Difference between revisions of "2014 AMC 12A Problems/Problem 6"
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\textbf{(E) }110</math> | \textbf{(E) }110</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Let the two digits be <math>a</math> and <math>b</math>. Then, <math>5a + 5b = 10a + b - 10b - a = 9a - 9b</math>, or <math>2a = 7b</math>. This yields <math>a = 7</math> and <math>b = 2</math> because <math>a, b < 10</math>. Then, <math>72 + 27 = \boxed{\textbf{(D) }99}.</math> | Let the two digits be <math>a</math> and <math>b</math>. Then, <math>5a + 5b = 10a + b - 10b - a = 9a - 9b</math>, or <math>2a = 7b</math>. This yields <math>a = 7</math> and <math>b = 2</math> because <math>a, b < 10</math>. Then, <math>72 + 27 = \boxed{\textbf{(D) }99}.</math> | ||
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| + | ==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> | ||
==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}} | ||
Revision as of 14:47, 2 January 2016
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 
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. 
