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.
The problem is obviously contrived. The last two digits of the longish number are 76, implying that the number is divisible by 4. 396/4 = 99 = 9×11. We need to check that, however the ten digits are distributed over the empty spaces, the resulting number is divisible by 9 and also by 11.
A number is divisible by 9 iff the sum of its digits is divisible by 9. The sum of the missing digits is
0 + 1 + 2 + ... + 9 = 45.
Which is divisible by 9. The sum of the present digits is
5 + 3 + 8 + 3 + 8 + 2 + 9 + 3 + 6 + 5 + 8 + 2 + 0 + 3 + 9 + 3 + 7 + 6 = 90
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 45 + 17 = 62. The sum of the remaining digits is the 90 - 17 = 73. The difference of the two sums is 73 - 62 = 11 which is divisible by 11, and we are finished.
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Copyright © 1996-2012 Alexander Bogomolny
