Difference between revisions of "2013 AMC 8 Problems/Problem 13"
m (→Solution) |
m (→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | Let the two digits be <math> | + | Let the two digits be <math>a</math> and <math>b</math>. |
The correct score was <math>10a+b</math>. Clara misinterpreted it as <math>10b+a</math>. The difference between the two is <math>|9a-9b|</math> which factors into <math>|9(a-b)|</math>. Therefore, the difference is a multiple of 9, the only answer choice that is a multiple of 9 is <math>\boxed{\textbf{(A)}\ 45}</math>. | The correct score was <math>10a+b</math>. Clara misinterpreted it as <math>10b+a</math>. The difference between the two is <math>|9a-9b|</math> which factors into <math>|9(a-b)|</math>. Therefore, the difference is a multiple of 9, the only answer choice that is a multiple of 9 is <math>\boxed{\textbf{(A)}\ 45}</math>. | ||
Revision as of 13:37, 29 November 2013
Problem
When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?
Solution
Let the two digits be 
and 
.
The correct score was 
. Clara misinterpreted it as 
. The difference between the two is 
which factors into 
. Therefore, the difference is a multiple of 9, the only answer choice that is a multiple of 9 is 
.
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
