Difference between revisions of "2001 AMC 8 Problems/Problem 25"
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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. | 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= | + | <math>5724/2=2862</math>, <math>5724/3=1908</math>. <math>7245/3=2415</math>. <math>7254/2=3627</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. |
~ M4ST3R0FM4TH | ~ M4ST3R0FM4TH | ||
Latest revision as of 14:40, 23 December 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?
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 
, 
to divide them. Even 
will leads to a solution start with 
which we don't need.
, 
. 
. 
, 
. 
. The answer is 
. You can obtain the answer in only 6 calculations.
~ M4ST3R0FM4TH
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. 
