Number Theory

[siddharth11.sharma]

How many factors does (36^2) have?

[jlucero]

This is a problem dealing with the possible combinations of different, smaller, prime factors of 36:

36^2 = (2x2x3x3)^2 = (2x2x2x2x3x3x3x3)

Now we can systematically go through and find combinations of 0, 1, 2, 3, etc. prime factors:

0 prime factors= 1
1 p.f. = 2, 3
2 p.f. = 4, 6, 9
3 p.f. = 8, 12, 18, 27
4 p.f. = 16, 24, 36, 54, 81
5 p.f. = 48, 72, 108, 162
6 p.f. = 144, 216, 324
7 p.f. = 432, 648
8 p.f. = 1296
This gives us a total of 25 unique factors.

Notice that we have a bell curve here where we have factors on either side that multiply to give us the total 1296. 1x1296, 2x648, 3x432, etc. If you can find your way to 4 factors, then realize you will be finding the other 'pair' for each factor you found, you can predict the second half of the bell curve.

There's also a shortcut method where you find the number of unique prime factors, find out how many of these prime factors there are, add one (to account for the fact that we don't need to include any prime factors in our factor), and multiply the numbers together. Easier to see than explain:

36^2 = 2^4 * 3^4
I can include 0, 1, 2, 3, or 4 twos (5 options)
I can include 0, 1, 2, 3, or 4 threes (5 options)
Therefore, there are 5*5 = 25 unique factors

Another example:
72 = 2^3 * 3^2
0-3 twos
0-2 threes
4*3 = 12 unique factors