# Factorials –> Factoring

I was working through this year’s MathCounts State Level exam, and found 2 problems involving factorials.  Your reflex when seeing a factorial problem is to see how you can factor the expression.  For example we are asked to find the value of this expression:

$\frac{5!+6!}{4!+3!}$

Notice that the 2 terms in the numerator are both products of $5!$ and the 2 terms in the denominator are both products of $3!$.  So we can factor these both out:

$\frac{5!+6!}{4!+3!}=\frac{5!(1+6)}{3!(4+1)}$

If you notice that $5!$ and $3!$ both have a factor of $3!$ then the fraction is much easier to evaluate.

Another problem on the same exam asks us to solve for $n$:

$(n+1)! - n! = 4320$

Again you can factor $n!$  from the 2 terms on the left hand side:

$(n+1)! - n! = n!(n+1 - 1)= 4320$

$n!\cdot n = 4320$

To finish out this problem we can find the prime factorization of $4320$ and match that up to the product of consecutive integers.

$4320 = 2^5 \cdot 3^3 \cdot 5 = 2 \cdot 3 \cdot 2^2 \cdot 5 \cdot (2 \cdot 3) \cdot (2 \cdot 3) = 6! \cdot 6$

$n = 6$