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==
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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 20:50, 20 July 2014

Problem

For how many $n$ in $\{1, 2, 3, ..., 100 \}$ is the tens digit of $n^2$ odd?

$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50$

Solution

Let $n=10a+b$. So $n^2=(10a+b)^2=100a^2+20ab+b^2$. The first digit of $100a^2$ it the hundreds place digit which means that $20ab+b^2$ is the tens and ones digit together. We note that $20ab$ will always be even. So in order for $20ab+b^2$ to have an odds tens digit, then $b^2$ must carry and have an odd tens digit. We note that $b$ can only equal either $4$ or $6$ for this to happen. The total number of positive integers less than or equal to $100$ that have $4$ or $6$ as the units digit is \[10\times 2=\boxed{\textbf{(B) }20.}\]

--Solution by TheMaskedMagician