# 59 arithmetic/digits/divisible.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.

# 59 arithmetic/digits/divisible.p

Find the least number using 0-9 exactly once that is evenly divisible by each

of these digits.

arithmetic/digits/divisible.s

Since the sum of the digits is 45, any permutation of the digits gives a

multiple of 9. To get a multiple of both 2 and 5, the last digit must

be 0, and thus to get a multiple of 8 (and 4), the tens digit must be

even, and the hundreds digit must be odd if the tens digit is 2 or 6,

and even otherwise. The number will also be divisible by 6, since it is

divisible by 2 and 3, so 7 is all we need to check. First, we will look

for a number whose first five digits are 12345; now, 1234500000 has a

remainder of 6 when divided by 7, so we have to arrange the remaining

digits to get a remainder of 1. The possible arrangements, in

increasing order, are

78960, remainder 0

79680, remainder 6

87960, remainder 5

89760, remainder 6

97680, remainder 2

98760, remainder 4

That didn't work, so try numbers starting with 12346; this is impossible

because the tens digit must be 8, and the hundreds digit cannot be even.

Now try 12347, and 1234700000 has remainder 2. The last five digits can

be

58960, remainder 6

59680, remainder 5, so this works, and the number is

1234759680.

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