Difference between revisions of "2016 AMC 8 Problems/Problem 11"

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11. Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is <math>132.</math>
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==Problem==
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Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is <math>132.</math>
  
 
<math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math>
 
<math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math>
  
  
==Solution==
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==Solutions==
{{solution}}
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===Solution 1===
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We can write the two digit number in the form of <math>10a+b</math>; reverse of <math>10a+b</math> is <math>10b+a</math>. The sum of those numbers is:
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<cmath>(10a+b)+(10b+a)=132</cmath><cmath>11a+11b=132</cmath><cmath>a+b=12</cmath>
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We can use brute force to find order pairs <math>(a,b)</math> such that <math>a+b=12</math>. Since <math>a</math> and <math>b</math> are both digits, both <math>a</math> and <math>b</math> have to be integers less than <math>10</math>. Thus our ordered pairs are <math>(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)</math> or <math>\boxed{\textbf{(B)} 7}</math> ordered pairs.
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===Solution 2 -SweetMango77===
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Since the numbers are “mirror images,” their average has to be <math>\frac{132}{2}=66</math>. The highest possible value for the tens digit is <math>9</math> because it is a two-digit number. <math>9-6=3</math>, and <math>6-3=3</math>, so our lowest tens digit is <math>3</math>. The numbers between <math>9</math> and <math>3</math> inclusive is <math>9-3+1=\boxed{\text{(B)}\;7}</math> total possibilities.
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==See Also==
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{{AMC8 box|year=2016|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 17:32, 21 April 2021

Problem

Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$

$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$


Solutions

Solution 1

We can write the two digit number in the form of $10a+b$; reverse of $10a+b$ is $10b+a$. The sum of those numbers is: \[(10a+b)+(10b+a)=132\]\[11a+11b=132\]\[a+b=12\] We can use brute force to find order pairs $(a,b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$. Thus our ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$ or $\boxed{\textbf{(B)} 7}$ ordered pairs.

Solution 2 -SweetMango77

Since the numbers are “mirror images,” their average has to be $\frac{132}{2}=66$. The highest possible value for the tens digit is $9$ because it is a two-digit number. $9-6=3$, and $6-3=3$, so our lowest tens digit is $3$. The numbers between $9$ and $3$ inclusive is $9-3+1=\boxed{\text{(B)}\;7}$ total possibilities.

See Also

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

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