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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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==Solution==
 
==Solution==
  
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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.
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We use a casework approach to solve the problem. These three digit numbers are of the form <math>\overline{xyx}</math>.(<math>\overline{abc}</math> denotes the number <math>100a+10b+c</math>).  We see that <math>x\neq 0</math> and <math>x\neq 5</math>, as <math>x=0</math> does not yield a three-digit integer and <math>x=5</math> yields a number divisible by 5.
  
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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.   
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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.   
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When <math>x=9</math>, <math>18+y<20</math> so <math>y<2</math>.  This yields <math>2</math> more numbers.   
 
When <math>x=9</math>, <math>18+y<20</math> so <math>y<2</math>.  This yields <math>2</math> more numbers.   
  
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Summing, we get <math>40 + 8 + 6 + 4 + 2 = 60</math>, <math>\textbf{(B)}</math>.
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Summing, we get <math>40 + 8 + 6 + 4 + 2 = \boxed{\textbf{(B) }60}</math>
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==See Also==
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{{AMC10 box|year=2013|ab=A|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 11:15, 21 January 2021

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 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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