Divisible by this year??? (Part 2: Factorials)

Number Theory Level 3

$n!$ or $n$ -factorial is the product of all integers from $1$ up to $n$ $(n! = 1 \times 2 \times 3 \times ... \times n)$ . Find the maximum integral value of $k$ such that $2014^k$ divides $2014!$

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This problem is part of the set " Symphony "

The answer is 38.

2 solutions

Jake Lai
Dec 14, 2014

In this question, we only need to consider the prime factors of 2014: 2, 19, and 53.

Since 2, 19, and 53 are pairwise coprime , we will not overcount and thus we can further simplify the question by looking at the maximal $k$ for which $53^k | 2014!$ .

We begin by checking all factors of $2014!$ of the form $53n$ ; they have 53 as a factor, and there are 38 such integers; hence, a lower bound for $k = 38$ .

Then, we check all factors of the form $53^{2}n$ ; however, because $53^{2} > 2014$ , there are no more factors to check and hence our answer is $k = \boxed{38}$ .

Riska Mulyani
Dec 9, 2014

2014 = 2.19.53 so 2014/53 = 38

Doesn't explain how you got to 38.

Jake Lai - 6 years, 6 months ago

Divisible by this year??? (Part 2: Factorials)

The answer is 38.

2 solutions

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