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 1

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. ~ M4ST3R0FM4TH

See Also

2001 AMC 8 (ProblemsAnswer KeyResources)
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