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# Sunday Times Teaser 2819 – An Age-Old Problem

### by Angela Newing

#### Published: 2 October 2016 (link)

Five men of different ages under fifty are celebrating their joint birthday in the pub. Alan’s age is the average of their five ages. When Bob is three times Alan’s age today, the sum of Alan’s and Colin’s ages will equal the sum of their five ages today. Furthermore, Derek will then be three times his age today, and Eric will be one year more than double Bob’s age today.

The landlord checked that the youngest of the five was allowed to buy alcohol.

Who is the youngest, and how old is he?

2 Comments Leave one →
1. Brian Gladman permalink

Here is a non programmed solution.

Let the current ages be $$a$$, $$b$$, $$c$$, $$d$$ and $$e$$. The average condition gives: $4a=b+c+d+e$ Moving $$x$$ years ahead the other conditions are: $b+x=3a$ $(a+x)+(c+x)=5a$ $d+x=3d$ $e+x=2b+1$These can be simplified by eliminating $$x$$ to show that: $b=3a-2d$ $c=4a-4d$ $2d=3a-b$ $e=2b-2d+1$ If we now substitute these expressions into the average above, we obtain: $11d-9a=1$ Multiplying this by 5 and taking the result mod 11 shows that $$a=6\mod11$$ so that $$a=11k+6$$ for some integer $$k$$. This allows all ages to be expressed in terms of $$k$$ as: $a=11k+6$ $b=15k+8$ $c=8k+4$ $d=9k+5$ $e=12k+7$

By inspection the oldest age is $$b$$, which must be less than 50, meaning that $$k$$ is less than 3.

The youngest is $$c$$, which is at least 18, meaning that $$k$$ is greater than 1.

Hence $$k = 2$$, so the youngest is Colin, who is 20.

2. Here’s my programmed solution. It runs in 44ms.

It prints the three possible sets of solutions, or if you increase the minimum age of the “men” to 13 (or more) you get a single solution.