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 Post subject: Challenging problem.
PostPosted: Sun, 9 Nov 2008 00:16:40 UTC 
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This was a question from APICS 2008 (3 weeks ago). The average on those exams are ~20%, so I hope this is challenging. I participated in APICS and scored 3/8 and couldn't figure out the following question.

By the way we had no calculators allowed.

Question:

What is the smallest number of 9 digits of base 10 that contains all the numbers from 1 to 9 and is divisible by 99?


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PostPosted: Sun, 9 Nov 2008 11:05:03 UTC 
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123475869

Spoiler:
The sum of the digits is 45, which is divisible by 9. So any such number will be divisible by 9. In order to be divisible by 11 (and therefore by 99), the sum of the even-numbered digits must differ from the sum of the odd-numbered digits by a multiple of 11. The only ways to write 45 in this way are 17+28 and 6+39. You can't have four or five distinct digits with sum 6, so we are looking for either four digits with sum 28 (so that the other five have sum 17), or five digits with sum 28 (so that the other four have sum 17). A bit of thought shows that the best you can do in the first case is 28=4+7+8+9 and 17=1+2+3+5+6, giving the number 142738596. In the second case you can have 28=1+3+7+8+9 and 17=2+4+5+6, giving the number 123475869.


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PostPosted: Mon, 17 Nov 2008 04:41:27 UTC 
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try the putnam. the median on the putnam is 0 or 1/120.

_________________
Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1


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 Post subject:
PostPosted: Sat, 22 Nov 2008 04:21:50 UTC 
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bugzpodder wrote:
try the putnam. the median on the putnam is 0 or 1/120.
The median score is usually one or two points possible out of 120 points possible.

Source: Wikipedia, William Lowell Putnam Mathematical Competition


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 Post subject:
PostPosted: Sat, 22 Nov 2008 21:15:23 UTC 
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It's getting harder these days ;)
Besides, I can just press edit on Wikipedia and change that to zero or one.

_________________
Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1


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