Consider a ten-digit number with the following properties: each digit from 0-9 occurs in the number exactly once. The number itself is divisible by ten. If you remove the last digit, the nine-digit number so formed is divisible by nine. If you remove the last two digits, the eight-digit number so formed is divisible by eight. If you remove the last three digits, the seven digit number so formed is divisible by seven, and in general the n-digit number formed from the first n digits is divisible by n. Call any number with all of these properties Pascalian. Now the question:
Part A: List all the Pascalian numbers, if any exist.
Part B: Show, by human-verifiable proof, that such a list is exhaustive.
You may wish to make use of the following information in solving this problem: http://en.wikipedia.org/wiki/Divisibility_rule
Good luck.
2006-06-11
13:25:18
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3 answers
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asked by
Pascal
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Science & Mathematics
➔ Mathematics
That's good superobotz, but can you prove that it's the only one?
2006-06-11
13:32:28 ·
update #1