Difference between revisions of "2014 AMC 12A Problems/Problem 6"
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==Solution 2 (Meta)== | ==Solution 2 (Meta)== | ||
We start like above. Let the two digits be <math>a</math> and <math>b</math>. Therefore, <math>5(a+b) = 10a+b-10b-a=9(a-b)</math>. Since we are looking for <math>10a+b+10b+a=11(a+b)</math> and we know that <math>a+b</math> must be a multiple of <math>9</math>, the only answer choice that works is <math>\boxed{\textbf{(D) }99}.</math> | We start like above. Let the two digits be <math>a</math> and <math>b</math>. Therefore, <math>5(a+b) = 10a+b-10b-a=9(a-b)</math>. Since we are looking for <math>10a+b+10b+a=11(a+b)</math> and we know that <math>a+b</math> must be a multiple of <math>9</math>, the only answer choice that works is <math>\boxed{\textbf{(D) }99}.</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/rJytKoJzNBY | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2014|ab=A|num-b=5|num-a=7}} | {{AMC12 box|year=2014|ab=A|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:15, 25 November 2020
Problem
The difference between a two-digit number and the number obtained by reversing its digits is times the sum of the digits of either number. What is the sum of the two digit number and its reverse?
Solution 1
Let the two digits be and . Then, , or . This yields and because . Then,
Solution 2 (Meta)
We start like above. Let the two digits be and . Therefore, . Since we are looking for and we know that must be a multiple of , the only answer choice that works is
Video Solution
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.