In the little village of Oldham, there are exactly 100 inhabitants. The oldest one was born in 1900, and all the people living in Oldham are born in a different year, but all the 1st of january.

In 1999, the sum of the four digits of Charles' year of birth is equal to his age.

How old is he, in this year of 1999?

  • $\begingroup$ We should assume thar Charles lives in Oldham. $\endgroup$ Aug 10, 2016 at 5:45

2 Answers 2


Charles was born in


so in 1999 he is

$23$ years old.


His date of birth is $19xy$ and we know that his age in $1999$ (which is $99 - (10x + y)$) is equal to the sum of digits $1 + 9 + x + y$.

This gives us:

$99 - 10x - y = 1 + 9 + x + y$
$11x + 2y = 89$
$x$ and $y$ must be between $0$ and $9$

We are left with 4 possibilities:

For $0 \le x \le 6 \rightarrow y > 9$
For $x = 7 \rightarrow y = 6$
For $x = 8 \rightarrow y = 0.5$
For $x = 9 \rightarrow y = -0.5$

The only feasible solution is:

$x = 7,\ y = 6$
Since $y$ has to be a positive integer, but cannot be greater than 9

  • $\begingroup$ That's the solution. You can also find it by considering that (89-2y) must be dividible by eleven, thus giving y. $\endgroup$
    – IAmInPLS
    Jul 28, 2016 at 16:22

Much like @Meiffert I determined that:

Charles was born in 1976.

The equations used are the same but rather than looking at the possibilities I used more maths to eliminate options.

So we start with:

$99 − 10x − y = 1 + 9 + x + y$
$11x + 2y = 89$

And then make the following deductions:

Since $11x+2y$ is odd and $2y$ is always even, $11x$ must be odd. So $x$ is odd.


Similarly since $2y<19$ then $89-11x<19$ so $11x>70$ so $x>6$.


Lastly $11x < 90$ must be true, or $y$ would be negative which doesn't make sense. Thus $x < 9$.


So $x$ is odd. $x$ is less than $9$. $x$ is greater than $6$. This leaves $7$ as the only possible value for $x$ and then we simply work out $y$ is $6$ for the answer.


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