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 | + | 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 in is the tens digit of odd?
Solution
Let . So . The term only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd. The term only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of which is even. So we see this term also does not affect whether the tens digit is odd. This means only can affect whether the tens digit is odd. We can quickly check and discover only for or does have an odd tens digit. The total number of positive integers less than or equal to that have or as the units digit is
--Solution by TheMaskedMagician
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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