
Q2. 2,5,7 when divided by 7,10 and 12 respectively
Q3. 1,2,3,4 with 3,4,5,7 respectively.
Q4. 6 with 7,8,9,10 and 3 with 11.
Q5. 3 with 6, 0 with 11, 3 with 5, 7 with 8
Q6. 2 with 5, 7 with 8, 3 with 4, 5 with 7&11
Q7. 1 with 11, 4 with 5, 9 with 10, 7 with 9.
factorial based questions asking no. of zeroes and max power of sum integer.
factorial based questions asking no. of zeroes and max power of sum integer.
Find the no. of zeroes at the right end of 300!
for every zero, we require 10..n every 10 is made up of 5x2.
in the expression 1x2x3...300, multiples of 2 wud obviously be more than the multiples of 5...so v need to find the maximum power of 5 in 300!
300/5 = 60 (because every fifth no. is a multiple of 5)
300/25 = 12(because every mutiple of 25 has two 5s in it) or, 60/5=12
300/125 = 3 (because multiples of 125 have three 5s in it) or,
12/5 = 2
now 2 cannot be further divided by 5 so add all the quotients...60 + 12 + 2 = 74.
we might also get the same type of questions in a different form,
500! is divisible by 1000^n...what is the max. integral value of n?
now every 1000 is made up of 3 5s and 3 2s....2s are redundant...we need to count no. of 5s....so find total no. of 5s and divide by 3
500/5 = 100
100/5 = 20
20/5 = 4
100 + 20 + 4 =124
124/3 = 41.33
max integral value is 41.
500! is divisible by 99^n...what is the max. integral value of n?
now every 99 is made of two 3s and one 11. obviously 11 will be the deciding factor. so count no. of 11s for the answer
500/11 = 45
45/11 = 4
ans will be 49.
so in such questions, just check which prime no. will be the deciding factor and count the no. of times it occurs. but please understand that highest prime no. is not necessarily always the deciding factor. see this example:
100! is divisible by 160^n...what is the max. integral value of n?
now 160 = 2^5 * 5^1. now although 5 is the biggest prime no. that 160 is made of, the deciding factor wud be 2. because five 2s occur less often than one 5 does. so we'll count the no. of 2s and divide by 5.
100/2 = 50
50/2 = 25
25/2 =12
12/2 = 6
6 /2 = 3
3/2 = 1
add 'em all...97.
97/5 = 19.
so the answer wud be 19
had v taken 5 as the deciding factor, the answer wud have been 100/5 + 100/25 = 24 which is more than 19...hence a wrong answer...
when in dilemma as to which prime no. wud be the deciding factor (e.g. a divisor like 144...its not possible to decide whether 3 or 2 will give the right answer) ....take out answer using both the prime nos...the one thats less is the right answer.
50! is divisible by 144^n...what is the max. integral value of n?
144 = 2^4 * 3^2...difficult to decide whether 3 or 2 will be the deciding factor...
count 2s
50/2=25
25/2=12
12/2=6
6/2=3
3/2=1
sum=47
answer = 47/4 = 11.
count 3s
50/3=17
17/3=5
5/3=1
sum = 23
23/2 = 11
a tie...else the smaller value wud have been the answer.
300! is divisible by (24!)^n. what is the max. possible integral value of n?
such questions are tricky...when u expand 24!...u get 1x2x3...24.
in this range the highest prime no. is 23...so maximum power of 23 in 300! will decide the max value of x...
when v expand 300!...v get a 23 in 23, 46,69,92....
total no of multiples of 23 in 300! will be 300/23 = 13,
forget the fractional part. so the maximum possible answer is 13. hope am clear...else, feel free to revert.
256! is expanded and expressed in base 576 . how many zeroes will this expression have on its right end?
such questions are same as finding maximum power of 576 in 256!
