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Difference between revisions of "2016 AMC 8 Problems/Problem 11"
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==Solution==
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==Solution 1==
  
 
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:
 
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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{{AMC8 box|year=2016|num-b=10|num-a=12}}
 
{{AMC8 box|year=2016|num-b=10|num-a=12}}
 
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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.

Revision as of 16:11, 4 November 2020

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 $132.$

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


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.

2016 AMC 8 (Problems • Answer Key • Resources)
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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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.

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