2004 AMC 12B Problems/Problem 15

From AoPSWiki

The following problem is from both the 2004 AMC 12B #15 and 2004 AMC 10B #17, so both problems redirect to this page.

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?

$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45$

Solution

Solution 1

If Jack's current age is $\overline{ab}=10a+b$ , then Bill's current age is $\overline{ba}=10b+a$ .

In five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$ .

We are given that $10a+b+5 = 2(10b+a+5)$ . Thus $8a = 19b + 5$ .

For $b=1$ we get $a=3$ . For $b=2$ and $b=3$ the value $\frac{19b+5}8$ is not an integer, and for $b\geq 4$ it is more than $9$ . Thus the only solution is $(a,b)=(3,1)$ , and the difference in ages is $31-13 = \boxed{18}$ .

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 $(10a+b)-(10b+a)=9(a-b)$ , hence it is a multiple of $9$ . Thus Bill's current age modulo $9$ must be $4$ .

Thus Bill's age is in the set $\{13,22,31,40,49,58,67,76,85,94\}$ .

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 $\{13,49,58,67\}$ .

Checking each of them, we see that only $13$ works, and gives the solution $31-13 = \boxed{18}$ .

2004 AMC 12B (Problems • 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

2004 AMC 10B (Problems • 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

Personal tools

Navigation

Search

Toolbox