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

(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...")
 
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Summing, we get <math>40 + 8 + 6 + 4 + 2 = 60</math>, <math>\textbf{(B)}</math>.
 
Summing, we get <math>40 + 8 + 6 + 4 + 2 = 60</math>, <math>\textbf{(B)}</math>.
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==See Also==
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{{AMC10 box|year=2013|ab=A|num-b=12|num-a=14}}

Revision as of 21:17, 7 February 2013

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 $40 + 8 + 6 + 4 + 2 = 60$, $\textbf{(B)}$.

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 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
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