factors of 36^2

36^2
= (2^2 * 3^2) ^ 2
= 2^4 * 3^4

So total number of factors = (4+1)*(4+1) = 25

This is a formula: first break the number to all possible prime numbers (along with the powers) Then add one to each power and multiple.

N = a^x*b^y*c^z where a b c are distinct prime numbers and x y and z are corresponding powers then
Total number of factors = (x+1)*(y+1)*(z+1)

Try the same for 36 = 2^2 * 3^2
Total factors = (2+1)*(2+1) = 9 which are [1,2,3,4,6,9,12,18,36]

You just cannot add 9+9 by doing so you are considering (1,2,3,4,6,9,12,18,36 ) twice but there are other factors such as 72 which is not covered in your list of 18.

Hope this makes sense and always remeber the formula at your disposal

replayyyy wrote:How many factors does 36^2 have ? The OA is 25 and it is reached by transforming 36 into a product of its prime exponents - 2 and 3 --> 2^4 * 3^4. However, I can`t understand why my approach is wrong - 1,2,3,4,6,9,12,18,36 (9 factors) are factors of 36; another 36 will have the same number of factors - which means a total of 18?

As Shovan noted, to determine the number of positive factors of an integer:

1) Prime-factorize the integer
2) Add 1 to each exponent
3) Multiply

Since 36^2 = 2^4 * 3^4, we get (4+1)*(4+1) = 25 factors.

Here's the reasoning. To determine how many factors can be created from 36^2 = 2^4 * 3^4, we need to determine the number of choices we have of each prime factor:

For 2, we can use 2^0, 2^1, 2^2, 2^3, or 2^4, giving us 5 choices.
For 3, we can use 3^0, 3^1, 3^2, 3^3, or 3^4, giving us 5 choices.

Multiplying, we get 5*5 = 25 possible factors.