Difference between revisions of "2006 AMC 8 Problems/Problem 11"

(Solution)
(Solution)
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There are <math> 4 </math> integers whose digits sum to <math> 4 </math>: <math>13, 22, 31, and 40</math>.
 
There are <math> 4 </math> integers whose digits sum to <math> 4 </math>: <math>13, 22, 31, and 40</math>.
  
There are <math> 9 </math> integers whose digits sum to <math> 9 </math>: <math>18, 27, 36, 45, 54, 63, 72, 81, and 90</math>.
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There are <math> 9 </math> integers whose digits sum to <math> 9 </math>: <math>18, 27, 36, 45, 54, 63, 72, 81, \text{and} 90</math>.
  
 
There are <math> 3</math> integers whose digits sum to <math> 16 </math>: <math>79, 88, and 97</math>.
 
There are <math> 3</math> integers whose digits sum to <math> 16 </math>: <math>79, 88, and 97</math>.

Revision as of 00:49, 5 November 2016

Problem

How many two-digit numbers have digits whose sum is a perfect square?

$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19$

Solution

There is $1$ integer whose digits sum to $1$: $10$.

There are $4$ integers whose digits sum to $4$: $13, 22, 31, and 40$.

There are $9$ integers whose digits sum to $9$: $18, 27, 36, 45, 54, 63, 72, 81, \text{and} 90$.

There are $3$ integers whose digits sum to $16$: $79, 88, and 97$.

Two digits cannot sum to $25$ or any greater square since the greatest sum of digits of a two-digit number is $9 + 9 = 18$.

Thus, the answer is $1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17}$.

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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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