1976 AHSME Problems/Problem 5: Difference between revisions
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=Problem | == 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> | ||
=Solution== | == 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>. | 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 
and less than 
, written in base- notation, are increased by 
when their digits are reversed?
Solution
Let our two digit number be 
, where 
is the tens digit, and 
is the ones digit. So, 
. When we reverse our digits, it becomes 
. So, 
. So, our numbers are 
.~MathJams
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