If n is a positive integer and N^2 is divisible by 72, then the largest possible integer that must divide n must be

[user2026]

In Tips and Tricks Session 4 (Tips Tricks and Shortcuts Session 4 - GregMat)

I am confused by the step he takes in solving this after rewriting the question/equation as m^2 / (2^3 * 3^2):

He says okay, if m^2 is divisible by 3 2’s and 2 3’s, then m must be divisible by two 2s and one 3. I get intuitively the one 3, but my mind wants to think that m must be divisible by just one 2 (since the square root of M^2 is M, the square root of of 2^3 is 2 rt(2).

Why is it 2 twos instead of one?

In an earlier problem (the "if M is a positive integer and m^2 is divisible by 48, he performs a similar step – m must have two 2s and one 3 if m^2 must be divisible by four 2s and one 3).

Any help would be much appreciated. Thanks!

[ganesh]

Hi, do you mind adding a screenshot of the question or a timestamp for where I can find it in the video?

[user2026]

[ganesh]

So, if we break down 72 into prime factors it would be 2^3 \times 3^2 right?

If n^2 is divisible by (and a multiple of 72, then we can say:

n^2 = 2^3 \times 3^2 \times SOMETHING

If we want to find n, we need to square root n^2, which means we need to divide every power by 2. Can we do that for 2^3?

Since n^2 is a square number, all its powers must be even, so if we know it contains 2^3 it must actually contain 2^4 at least.

Does that make sense so far?