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.