# Probability of Divisibility

Find the probability that if the digits 0, 1, 2, ..., 9 be placed in random order in the blank spaces of

5_383_8_2_936_5_8_203_9_3_76

the resulting number will be divisible by 396.

### Reference

- C. W. Trigg,
*Mathematical Quickies*, Dover, 1985, #56

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Copyright © 1996-2017 Alexander Bogomolny

1. You can check the solution.

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Copyright © 1996-2017 Alexander Bogomolny

The problem is obviously contrived. The last two digits of the longish number are 76, implying that the number is divisible by 4.

A number is divisible by 9 iff the sum of its digits is divisible by 9. The sum of the missing digits is

Which is divisible by 9. The sum of the present digits is

Which is also divisible by 9. It follows that the resulting number is divisible by 9 regardless of the placement of the missing digits.

To check whether a number is divisible by 11 we compute two sums: that of the evenly placed digits and the sum of the oddly placed digits. All the missing digits come together along with 8, 3, 0, and 6, which add up to

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