Question

How many pairs (A,B) of positive integers are there such that A and B are factors of 6^6 and a is A factor of B

Answer 1

Well \[
6^6 = 2^63^6
\]Any factors will be in the form: \[
2^n3^m
\]

Answer 2

WE have \(n\) and \(m\) range from 0 to 6, so we have 7 choices for n and 7 choices for m.

Answer 3

So simply considering choices for B, we'd have \(7^2\). But each of them has their own way of getting values of A.

Answer 4

If we say \[
B = 2^n3^m
\]Then the combinations we could expect to have for A would be \((n+1)\cdot (m+1)\).

Answer 5

So I get \[
\sum_{m=0}^7\sum_{n=0}^7(m+1)(n+1)
\]

Answer 6

Which is \[
\sum_{m=0}^7 (m+1)\sum_{n=0}^7 n +1 = \sum_{m=0}^7 (m+1)\cdot 36 = \sum_{m=0}^7 (m+1)\cdot 36 = 36\cdot 36 = 1296
\]

Answer 7

Does this answer check out or not?

Answer 8

for my question.......answer is never to be over 1000
its only 700+

Answer 9

Hmmm, I guess that \(0\) doesn't count?

Answer 10

Though, I don't think my answer will have \(0\).

Answer 11

yea i think (1,6) is counted

Answer 12

is (1,1) counted?

Answer 13

yes

Answer 14

Do you understand when I say: \[
B= 2^n3^m
\]

Answer 15

Then we say \[
A = 2^{n'}3^{m'}
\]where \(n'\leq n\) and \(m'\leq m\)

Answer 16

Then we know that A is a factor of B.

Answer 17

oh wait!! I put 7 instead of 6!
So I should get: \[
\sum_{m=0}^6\sum_{n=0}^6(m+1)(n+1) = \sum_{m=0}^6 (m+1)\sum_{n=0}^6 (n+1) = 28\cdot 28 = 784
\]

Answer 18

???