Problem 5=

How many integers greater than $10$ and less than $100$ , written in base- $10$ notation, are increased by $9$ when their digits are reversed?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 10$

Solution=

Let our two digit number be $\overline{ab}$ , where $a$ is the tens digit, and $b$ is the ones digit. So, $\overline{ab}=10a+b$ . When we reverse our digits, it becomes $10b+a$ . So, $10a+b+9=10b+a\implies a-b=1$ . So, our numbers are $12, 23, 34, 45, 56, 67, 78, 89\Rightarrow \textbf{(C)}$ .

1976 AHSME (Problems • Answer Key • Resources)
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1976 AHSME Problems/Problem 5

Problem 5=

Solution=