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

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==Solution==
 
==Solution==
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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{(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 08:23, 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?

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

Solution

Let's say that the number that Clara reversed is $10a+b$. Then she misinterpreted it as $10b+a$. The difference between the two is $|9a-9b|$. Either way, this is a multiple of 9, so the answer is $\boxed{(A) 45}$.

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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