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

Latest revision as of 13:24, 29 October 2024

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)
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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 Thus, the answer is <math> 1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17} </math>.
+==Video Solution by WhyMath==
+https://youtu.be/Ke78CmjlqgM
 ==See Also==
 {{AMC8 box|year=2006|num-b=10|num-a=12}}
 {{MAA Notice}}