# Math counts problem

• I did a hard math counts problem recently, and I don’t know why the answer is 72, can someone please explain it for me? Thanks a lot!

How many perfect squares are divisors of the product 1! ∙ 2! ∙ 3! ∙ 4! ∙ 5! ∙ 6! ∙ 7! ?

• Hey @modestwallaby !

Whenever tackling a divisors problem, oftentimes it's a great strategy to first consider the prime factorization.

The prime factorization of the product is $$2^{16}\cdot3^{7}\cdot5^{3}\cdot7$$. Since we're dealing with perfect squares, let's think about what the prime factorization of a perfect square looks like-- in a perfect square, all of the exponents are even! For example, $$2^4\cdot3^2$$ is a perfect square, so is $$2^2 \cdot 5^2$$, so is $$3^6$$, but not $$2^3\cdot3^2$$ or $$3^4\cdot 5$$.

So, using $$2^{16}\cdot3^{7}\cdot5^{3}\cdot7$$, we have to find all of the possibilities to make a perfect square using each of these factors. Let's tackle each of the factors individually!

Every perfect square that is a divisor of that product is either going to have $$2^0,2^2,2^4,2^6,2^8,2^{10},2^{12},2^{14},$$ or $$2^{16}$$ in its prime factorization. This gives us $$9$$ possibilities for the power of $$2$$.

Then, using the same logic, every perfect square that is a divisor of that product has a power of $$3$$ that's either $$3^0,3^2,3^4$$ or $$3^{6}$$ in its prime factorization. This gives us $$4$$ possibilities for the power of $$2$$.

Finally, there are only $$2$$ possibilities for a perfect square power of $$5$$: $$5^0$$ and $$5^2$$. We can stop here since we have no other options for any factors greater than 5.

So, multiplying our possibilities together, we have $$9\cdot4\cdot2=\boxed{72}$$ in total. Let me know if this made sense and if you still have any questions!

• @quacker88 Thanks for your help!