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.