Find $n$ such that $2^5 \cdot 3^2 \cdot n = 8!$.
I notice that the left-hand side of the equation has powers of 2 and 3, but the right-hand side has a factorial. I wonder if I can rewrite the factorial as a product of powers of 2 and 3 and other primes. I recall that 8! means 8 times 7 times 6 and so on down to 1. I can write each of these factors as a product of primes using a factor tree or a divisibility test. For example, 8 is 2 cubed, 7 is prime, 6 is 2 times 3, 5 is prime, 4 is 2 squared, 3 is prime, 2 is prime, and 1 is 1. So 8! is equal to 2 cubed times 7 times 2 times 3 times 5 times 2 squared times 3 times 2 times 1. I can group the powers of 2 and 3 together and write 8! as 2 to the seventh power times 3 squared times 5 times 7. Now I can compare this with the left-hand side of the equation and see what n must be. I see that the left-hand side has 2 to the fifth power, but the right-hand side has 2 to the seventh power. So n must have 2 squared as a factor to make up the difference. I also see that the left-hand side has 3 squared, which matches the right-hand side. So n does not need any more factors of 3. I also see that the left-hand side does not have any factors of 5 or 7, but the right-hand side does. So n must have 5 and 7 as factors to make up the difference. Therefore, n must be 2 squared times 5 times 7, which is 140. 