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# Difference between revisions of "2013 AMC 10A Problems/Problem 13"

## Problem

How many three-digit numbers are not divisible by $5$, have digits that sum to less than $20$, and have the first digit equal to the third digit?

$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$

## Solution

We use a casework approach to solve the problem. These three digit numbers are of the form $\overline{xyx}$.($\overline{abc}$ denotes the number $100a+10b+c$). We see that $x\neq 0$ and $x\neq 5$, as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5.

The second condition is that the sum $2x+y<20$. When $x$ is $1$, $2$, $3$, or $4$, $y$ can be any digit from $0$ to $9$, as $2x<10$. This yields $10(4) = 40$ numbers.

When $x=6$, we see that $12+y<20$ so $y<8$. This yields $8$ more numbers.

When $x=7$, $14+y<20$ so $y<6$. This yields $6$ more numbers.

When $x=8$, $16+y<20$ so $y<4$. This yields $4$ more numbers.

When $x=9$, $18+y<20$ so $y<2$. This yields $2$ more numbers.

Summing, we get $40 + 8 + 6 + 4 + 2 = \boxed{\textbf{(B) }60}$

## See Also

 2013 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 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 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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