A challenging factorial problem

Yesterday my older son and I worked through a a challenging algebra problem. Today’s challenging problem involved factorials. The problem is #23 from the 2015 AMC 10b:

Problem 23 from the 2015 AMC 10b

Here’s the problem:

Let $n$ be a positive integer greater than 4 such that the decimal representation of n! ends in k zeros and the decimal representation of (2n)! ends in 3k zeros. Let s denote the sum of the four least possible values of n. What is the sum of the digits of s?

We started by just talking through the problem and coming up with a plan:

Next he implemented that plan and did a great job working through to the end of the problem:

Finally, we went to Wolfram alpha and double checked that the numbers he found were indeed solutions to the problem (and sorry for the stumbling around by me in this part!)

So, a challenging problem and a good solution from my son. We continue to work on ideas about problem solving. It is always nice when everything comes together like it did for today’s project!



One Comment so far. Leave a comment below.
  1. I always like asking questions about structure, so:
    Why did they ask for the sum of the digits of s?

    Basically, there are two ideas that could be at play, would be curious if there are more:
    (a) forming s and summing its digits reduces information, so it obscures the 4 smallest values of n (meaning that the choices don’t help you find the answer)
    (b) there is a short-cut that let’s you get to the sum of the digits without knowing s or the 4 smallest numbers

    In this case, (a) seems the most likely reason. If the final answer had been 9, then case (b) would be intriguing as it would suggest we should look for reasons that the sum would be divisible by 9.

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