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| Mission impossible (Posted on 2011-04-18) |
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I was asked to erase one number from a series 3,13,23,…103 and then to erase two members of the remaining sequence, then three and finally four.
I was requested to make sure that after each step of erasures the sum of the remaining members will be divisible by eleven.
Rather than trying to find the "right" order of erasures prove that it cannot be done.
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Submitted by Ady TZIDON
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Solution:
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The total the series = 583 = 11*53.
33 is the only term of the series divisible by 11, so it has to be removed first.
But, after removing 1+2+3+4 = 10 members, 33 is the only one to be left as the last to be emoved.
This contradiction makes it mission impossible.
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