2006 AMC 8 Problems/Problem 11: Difference between revisions

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

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

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

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

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 $16$ : $79, 88, \text{and } 97$ .

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 $1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17}$ .

2006 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

@@ Line 7: / Line 7: @@
 There is <math> 1 </math> integer whose digits sum to <math> 1 </math>: <math>10</math>.
-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, \text{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, \text{and} 90</math>.
+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, \text{and } 97</math>.
 Two digits cannot sum to <math>25</math> or any greater square since the greatest sum of digits of a two-digit number is <math> 9 + 9 = 18 </math>.