Find the greatest common divisor of $7!$ and $(5!)^2.$
I need to find the largest positive integer that divides both $7!$ and $(5!)^2$ evenly. One way to do this is to factor both numbers into prime factors and look for the common ones. I know that $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 7 \times (2 \times 3) \times 5 \times (2 \times 2) \times 3 \times 2 = 2^4 \times 3^2 \times 5 \times 7.$ I also know that $(5!)^2 = (5 \times 4 \times 3 \times 2)^2 = (5 \times (2 \times 2) \times 3 \times 2)^2 = (2^3 \times 3 \times 5)^2 = 2^6 \times 3^2 \times 5^2.$ Now I can see that the common prime factors are $2^4, 3^2,$ and $5.$ To find the greatest common divisor, I need to multiply these common factors together. So the greatest common divisor is $2^4 \times 3^2 \times 5 = 16 \times 9 \times 5 = 720.$ 