576 = 2^6 x 3^2
to get six 2s i have to travel eight places...1x2x3x4x5x6x7x8 has seven 2s. but to two 3s i have to travel only six places...1x2x3...6 has two 3s...hence 2 will be the constrain.
total 2s in 256! = 255
hence, no. of zeroes = 256/6 = 42.
just to check...3s = 126, 126/2 = 63>42
ans-42
Questions based on this concept
400! is divisible by x^n. what is the max. possible integral value of n if the value of x is:
Q1. 300
Q2. 99
Q3. 500
Q4. 320
Q5. 770
Q6. 5200
Q7. 270
Q8. 686
Q9. 338
Q10. 13000
Answers... 49, 39, 33, 66, 39, 32, 65, 22, 16, 32 (the answer is not 33, this one is actually tricky! )
200! is divisible by (x!)^n...whats the max. possible value of n when x =
Q11. 25
12. 35
13. 50
14. 100
15. 70
16. 300
17.15
answers... 8, 6, 4, 2, 3, 0, 16
300! is expanded and expressed in base x. find the no. of zeroes at the right end of this expression when x=
18. 25
19. 15
20. 35
21. 39
22. 98
answers...37, 74, 48, 23, 24
Reminders
HERE WE START WITH REMAINDERS!
Remainder questions can be broadly divided into 2 types...
1). LCM based questions. e.g. a no. leaves remainders 3,2 when successively divided by 5,6...what will be the remainder when this no. is divided by 30?
2). Power based questions. e.g. remainder when (31)^[(373)^(432)] is divided by 7.
we have already discussed type 1 in two parts...those who have missed it or wish to revisit the concept may use the link below.
first half
http://www.pagalguy.com/forum/quanti...-fundas-2.html (Concepts...total fundas!!)
second half
http://www.pagalguy.com/forum/quanti...-fundas-5.html (Concepts...total fundas!!)
so what we're left to discuss is type 2 mentioned above...i think there are 3 ways of solving these questions...v gotto use our own sensibility to c which approach suits where...
a) Using binomial theorem
b) using cyclicity with remainders
c) using euler's theorem.
for better understanding, i'll divide this post into 3 parts...n will discuss one approach in each part...
Finding remainders using binomial theorem.
(x+y)^n can be expressed as :
nC0 x^n + nC1 x^(n-1)y^1 + nC2 x^(n-2)y^2 + .......... nCr x^(n-r)y^r + ..... nCn y^n
concept:
first term, i.e nC0 x^n is the only term that is independent of y and last term, i.e nCn y^n is the only term that is independent of x. rest, all the terms are divisible by both x & y.
we'll leverage this property to solve complex problems.
e.g. 28^37 % 9 = ?
see...v can express it as (27 + 1 )^37 % 9. now, since 27 is a multiple of 9, the only term that'll be independent of 9 will be the last term i.e. nCn 27^0 * 1^n = 1.
hence, the remainder is 1%9 = 1.
important observation:
we know 1^ (any damn thing...even infinity) = 1 and (-1)^(anything) = 1 or -1 with even and odd values of power respectively. so we'll try getting a form of (nD +/-1)^N. so that v r ultimately left with (+/-1)^N.
342^423 % 7 = ?
step1:
342%7 = 6.
so the question becomes...
6^423 % 7.
express 6 as 7 - 1.
so the expression becomes...
(7-1)^423.
the only terms independent of 7 is ... -1^423 = -1.
hence, the remainder is -1 + 7 = 6.
523^325 % 7 =?
=> 5^325 % 7.
now, i seek a remainder of 1 or 7-1=6 with a power of 5.
5^1 % 7 = 5....wont work
5^2 % 7 = 4....wont work.
5^3 % 7 = 6...will work.
so, (5^3)^ (325/3) = (5^3)^ (108 ) x 5
= (-1)^108 x 5
= 1x5 =5...the remainder.
529^700000 % 7
= 4^700000
4^3%7 = 1. hence, we'll express the given expression in terms of 4^3
4^3 ^ (700000/3)...c how to save time...700000%3 = (7 + 0 + 0 + 0 + 0 + 0) % 3 = 1. hence (700000-1)/3 will be an integer...dont evaluate its value... bcoz 1^ anything = 1.
hence v have 1^I x 4^1 = 1 x 4 = 4
therefore, the remainder is 4.
(31)^[(373)^(432)] % 7 = ?
31 % 7 = 3
hence,
=(3)^[(373)^(432)]
3^3 = 27 = 28 -1. hence, expression shud be in terms of 3^3
= 3^3 ^ [(373^432)/3]
now treat [(373^432)/3] as a new, different question...
