Number 31:
An integer n is divisible by both a and b iff n is
divisible by LCM (a,b). Since LCM(7,17) = 119, we need to find an
integer n such that both n and the sum of its digits are
multiples of 119. If we write 
where the digits 119 are repeated one hundred and nineteen times,
we have a positive integer N such that both N and 
, the
sum of the digits of N, are divisible by both 7 and 17, but this
N, with 357 digits, is certainly not the smallest solution.
The smallest positive integer K such that 
is (a multiple
of) 119 is readily seen to be K=29,999,999,999,999, a 14-digit
number; but long division reveals that 
(mod 119).
Our procedure is to increase K as little as possible to a new
integer 
, where 
, but such that 
(mod 119). Specifically, we will increase the leading
digit of K by the smallest positive integer r such that we
can decrease other digits of K by a total amount r,
thus preserving 
, while obtaining 
(mod 119). To assist in this task, we calculate 
(mod
119) for 
:
If we increase the leading 
digit of K by 1, we
add 45 to its remainder modulo 119, getting a total of 
(modulo 119). From the Table, we see that
there is no digit of K, which, if decreased by 1, will add
59 (or, subtract 60), modulo 119.
Next, we increase the leading 
digit of K by 2,
giving a remainder of 
(mod
119). We can reduce other digits of K by a total of 2 (keeping
) to add the needed 14 (mod 119) in exactly one way:
Thus, we take
where 
, and 
; and 
is the smallest positive integer such that both 
and
are multiples of both 7 and 17.