Art of Problem Solving

AoPS Online

2013 AMC 10A Problems/Problem 13

Revision as of 19:39, 7 February 2013 by Countingkg (talk | contribs) (Created page with "==Problem== How many three-digit numbers are not divisible by <math>5</math>, have digits that sum to less than <math>20</math>, and have the first digit equal to the third digi...")

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

These three digit numbers are of the form $xyx$ . We see that $x\ne0$ and $x\ne5$ , 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 $40 + 8 + 6 + 4 + 2 = 60$ , $\textbf{(B)}$ .

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_13&oldid=50856"

© 2024 AoPS Incorporated