2019 AMC 10A Problems/Problem 2

Problem

What is the hundreds digit of $(20!-15!)?$

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

Solution

The last three digits of $n!$ for all $n\geq15$ are $000$ , because there are at least three $2$ s and three $5$ s in its prime factorization. Because $0-0=0$ , the answer is $\boxed{\textbf{(A) }0}$ .

Solution 2

20 and 15 are both greater than 10, therefore they are divisible by 100 because of powers of 5 and powers of 2, so the hundreds digit is $\boxed{\textbf{(A) }0}$ . ~peppapig_

2019 AMC 10A (Problems • Answer Key • Resources)
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2019 AMC 10A Problems/Problem 2

Contents

Problem

Solution

Solution 2

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