Question

The digits 1,2,3,4,5,6,7,8 and 9 are written in random order to form a nine digit number. Let p be the probability that this number is divisible by 36, find 9p

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Solution

Since 1+2+...+9=36, a number consisting all these digits will be divisible by 9. Thus, the number will be divisible by 36 if and only if it is divisible by 4.There are 9P9=9! ways of arranging the nine digits to form a nine-digit number. If this number is to be divisible by 4, then its last digit must be even and the number formed by its last two digits must be divisible by 4. This limits the possible values of the last pair to the following:12,32,52,72,92,24,64,84,16,36,56,76,96,28,48,68.That is, there are 16 ways of choosing the last two digits. The remaining digits can be arranged in 7P7=7! ways. Therefore the number of favorable arrangements is (7!)(16) out of a total of 9!, so that the required probability is(16)(7!)9!=169×8=29.

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