Difference between revisions of "2016 AMC 8 Problems/Problem 11"
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==Solutions== | ==Solutions== | ||
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===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: | 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> | <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. |
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+ | -kindlymath55532 | ||
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+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/G_0KQJhZKGY | ||
− | + | ~Education, the Study of Everything | |
− | |||
==Video Solution== | ==Video Solution== |
Latest revision as of 17:45, 7 July 2024
Contents
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
Solutions
Solution 1
We can write the two digit number in the form of ; reverse of
is
. The sum of those numbers is:
We can use brute force to find order pairs
such that
. Since
and
are both digits, both
and
have to be integers less than
. Thus, our ordered pairs are
; or
ordered pairs.
-kindlymath55532
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.