Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 8 Problems/Problem 25"

(→‎Problem)
(→‎Solution)
 
Line 11: Line 11:
  
 
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 <math> 7425/3 = 2475,  \boxed{\textbf{(D)}}</math>
 
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 <math> 7425/3 = 2475,  \boxed{\textbf{(D)}}</math>
 +
 +
==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 <math>2</math>, <math>3</math> to divide them. Even <math>5</math> will leads to a solution start with <math>1</math> which we don't need.
 +
 +
<math>5724/2=2862</math>, <math>5724/3=1908</math>. <math>7245/3=2415</math>. <math>7254/2=3612</math>, <math>7254/3=2418</math>. <math>7425/3=2475</math>. The answer is <math>\boxed{\textbf{(D)}}</math>. You can obtain the answer in only 6 calculations.
  
 
==See Also==
 
==See Also==
 
  {{AMC8 box|year=2001|num-b=24|after=Last<br />Question}}
 
  {{AMC8 box|year=2001|num-b=24|after=Last<br />Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 03:07, 12 March 2024

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 $7425/3 = 2475, \boxed{\textbf{(D)}}$

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.

See Also

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last
Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_25&oldid=216913"