d, e and f each have nine digits when written in base 10. Each of the nine numbers formed from d by replacing one of its digits by the corresponding digit of e is divisible by 7. Similarly, each of the nine numbers formed from e by replacing one of its digits by the corresponding digit of f is divisible by 7. Show that each of the nine differences between corresponding digits of d and f is divisible by 7.
Solution
Let d = d1d2... d9 and similarly for e and f. We have are given that (d + e9 - d9), (d + 10e8 - 10d8), ... , (d + 108e1 - 108d1) are all multiples of 7. Adding gives that (9d + e - d) is a multiple of 7 and hence also (d + e).
(d + e9 - d9) and (e + f9 - e9) are multiples of 7. So, adding, (d + e + f9 - d9) is a multiple of 7 and hence also (f9 - d9). Similarly, (d + 10e8 - 10d8) + (d + 10f8 - 10e8) - (d + e) = 10(f8 - d8) is a multiple of 7, and hence also (f8 - d8). Similarly, for the other terms.
© John Scholes
jscholes@kalva.demon.co.uk
12 Dec 1998