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

Revision as of 21:34, 6 September 2011

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, and 40.

There are $9$ integers whose digits sum to $9$ : 18, 27, 36, 45, 54, 63, 72, 81, 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 $1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17}$ .

Revision as of 21:33, 6 September 2011 (view source) Math Kirby (talk \| contribs) (Created page with "== Problem == How many two-digit numbers have digits whose sum is a perfect square? <math> \textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qq...")		Revision as of 21:34, 6 September 2011 (view source) Math Kirby (talk \| contribs) (→‎Solution) Newer edit →
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	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 25 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{\~~textbh~~{(C)} 17} </math>.	+	Thus, the answer is <math> 1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17} </math>.