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

Latest revision as of 16:45, 7 July 2024

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.

-kindlymath55532

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/G_0KQJhZKGY

~Education, the Study of Everything

Video Solution

https://youtu.be/lbfbJWcVsEE

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
+==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 <math>132.</math>
 <math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math>
+==Solutions==
-==Solution 1==
+===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:
 <cmath>(10a+b)+(10b+a)=132</cmath><cmath>11a+11b=132</cmath><cmath>a+b=12</cmath>
-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.
+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.
+-kindlymath55532
+==Video Solution (CREATIVE THINKING!!!)==
+https://youtu.be/G_0KQJhZKGY
+~Education, the Study of Everything
+==Video Solution==
+https://youtu.be/lbfbJWcVsEE
-==Solution 2 -SweetMango77==
+~savannahsolver
-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.
+==See Also==
 {{AMC8 box|year=2016|num-b=10|num-a=12}}
 {{MAA Notice}}

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