Maths Olympiad Prep

Track / Stage 3 / 212 of 260 #212 of 1964

Problem 212

AMC 10/12, early questions
Number theory Difficulty 3.8 Find the answer

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?

Pick one

Official 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 2457×2=49142457\times2=4914, since 24572457 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 22, it is less than or equal to 75427542, the largest number in the set. This happens to be 2754×2=55082754\times2=5508. Therefore, the number would have to be between 49144914 and 55085508, and also even. The only even numbers in the set and in this range are 54725472 and 52745274. 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 2457×4=98282457\times4=9828, well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is 2457×3=73712457\times3=7371 and the greatest is 2475×3=74252475\times3=7425, since any higher number in the set multiplied by 33 would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is 2475×3=7425,(D)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,(D)7425/3 = 2475, \boxed{\textbf{(D)}}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.