(373^432)%3 = 1.
hence, the term becomes 3^3^(I) x 3^1
I = (373^432 - 1)/3 and is of the form (odd-odd) / odd = even/odd = even for sure!!
hence, the expression becomes....
(3^3)^(even I) x 3 % 7
= (-1)^even integer x 3 = 1 x 3 = 3
hence, remainder = 3.
case when divisor is a large no.
in such questions, v try to reduce power by increasing the value of base and bringing it close to a multiple of divisor.
e.g. 2^35 % 61 = ?
v know 2^6 = 64
hence, (2^6)^5 x 2^5
2^6 % 61 = 3.
hence, 3^5 x 2^5
= 3^4 x 3 x 2^5
=3^4 % 61 = 20.
3x 2^5 = 96, 96%61 = 35.
hence, 20 x 35 % 61 = 700%61
= 29, the reqd answer.
Remainder questions can be broadly divided into 2 types...
1). LCM based questions. e.g. a no. leaves remainders 3,2 when successively divided by 5,6...what will be the remainder when this no. is divided by 30?
2). Power based questions. e.g. remainder when (31)^[(373)^(432)] is divided by 7.
we have already discussed type 1 in two parts...those who have missed it or wish to revisit the concept may use the link below.
first half
http://www.pagalguy.com/forum/quanti...-fundas-2.html (Concepts...total fundas!!)
second half
http://www.pagalguy.com/forum/quanti...-fundas-5.html (Concepts...total fundas!!)
so what we're left to discuss is type 2 mentioned above...i think there are 3 ways of solving these questions...v gotto use our own sensibility to c which approach suits where...
a) Using binomial theorem
b) using cyclicity with remainders
c) using euler's theorem.
for better understanding, i'll divide this post into 3 parts...n will discuss one approach in each part...
Finding remainders using binomial theorem.
(x+y)^n can be expressed as :
nC0 x^n + nC1 x^(n-1)y^1 + nC2 x^(n-2)y^2 + .......... nCr x^(n-r)y^r + ..... nCn y^n
concept:
first term, i.e nC0 x^n is the only term that is independent of y and last term, i.e nCn y^n is the only term that is independent of x. rest, all the terms are divisible by both x & y.
we'll leverage this property to solve complex problems.
e.g. 28^37 % 9 = ?
see...v can express it as (27 + 1 )^37 % 9. now, since 27 is a multiple of 9, the only term that'll be independent of 9 will be the last term i.e. nCn 27^0 * 1^n = 1.
hence, the remainder is 1%9 = 1.
important observation:
we know 1^ (any damn thing...even infinity) = 1 and (-1)^(anything) = 1 or -1 with even and odd values of power respectively. so we'll try getting a form of (nD +/-1)^N. so that v r ultimately left with (+/-1)^N.
342^423 % 7 = ?
step1:
342%7 = 6.
so the question becomes...
6^423 % 7.
express 6 as 7 - 1.
so the expression becomes...
(7-1)^423.
the only terms independent of 7 is ... -1^423 = -1.
hence, the remainder is -1 + 7 = 6.
523^325 % 7 =?
=> 5^325 % 7.
now, i seek a remainder of 1 or 7-1=6 with a power of 5.
5^1 % 7 = 5....wont work
5^2 % 7 = 4....wont work.
5^3 % 7 = 6...will work.
so, (5^3)^ (325/3) = (5^3)^ (108 ) x 5
= (-1)^108 x 5
= 1x5 =5...the remainder.
529^700000 % 7
= 4^700000
4^3%7 = 1. hence, we'll express the given expression in terms of 4^3
4^3 ^ (700000/3)...c how to save time...700000%3 = (7 + 0 + 0 + 0 + 0 + 0) % 3 = 1. hence (700000-1)/3 will be an integer...dont evaluate its value... bcoz 1^ anything = 1.
hence v have 1^I x 4^1 = 1 x 4 = 4
therefore, the remainder is 4.
(31)^[(373)^(432)] % 7 = ?
31 % 7 = 3
hence,
=(3)^[(373)^(432)]
3^3 = 27 = 28 -1. hence, expression shud be in terms of 3^3
= 3^3 ^ [(373^432)/3]
now treat [(373^432)/3] as a new, different question...
