 # 67 arithmetic/digits/nine.digits.p

## Description

This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

# 67 arithmetic/digits/nine.digits.p

Form a number using 0-9 once with its first n digits divisible by n.

arithmetic/digits/nine.digits.s

First, reduce the sample set. For each digit of ABCDEFGHI, such that the last
digit, (current digit), is the same as a multiple of N :

A: Any number 1-9
B: Even numbers 2,4,6,8 (divisible by 2).
C: Any number 1-9 (21,12,3,24,15,6,27,18,9).
D: Even numbers 2,4,6,8 (divisible by 4, every other even).
E: 5 (divisible by 5 and 0 not allowed).
F: Even numbers (12,24,6,18)
G: Any number 1-9 (21,42,63,14,35,56,7,28,49).
H: Even numbers (32,24,16,8)
I: Any number 1-9 (81,72,63,54,45,36,27,18,9)

Since E must be 5, I can eliminate it everywhere else.
Since I will use up all the even digits, (2,4,6,8) filling in those spots
that must be even. Any number becomes all odds, except 5.

A: 1,3,7,9
B: 2,4,6,8
C: 1,3,7,9
D: 2,4,6,8
E: 5
F: 2,4,6,8
G: 1,3,7,9
H: 2,4,6,8
I: 1,3,7,9

We have that 2C+D=0 (mod 4), and since C is odd,
this implies that D is 2 or 6; similarly we find that H is 2 or 6 ==>
{B,F} = {4,8}. D+5+F=0 (mod 3) ==> if D=2 then F=8, if D=6 then F=4.

We have two cases.

Assume our number is of the form A4C258G6I0. Now the case n=8 ==>
G=1,9; case n=3 ==> A+1+C=0 (mod 3) ==> {A,C}={1,7} ==> G=9, I=3.
The two numbers remaining fail for n=7.

Assume our number is of the form A8C654G2I0. The case n=8 ==> G=3,7.
If G=3, we need to check to see which of 1896543, 9816543, 7896543,
and 9876543 are divisible by 7; none are.

If G=7, we need to check to see which of 1896547, 9816547, 1836547,
and 3816547 are divisible by 7; only the last one is, which yields
the solution 3816547290.

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