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

(Created page with "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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==Solution==
 
==Solution==
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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 are 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

Revision as of 09:43, 23 November 2016

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 $132.$

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


Solution

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 are ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$ or $\boxed{\textbf{(B)} 7}$ ordered pairs

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