Comments on: Probability of Divisibility http://www.mathteacherctk.com/blog/2009/05/probability-of-divisibility/ Thoughts on math education and related tidbits Fri, 20 Jan 2012 11:38:59 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: admin http://www.mathteacherctk.com/blog/2009/05/probability-of-divisibility/comment-page-1/#comment-1805 admin Tue, 15 Dec 2009 13:34:38 +0000 http://www.mathteacherctk.com/blog/?p=62#comment-1805 Quite 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. Quite 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.

]]> By: Sol Lederman http://www.mathteacherctk.com/blog/2009/05/probability-of-divisibility/comment-page-1/#comment-1804 Sol Lederman Tue, 15 Dec 2009 04:45:46 +0000 http://www.mathteacherctk.com/blog/?p=62#comment-1804 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. 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.

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