(373^432)%3 = 1.
hence, the term becomes 3^3^(I) x 3^1
I = (373^432 - 1)/3 and is of the form (odd-odd) / odd = even/odd = even for sure!!
hence, the expression becomes....
(3^3)^(even I) x 3 % 7
= (-1)^even integer x 3 = 1 x 3 = 3
hence, remainder = 3.
case when divisor is a large no.
in such questions, v try to reduce power by increasing the value of base and bringing it close to a multiple of divisor.
e.g. 2^35 % 61 = ?
v know 2^6 = 64
hence, (2^6)^5 x 2^5
2^6 % 61 = 3.
hence, 3^5 x 2^5
= 3^4 x 3 x 2^5
=3^4 % 61 = 20.
3x 2^5 = 96, 96%61 = 35.
hence, 20 x 35 % 61 = 700%61
= 29, the reqd answer.
Finding remainders using cyclicicty with remainders:
This approach is useful when the divisor is small or at times when it is a factor of 100.
3^327%7 = ?
3^1 % 7 = 3
3^2 % 7 = 2
3^3 % 7 = 6
3^4 % 7 = 4
3^5 % 7 = 4x3 % 7 = 5
3^6 % 7 = 5 x 3 % 7 = 1
3^7 % 7 = 1 x 3 % 7 = 3
remainder with first power is same as remainder with 7th power...hence v can say that cyclicity in remainders is 7-1 = 6.
so, 327 % 6 = 3,
hence, effectively, the remainder is 3^3 % 7 = 6
326^524 % 9 =?
326 % 9 = 2
hence, 2^524
now,
2^1 % 9 = 2
2^2 % 9 = 4
2^3 % 9 = 8
2^4 % 9 = 7
2^5 % 9 = 5
2^6 % 9 = 1.
2^7 % 9 = 2
hence cyclicity in remainder = 7-1 = 6.
524 % 6 = 2
hence, the remainder is 2^2 % 9 = 4.
81^502 % 100
81^1 % 100 = 81
81^2 % 100 = 61
81^3 % 100 = 41
81^4 % 100 = 21
81^5 % 100 = 01
81^6 % 100 = 81
hence, cyclicity = 6-1 = 5.
502 % 100 = 2.
so the reqd remainder is same as that with 81^2 = 61.
similarly...this method can be effectively used when remainders are other factors of 100.
viz, when the factors are 20,25,50,100...knowing last 2 digits wud suffice knowing the remainders...we've already discussed this concept while discussing cyclicity...
important: dont try this method when the divisors are complex...viz 37,73 etc...the cylicty wud come very late n calculations will grow cumbersome...when divisors are complex, there must be sum other catch in the question...look for that catch...e.g. last type discussed in the binomial method...
This approach is useful when the divisor is small or at times when it is a factor of 100.
3^327%7 = ?
3^1 % 7 = 3
3^2 % 7 = 2
3^3 % 7 = 6
3^4 % 7 = 4
3^5 % 7 = 4x3 % 7 = 5
3^6 % 7 = 5 x 3 % 7 = 1
3^7 % 7 = 1 x 3 % 7 = 3
remainder with first power is same as remainder with 7th power...hence v can say that cyclicity in remainders is 7-1 = 6.
so, 327 % 6 = 3,
hence, effectively, the remainder is 3^3 % 7 = 6
326^524 % 9 =?
326 % 9 = 2
hence, 2^524
now,
2^1 % 9 = 2
2^2 % 9 = 4
2^3 % 9 = 8
2^4 % 9 = 7
2^5 % 9 = 5
2^6 % 9 = 1.
2^7 % 9 = 2
hence cyclicity in remainder = 7-1 = 6.
524 % 6 = 2
hence, the remainder is 2^2 % 9 = 4.
81^502 % 100
81^1 % 100 = 81
81^2 % 100 = 61
81^3 % 100 = 41
81^4 % 100 = 21
81^5 % 100 = 01
81^6 % 100 = 81
hence, cyclicity = 6-1 = 5.
