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

Revision as of 15:01, 15 January 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=== by kindlymath55532

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.

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/G_0KQJhZKGY

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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 6: / Line 6: @@
 ==Solutions==
-===Solution 1===
+===Solution 1=== by kindlymath55532
 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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