Difference between revisions of "2013 AMC 8 Problems/Problem 13"

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==Solution==
 
==Solution==
Let's say that the number that Clara reversed is <math>10a+b</math>. Then she misinterpreted it as <math>10b+a</math>. The difference between the two is <math>|9a-9b|</math>. Either way, this is a multiple of 9, so the answer is <math>\boxed{\textbf{(A)}\ 45}</math>.
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Let the two digits be <math>a</math> and <math>b</math>.
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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>.
  
 
==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 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?

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49$

Solution

Let the two digits be $a$ and $b$.

The correct score was $10a+b$. Clara misinterpreted it as $10b+a$. The difference between the two is $|9a-9b|$ which factors into $|9(a-b)|$. Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is $\boxed{\textbf{(A)}\ 45}$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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

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