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

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These three digit numbers are of the form <math>xyx</math>.  We see that <math>x\ne0</math> and <math>x\ne5</math>, as <math>x=0</math> does not yield a three-digit integer and <math>x=5</math> yields a number divisible by 5.
 
These three digit numbers are of the form <math>xyx</math>.  We see that <math>x\ne0</math> and <math>x\ne5</math>, as <math>x=0</math> does not yield a three-digit integer and <math>x=5</math> yields a number divisible by 5.
  
The second condition is that the sum <math>2x+y<20</math>.  When <math>x</math> is <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math>, y can be any digit from <math>0</math> to <math>9</math>, as <math>2x<10</math>.  This yields <math>10(4) = 40</math> numbers.   
+
The second condition is that the sum <math>2x+y<20</math>.  When <math>x</math> is <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math>, <math>y</math> can be any digit from <math>0</math> to <math>9</math>, as <math>2x<10</math>.  This yields <math>10(4) = 40</math> numbers.   
  
 
When <math>x=6</math>, we see that <math>12+y<20</math> so <math>y<8</math>.  This yields <math>8</math> more numbers.   
 
When <math>x=6</math>, we see that <math>12+y<20</math> so <math>y<8</math>.  This yields <math>8</math> more numbers.   

Revision as of 22:30, 20 January 2016

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

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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