Difference between revisions of "2013 AMC 8 Problems/Problem 13"
(→Solution) |
m (→Solution) |
||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
− | Let | + | Let the two digits be <math>\text{ab}</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>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=12|num-a=14}} | {{AMC8 box|year=2013|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:19, 27 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 .
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.