Difference between revisions of "2001 AMC 8 Problems/Problem 25"
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==Solution== | ==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 <math> 2457\times2=4914 </math>, since <math> 2457 </math> 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 <math> 2 </math>, it is less than or equal to <math> 7542 </math>, the largest number in the set. This happens to be <math> 2754\times2=5508 </math>. Therefore, the number would have to be between <math> 4914 </math> and <math> 5508 </math>, and also even. The only even numbers in the set and in this range are <math> 5472 </math> and <math> 5274 </math>. 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 <math> 2457\times4=9828 </math>, well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is <math> 2457\times3=7371 </math> and the greatest is <math> 2475\times3=7425 </math>, since any higher number in the set multiplied by <math> 3 </math> would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is <math> 2475\times3=7425, \boxed{\ | + | 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 <math> 2457\times2=4914 </math>, since <math> 2457 </math> 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 <math> 2 </math>, it is less than or equal to <math> 7542 </math>, the largest number in the set. This happens to be <math> 2754\times2=5508 </math>. Therefore, the number would have to be between <math> 4914 </math> and <math> 5508 </math>, and also even. The only even numbers in the set and in this range are <math> 5472 </math> and <math> 5274 </math>. 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 <math> 2457\times4=9828 </math>, well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is <math> 2457\times3=7371 </math> and the greatest is <math> 2475\times3=7425 </math>, since any higher number in the set multiplied by <math> 3 </math> would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is <math> 2475\times3=7425, \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{\ | + | 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> |
==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}} | ||
Revision as of 12:58, 31 October 2022
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?
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 
, since 
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 
, it is less than or equal to 
, the largest number in the set. This happens to be 
. Therefore, the number would have to be between 
and 
, and also even. The only even numbers in the set and in this range are 
and 
. 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 
, well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is 
and the greatest is 
, since any higher number in the set multiplied by 
would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is 
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 
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. 
