Difference between revisions of "2013 AMC 8 Problems/Problem 13"
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<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49</math> | <math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49</math> | ||
| − | ==Solution== | + | |
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=9FkjSCcdTqY ~David | ||
| + | |||
| + | https://youtu.be/KBM2YN4kKGA ~savannahsolver | ||
| + | |||
| + | ==Solution(educated guess)== | ||
Let the two digits be <math>a</math> and <math>b</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, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is <math>\boxed{\textbf{(A)}\ 45}</math>. |
| + | |||
| + | ==Solution 2(brute force elimination) == | ||
| + | |||
| + | We can simply test out numbers to see which one works. We can see that Clara’s score can’t be a multiple of ten because the reverse of the score is a one digit number, to small for the answer choices. After testing multiples, the answer should be <math>\textbf{A}</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}} | ||
Latest revision as of 20:40, 3 January 2024
Contents
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?
Video Solution
https://www.youtube.com/watch?v=9FkjSCcdTqY ~David
https://youtu.be/KBM2YN4kKGA ~savannahsolver
Solution(educated guess)
Let the two digits be 
and 
.
The correct score was 
. Clara misinterpreted it as 
. The difference between the two is 
which factors into 
. Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is 
.
Solution 2(brute force elimination)
We can simply test out numbers to see which one works. We can see that Clara’s score can’t be a multiple of ten because the reverse of the score is a one digit number, to small for the answer choices. After testing multiples, the answer should be 
.
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. 
