Difference between revisions of "1994 AHSME Problems/Problem 15"
(Created page with "==Problem== For how many <math>n</math> in <math>\{1, 2, 3, ..., 100 \}</math> is the tens digit of <math>n^2</math> odd? <math> \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\t...") |
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<math> \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50 </math> | <math> \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50 </math> | ||
==Solution== | ==Solution== | ||
+ | Let <math>n=10a+b</math>. So <math>n^2=(10a+b)^2=100a^2+20ab+b^2</math>. The first digit of <math>100a^2</math> it the hundreds place digit which means that <math>20ab+b^2</math> is the tens and ones digit together. We note that <math>20ab</math> will always be even. So in order for <math>20ab+b^2</math> to have an odds tens digit, then <math>b^2</math> must carry and have an odd tens digit. We note that <math>b</math> can only equal either <math>4</math> or <math>6</math> for this to happen. The total number of positive integers less than or equal to <math>100</math> that have <math>4</math> or <math>6</math> as the units digit is <cmath>10\times 2=\boxed{\textbf{(B) }20.}</cmath> | ||
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+ | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] |
Revision as of 19:50, 20 July 2014
Problem
For how many in is the tens digit of odd?
Solution
Let . So . The first digit of it the hundreds place digit which means that is the tens and ones digit together. We note that will always be even. So in order for to have an odds tens digit, then must carry and have an odd tens digit. We note that can only equal either or for this to happen. The total number of positive integers less than or equal to that have or as the units digit is
--Solution by TheMaskedMagician