Difference between revisions of "2004 AMC 12B Problems/Problem 15"
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<math> \mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45 </math> | <math> \mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45 </math> | ||
| − | + | == Solution 1 == | |
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If Jack's current age is <math>\overline{ab}=10a+b</math>, then Bill's current age is <math>\overline{ba}=10b+a</math>. | If Jack's current age is <math>\overline{ab}=10a+b</math>, then Bill's current age is <math>\overline{ba}=10b+a</math>. | ||
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In five years, Jack's age will be <math>10a+b+5</math> and Bill's age will be <math>10b+a+5</math>. | In five years, Jack's age will be <math>10a+b+5</math> and Bill's age will be <math>10b+a+5</math>. | ||
| − | We are given that <math>10a+b+5 = 2(10b+a+5)</math>. Thus <math>8a = 19b + 5</math>. | + | We are given that <math>10a+b+5=2(10b+a+5)</math>. |
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| + | Thus <math>8a=19b+5 \Rightarrow a=\dfrac{19b+5}{8}</math>. | ||
| − | For <math>b=1</math> we get <math>a=3</math>. For <math>b=2</math> and <math>b=3</math> the value <math>\frac{19b+5}8</math> is not an integer, and for <math>b\geq 4</math> | + | For <math>b=1</math> we get <math>a=3</math>. For <math>b=2</math> and <math>b=3</math> the value <math>\frac{19b+5}8</math> is not an integer, and for <math>b\geq 4</math>, <math>a</math> is more than <math>9</math>. Thus the only solution is <math>(a,b)=(3,1)</math>, and the difference in ages is <math>31-13=\boxed{\mathrm{(B)\ }18}</math>. |
| − | + | == Solution 2 == | |
Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference. | Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference. | ||
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As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options <math>\{13,49,58,67\}</math>. | As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options <math>\{13,49,58,67\}</math>. | ||
| − | Checking each of them, we see that only <math>13</math> works, and gives the solution <math>31-13 = \boxed{18}</math>. | + | Checking each of them, we see that only <math>13</math> works, and gives the solution <math>31-13=\boxed{\mathrm{(B)\ }18}</math>. |
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== See also == | == See also == | ||
Latest revision as of 08:47, 17 October 2020
- The following problem is from both the 2004 AMC 12B #15 and 2004 AMC 10B #17, so both problems redirect to this page.
Contents
Problem
The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
Solution 1
If Jack's current age is 
, then Bill's current age is 
.
In five years, Jack's age will be 
and Bill's age will be 
.
We are given that 
.
Thus 
.
For 
we get 
. For 
and 
the value 
is not an integer, and for 
, 
is more than 
. Thus the only solution is 
, and the difference in ages is 
.
Solution 2
Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.
The age difference is 
, hence it is a multiple of . Thus Bill's current age modulo must be 
.
Thus Bill's age is in the set 
.
As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options 
.
Checking each of them, we see that only 
works, and gives the solution .
See also
| 2004 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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 | |
| 2004 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
