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Topic: An Even Decimal Digit Puzzle (Read 651 times) |
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K Sengupta
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An Even Decimal Digit Puzzle
« on: Jun 22nd, 2007, 2:14am » |
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Let X be a 17 digit positive decimal integer with no leading zeroes. The number Y is obtained by considering the decimal digits of X in the reverse order.
Is it always the case that at least one digit of the sum (X+Y) is even?
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| « Last Edit: Jun 22nd, 2007, 2:16am by K Sengupta » |
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Grimbal
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Re: An Even Decimal Digit Puzzle
« Reply #1 on: Jun 22nd, 2007, 5:05am » |
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yes
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FiBsTeR
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Re: An Even Decimal Digit Puzzle
« Reply #2 on: Jun 22nd, 2007, 8:14am » |
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Although the question did not ask for proof, here's my attempt at doing it anyway: When adding x and y vertically, the 9th digit of x and y are equal, so in x+y, adding the two digits would give you an even digit. If, however, there is carrying that preceeds it that causes this 9th digit to be odd, there must be another carry on the other side of the 9th digit. In general, carrying that occurs before adding the 9th digits of x and y would also exist after adding the 9th digits of x and y, since the pairs of digits being added on either side of the 9th digits of x and y are equal. There must therefore be an even number of carrying, although there are an odd number of digit sums to execute. Therefore there will exist a pair of digits, one from x and one from y, that, when added, gets a +1 carry, whereas on the other side of the 9th digit does not get a +1 carry. One of these digits will be odd and the other even. Thus there is at least one even digit.
EDIT: on rereading, i think i started handwaving at the end 
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| « Last Edit: Jun 23rd, 2007, 7:24am by FiBsTeR » |
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K Sengupta
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Re: An Even Decimal Digit Puzzle
« Reply #3 on: Jun 23rd, 2007, 1:11am » |
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on Jun 22nd, 2007, 8:14am, FiBsTeR wrote:| Although the question did not ask for proof, here's my attempt at doing it anyway |
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Nice Proof.
Well Done!
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