## 22. Age Difference

The sums of the digits that make up the birth years of Jean and Jacques are equal to each other, and the age of each of them starts with the same digit. Could you determine the difference in their ages?
22. Let $(m, c, d, u)$ be the number of thousands, hundreds, tens, and units in Jean's birth year, and let $\left(m^{\prime}, c^{\prime}, d^{\prime}, u^{\prime}\right)$ be the corresponding digits of Jacques' birth year. Then Jean's age is

$$
1979-(1000 m+100 c+10 d+u)
$$

Jacques' age is

$$
1979-\left(1000 m^{\prime}+100 c^{\prime}+10 d^{\prime}+u^{\prime}\right)
$$

The difference in their ages is

$$
1000\left(m-m^{\prime}\right)+100\left(c-c^{\prime}\right)+10\left(d-d^{\prime}\right)+\left(u-u^{\prime}\right)
$$

But according to the problem statement,

$$
m+c+d+u=m^{\prime}+c^{\prime}+d^{\prime}+u^{\prime}
$$

i.e.,

$$
\left(m-m^{\prime}\right)+\left(c-c^{\prime}\right)+\left(d-d^{\prime}\right)+\left(u-u^{\prime}\right)=0
$$

Subtracting the last expression from the difference in Jean's and Jacques' ages term by term, we get the number

$$
999\left(m-m^{\prime}\right)+99\left(c-c^{\prime}\right)+9\left(d-d^{\prime}\right)
$$

which is clearly divisible by 9.

But this number is less than 10 (since Jean's and Jacques' ages are written as numbers starting with the same digit), and therefore it equals 9.

Thus, the difference in Jean's and Jacques' ages is 9 years.