Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solutions)
 
(10 intermediate revisions by 8 users not shown)
Line 5: Line 5:
 
<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>
  
−
==Solutions==
+
==Solution 1==
  
−
===Solution 1 - kindlymath55532===
+
We can see that the original number can be written as <math>10a+b</math>, where <math>a</math> represents the tens digit and <math>b</math> represents the units digit. When this number is added to the number obtained by reversing its digits, which is <math>10b+a</math>, the sum would be <math>11a+11b</math>. From this, we can construct the equation <math>11a+11b=132</math>, which simplifies to <math>a+b=12</math>. Since there are 7 pairs of such digits <math>a</math> and <math>b</math>, <math>(3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)</math>, the answer would be <math>\boxed{\textbf{(B) } 7}.</math>
  
−
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:
+
~Aqf243
−
<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.
+
==Solution 2==
 +
 
 +
We can set the number as ab where a is the tens digit and b is the ones digit. So now the equation will be ab+ba=132. a+b has to have a remainder of 2 when divided by 10 so it will be \(a+b\equiv 2\quad (\bmod 10)\). We also know a<10 and b<10. So a+b can either be 2 or 12. a+b cannot be 2 because then there will be only 3 numbers that work and that isn't in the answer choice. So a+b=12 . To check this we can do ab+ba=132 which equals to (a+b)0+(a+b)=132 and since we said a+b=12 we get 120+12=132 which is true. So we have a+b=12 and a<10 and b<10.If a is 9 then b=3 and if a=3 then b=9 calculating how many pairs are in between you get 7. So the answer is <math>\boxed{\textbf{(B) } 7}.</math>
  
 
==Video Solution (CREATIVE THINKING!!!)==
 
==Video Solution (CREATIVE THINKING!!!)==
Line 19: Line 21:
  
 
==Video Solution==
 
==Video Solution==
−
https://youtu.be/lbfbJWcVsEE
+
https://youtu.be/lbfbJea43ldk
  
 
~savannahsolver
 
~savannahsolver
Line 27: Line 29:
 
{{AMC8 box|year=2016|num-b=10|num-a=12}}
 
{{AMC8 box|year=2016|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 13:30, 28 July 2025

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 $132.$

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

Solution 1

We can see that the original number can be written as $10a+b$, where $a$ represents the tens digit and $b$ represents the units digit. When this number is added to the number obtained by reversing its digits, which is $10b+a$, the sum would be $11a+11b$. From this, we can construct the equation $11a+11b=132$, which simplifies to $a+b=12$. Since there are 7 pairs of such digits $a$ and $b$, $(3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)$, the answer would be $\boxed{\textbf{(B) } 7}.$

~Aqf243

Solution 2

We can set the number as ab where a is the tens digit and b is the ones digit. So now the equation will be ab+ba=132. a+b has to have a remainder of 2 when divided by 10 so it will be \(a+b\equiv 2\quad (\bmod 10)\). We also know a<10 and b<10. So a+b can either be 2 or 12. a+b cannot be 2 because then there will be only 3 numbers that work and that isn't in the answer choice. So a+b=12 . To check this we can do ab+ba=132 which equals to (a+b)0+(a+b)=132 and since we said a+b=12 we get 120+12=132 which is true. So we have a+b=12 and a<10 and b<10.If a is 9 then b=3 and if a=3 then b=9 calculating how many pairs are in between you get 7. So the answer is $\boxed{\textbf{(B) } 7}.$

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/G_0KQJhZKGY

~Education, the Study of Everything

Video Solution

https://youtu.be/lbfbJea43ldk

~savannahsolver

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_11&oldid=255783"