Difference between revisions of "2006 AMC 8 Problems/Problem 11"
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− | There is <math> 1 </math> integer whose digits sum to <math> 1 </math>: 10. | + | 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>: 13, 22, 31, and 40. | + | 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>: 18, 27, 36, 45, 54, 63, 72, 81, and 90. | + | 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>. |
− | There are <math> 3</math> integers whose digits sum to <math> 16 </math>: 79, 88, and 97. | + | There are <math> 3</math> integers whose digits sum to <math> 16 </math>: <math>79, 88, and 97</math>. |
− | Two digits cannot sum to 25 or any greater square since the greatest sum of digits of a two-digit number is <math> 9 + 9 = 18 </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>. |
Thus, the answer is <math> 1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17} </math>. | Thus, the answer is <math> 1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17} </math>. |
Revision as of 21:10, 25 October 2016
Problem
How many two-digit numbers have digits whose sum is a perfect square?
Solution
There is integer whose digits sum to : .
There are integers whose digits sum to : .
There are integers whose digits sum to : .
There are integers whose digits sum to : .
Two digits cannot sum to or any greater square since the greatest sum of digits of a two-digit number is .
Thus, the answer is .
See Also
2006 AMC 8 (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.