Find the probability that if each of the digits 0, 1, 2, ..., 9 be placed exactly once 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.

396 = 4*9*11

The constant end 76 assures that the number will be divisible by 4.

The digits that are already present add up to 90 and the digits from 0 to 9 add to 45 and 90+45=135 is divisible by 9, so the number in question will be divisible by 9.

The blank positions all have the same parity (2, 6, 8, ... from the left, or 3, 5, 7, ... counting from the right). The alternating set that does not contain blanks add to 73; the set with the blanks sums to 17 and with the addition of the digits 0 to 9 come out to 62. The difference between the sums of odd and even positions therefore comes out to be 11. The number is also assured of being a multiple of 11.

The probability is therefore 1 that the number is divisible by 4 and by 9 and by 11, and therefore by 396.

 

Posted by Charlie on 2013-08-19 12:27:31