502 % 100 = 2.
so the reqd remainder is same as that with 81^2 = 61.
similarly...this method can be effectively used when remainders are other factors of 100.
viz, when the factors are 20,25,50,100...knowing last 2 digits wud suffice knowing the remainders...we've already discussed this concept while discussing cyclicity...
important: dont try this method when the divisors are complex...viz 37,73 etc...the cylicty wud come very late n calculations will grow cumbersome...when divisors are complex, there must be sum other catch in the question...look for that catch...e.g. last type discussed in the binomial method...
Finding remainders using Euler's theorem.
(special thanks to junoonmba for this)
This method is very useful when the divisor and dividend are relatively prime numbers...
step 1: To calculate euler's no. of a divisor.
euler's no. can be practically taken as cyclicity in remainders by a divisor..
to find euler's no, express the divisor in terms of prime factors...
100 = 2^2 x 5^2.
powers of the prime nos. have no significance...its jus the prime no. that matters...
euler's no (e for convenience) = divisor x (1-1/first prime factor) x (1-1/second prime factor) x ... (1-1/last prime factor)
so, for 100, e = 100 x (1-1/2) x (1-1/5) = 100 x 1/2 x 4/5
= 40.
that means e for 100 = 40. or, in other words, 100 divisor will definetly show a cylicity of 40 in the remainders.
whenever the power of a relatively prime no. will be a multiple of 40, the expression wud show a remainder 1 with 100.
e.g. 3^120 % 100 = ?
we know e for 100 = 40.
3 n 100 are relatively prime nos.
hence, 3^40 % 100 = 1.
hence 3^120 % 100 = (3^40)^3 % 100 = 1^3 = 1.
7^100 % 45 = ?
45 = 3x3x5
e for 45 = 45 x (1-1/3) x (1-1/5) = 24
hence, 7^24 % 45 = 1
hence, 7^100 % 45 = 1^4 x 7^4 % 45
= 2401 % 45
= 16, the required answer...
(special thanks to junoonmba for this)
This method is very useful when the divisor and dividend are relatively prime numbers...
step 1: To calculate euler's no. of a divisor.
euler's no. can be practically taken as cyclicity in remainders by a divisor..
to find euler's no, express the divisor in terms of prime factors...
100 = 2^2 x 5^2.
powers of the prime nos. have no significance...its jus the prime no. that matters...
euler's no (e for convenience) = divisor x (1-1/first prime factor) x (1-1/second prime factor) x ... (1-1/last prime factor)
so, for 100, e = 100 x (1-1/2) x (1-1/5) = 100 x 1/2 x 4/5
= 40.
that means e for 100 = 40. or, in other words, 100 divisor will definetly show a cylicity of 40 in the remainders.
whenever the power of a relatively prime no. will be a multiple of 40, the expression wud show a remainder 1 with 100.
e.g. 3^120 % 100 = ?
we know e for 100 = 40.
3 n 100 are relatively prime nos.
hence, 3^40 % 100 = 1.
hence 3^120 % 100 = (3^40)^3 % 100 = 1^3 = 1.
7^100 % 45 = ?
45 = 3x3x5
e for 45 = 45 x (1-1/3) x (1-1/5) = 24
hence, 7^24 % 45 = 1
hence, 7^100 % 45 = 1^4 x 7^4 % 45
= 2401 % 45
= 16, the required answer...
Some Questions
1). 4^32 % 36
2) 312^[(124)^(312)] % 7
3) 532^[(325)^(534)] % 7
4) 7^200 % 11
1.Remainder of 26^16 divided by 125
2. Remainder of 3^94 divided by 125
3.Reamainder of 2^13 divided by 25
4. Remainder of 10^14 divided by 37
5 2^1990%1990
2) 312^[(124)^(312)] % 7
3) 532^[(325)^(534)] % 7
4) 7^200 % 11
1.Remainder of 26^16 divided by 125
2. Remainder of 3^94 divided by 125
3.Reamainder of 2^13 divided by 25
4. Remainder of 10^14 divided by 37
5 2^1990%1990
2^2!^3!^4!...100! %
1. 7
2. 9
3. 5^32 % 5000
4. 2^60 % 130
5. 7^50 % 1001
1. 7
2. 9
3. 5^32 % 5000
4. 2^60 % 130
5. 7^50 % 1001
* source pagalguy
Subscribe to:
Posts (Atom)