A number theory problem by Aryan

You can split 7! into 1 x 2 x 3 x 4 x 5 x 6 x 7. When you factor each number out, you get $2^4$ x $3^2$ x 5 x 7. Adding one to all exponents yield the exponents 5, 3, 2, and 2 of the bases 2, 3, 5, and 7 respectively. We add one to all these exponents to account for the combinations of prime factors that do not use all the bases. For example 210 is a factor of 5040 whose prime factorization includes all the primes in the prime factorization of 5040 as 2 x 3 x 5 x 7. But 180 is also a factor of 5040, and it only uses the primes 2, 3, and 5. The seven is present as $7^0$ . Thus, if there is the prime 5 without any exponents, there are two possible exponents that we can use in creating factors, 0 and 1. Therefore 0 is the additional exponent that can be used for each prime in the prime factorization. If it is $2^4$ , then we can have $2^0$ , $2^1$ , $2^2$ , $2^3$ , and $2^4$ . There are 5 values and 5 = 4 + 1. That is why we add one to all the exponents. Multiplying these exponents of the prime factorization of 5040 gets 5 x 3 x 2 x 2 = $\boxed{60}$ .