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

 

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