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>: <math>10</math>. | 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>. | 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>. | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=10|num-a=12}} | {{AMC8 box|year=2006|num-b=10|num-a=12}} | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:44, 25 April 2021
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. 
