26 May
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.
As a hint, the question is more about divisibility than that of probability.
Reference
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The probability is 1.
Let n be the big number.
396 = 4*9*11.
Since 4, 9 and 11 are relatively prime, if 4 | n and 9 | n and 11 | n then 396 | n.
4 | n iff 4 | 76, which it does.
9 | n iff the sum of the digits of n is a multiple of 9. The sum = 135 which is a multiple of 9.
11 | n iff the alternating sum of the digits of n is a multiple of 11.
If the missing digits are d1, d2, ..., d10 then the alternating sum of the digits of n is:
5-d1+3-8+3-d2+8-d3+2-d4+9-3+6-d5+5-d6+8-d7+2-0+3-d8+9-d9+3-d10+7-6 = 5+3+3+8+2+9+6+5+8+2+3+9+3+7-(d1+8+d2+d3+d4+3+d5+d6+d7+0+d8+d9+d10+6) = 73 - 17 - sum(d1 .. d10) = 56-45 = 11.
So, because the alternating sum of the digits of 9 is 11, which is a multiple of 11, and because 4 | n and 9 | n, 396 | n.
December 14th, 2009 at 11:45 pmQuite right. Just to make it clear, a possibly intimidating problem of placing 10 digits into empty 10 spaces to obtain a 28-digit number, proves to be rather a dud: however the digits are placed the resulting number will always be divisible by 396. But such a realization may only be based on a solid understanding of the divisibility criteria.
December 15th, 2009 at 8:34 am