180 product of a prime factor using indices

Start with 180.

Is 180 divisible by 2? Yes, so write "2" as one of the prime factors, and then work with the quotient, 90.

Is 90 divisible by 2? Yes, so write "2" (again) as another prime factor, then work with the quotient, 45.

Is 45 divisible by 2? No, so try a bigger divisor.

Is 45 divisible by 3? Yes, so write "3" as a prime factor, then work with the quotient, 15

Is 15 divisible by 3? [Note: no need to revert to "2", because we've already divided out all the 2's] Yes, so write "3" (again) as a prime factor, then work with the quotient, 5.

Is 5 divisible by 3? No, so try a bigger divisor.

Is 5 divisible by 4? No, so try a bigger divisor (actually, we know it can't be divisible by 4 becase it's not divisible by 2)

Is 5 divisible by 5? Yes, so write "5" as a prime factor, then work with the quotient, 1

Once you end up with a quotient of "1" you're done.

In this case, you should have written down, "2 * 2 * 3 * 3 * 5"

The answer is 2•2•3•3•5