# 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

We begin by narrowing down the possibilities. If the larger number were twice the smaller number, then the smallest possibility for the larger number is $2457\times2=4914$, since $2457$ is the smallest number in the set. The largest possibility would have to be twice the largest number in the set such that when it is multiplied by $2$, it is less than or equal to $7542$, the largest number in the set. This happens to be $2754\times2=5508$. Therefore, the number would have to be between $4914$ and $5508$, and also even. The only even numbers in the set and in this range are $5472$ and $5274$. A quick check reveals that neither of these numbers is twice a number in the set. The number can't be quadruple or more another number in the set since $2457\times4=9828$, well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is $2457\times3=7371$ and the greatest is $2475\times3=7425$, since any higher number in the set multiplied by $3$ would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is $2475\times3=7425, \boxed{\textbf{(D)}}$

Or, since the greatest number possible divided by the smallest number possible is slightly greater than 3, divide all of the choices by 2, then 3 and see if the resulting answer contains 2,4,5 and 7. Doing so, you find that $7425/3 = 2475, \boxed{\textbf{(D)}}$

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