2001 AMC 8 Problems/Problem 25

Problem

There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?

$\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$

Solution 2

There are only 5 options for the problem so we can just try them. It is easy since that we only need try to use $2$ , $3$ to divide them. Even $5$ will leads to a solution start with $1$ which we don't need.

$5724/2=2862$ , $5724/3=1908$ . $7245/3=2415$ . $7254/2=3612$ , $7254/3=2418$ . $7425/3=2475$ . The answer is $\boxed{\textbf{(D)}}$ . You can obtain the answer in only 6 calculations.

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Question
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All AJHSME/AMC 8 Problems and Solutions

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2001 AMC 8 Problems/Problem 25

Problem

Solution 2

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