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

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

2013 AMC 10A (Problems • Answer Key • Resources)
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

@@ Line 21: / Line 21: @@
 Summing, we get <math>40 + 8 + 6 + 4 + 2 = 60</math>, <math>\textbf{(B)}</math>.
+==See Also==
+{{AMC10 box|year=2013|ab=A|num-b=12|num-a=14}}

Page

Toolbox

Search

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

Revision as of 21:17, 7 February 2013

Problem

Solution

See Also