Difference between revisions of "1994 AHSME Problems/Problem 15"

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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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Let <math>n=10a+b</math>. So <math>n^2=(10a+b)^2=100a^2+20ab+b^2</math>. The term <math>100a^2</math> only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd.  The term <math>20ab</math> only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of <math>2ab</math> which is even. So we see this term also does not affect whether the tens digit is odd.  This means only <math>b^2</math> can affect whether the tens digit is odd.  We can quickly check <math>1^2=1, \dots, 9^2=81</math> and discover only for <math>b=4</math> or <math>b=6</math> does <math>b^2</math> have an odd tens digit. 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>
  
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]

Latest revision as of 02:53, 28 May 2021

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 term $100a^2$ only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd. The term $20ab$ only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of $2ab$ which is even. So we see this term also does not affect whether the tens digit is odd. This means only $b^2$ can affect whether the tens digit is odd. We can quickly check $1^2=1, \dots, 9^2=81$ and discover only for $b=4$ or $b=6$ does $b^2$ have an odd tens digit. 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

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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