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Difference between revisions of "1976 AHSME Problems/Problem 5"

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=Problem 5=
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== Problem ==
 
How many integers greater than <math>10</math> and less than <math>100</math>, written in base-<math>10</math> notation, are increased by <math>9</math> when their digits are reversed?
 
How many integers greater than <math>10</math> and less than <math>100</math>, written in base-<math>10</math> notation, are increased by <math>9</math> when their digits are reversed?
  
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 10</math>
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 10</math>
  
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=Solution=
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== Solution ==
 
Let our two digit number be <math>\overline{ab}</math>, where <math>a</math> is the tens digit, and <math>b</math> is the ones digit. So, <math>\overline{ab}=10a+b</math>. When we reverse our digits, it becomes <math>10b+a</math>. So, <math>10a+b+9=10b+a\implies a-b=1</math>. So, our numbers are <math>12, 23, 34, 45, 56, 67, 78, 89\Rightarrow \textbf{(C)}</math>.~MathJams
 
Let our two digit number be <math>\overline{ab}</math>, where <math>a</math> is the tens digit, and <math>b</math> is the ones digit. So, <math>\overline{ab}=10a+b</math>. When we reverse our digits, it becomes <math>10b+a</math>. So, <math>10a+b+9=10b+a\implies a-b=1</math>. So, our numbers are <math>12, 23, 34, 45, 56, 67, 78, 89\Rightarrow \textbf{(C)}</math>.~MathJams
 
{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}}
 
{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}}

Latest revision as of 12:36, 16 July 2024

Problem

How many integers greater than $10$ and less than $100$, written in base-$10$ notation, are increased by $9$ when their digits are reversed?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 10$

Solution

Let our two digit number be $\overline{ab}$, where $a$ is the tens digit, and $b$ is the ones digit. So, $\overline{ab}=10a+b$. When we reverse our digits, it becomes $10b+a$. So, $10a+b+9=10b+a\implies a-b=1$. So, our numbers are $12, 23, 34, 45, 56, 67, 78, 89\Rightarrow \textbf{(C)}$.~MathJams

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