A Hard(ish) ages problem

Algebra Level pending

There are four people, called Amy, Bob, Carl and Don. You are told the following things about their ages:

14 years ago, Don had not yet been born, and Amy, Bob and Carl's ages were in geometric progression.
In 15 years time, Carl's age will the same as Amy's and Bob's ages combined.
The product of Amy's and Bob's ages is twice the product of Carl's and Don's ages.

How old are Amy, Bob, Carl and Don?

NOTE : when inputting you answer, write is as their ages put together to form one number. For example, If Amy is 99, Bob is 5, Carl is 17, and Don is 68, then write your answer as 9951768.

The answer is 1520503.

1 solution

Freddie Hand
Feb 16, 2017

Say Amy, Bob, Carl and Don have ages A, B, C and D respectively.

From the second fact, we can work out that $C+15=A+15+B+15$ , so $C-15=A+B$ .

This means Carl is older than Amy and Bob, so Amy is also younger than Bob (because their ages were in geometric progression 14 years ago)

From the first fact, we can deduce that $D<14$ , $A=A$ , $B=(A-14)k+14$ , and $C=(A-14)k^{2}+14$ , for some integer k so that $k\geq 2$ .

We can substitute this to get $A+(A-14)k+14=(A-14)k^{2}-1$

rearranging, we get $A+15=(A-14)(k^{2}-k)$

$\frac{A+15}{A-14}=k^{2}-k$

Therefore, $\frac{A+15}{A-14}$ must be an integer, so $\frac{A+15}{A-14}-\frac{A-14}{A-14}=\frac{29}{A-14}$ must also be an integer.

As 29 is prime, this means $A=15$ or $A=43$ . We can quickly verify that Don's age cannot be an integer or is more than 16 for $A=43$

Therefore, $A=15$ , so $k=6$ . This means $B=20$ , $C=50$ , and $D=3$

A Hard(ish) ages problem

The answer is 1520503.

1 solution

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