wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> An Even Decimal Digit Puzzle
        
(Message started by: K Sengupta on Jun 22^nd, 2007, 2:14am)

Title: An Even Decimal Digit Puzzle
Post by K Sengupta on Jun 22^nd, 2007, 2:14am

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?

Title: Re: An Even Decimal Digit Puzzle
Post by Grimbal on Jun 22^nd, 2007, 5:05am

[hide]yes[/hide]

Title: Re: An Even Decimal Digit Puzzle
Post by FiBsTeR on Jun 22^nd, 2007, 8:14am

Although the question did not ask for proof, here's my attempt at doing it anyway: [hide]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.[/hide]

EDIT: on rereading, i think i started handwaving at the end :-[

Title: Re: An Even Decimal Digit Puzzle
Post by K Sengupta on Jun 23^rd, 2007, 1:11am

on 06/22/07 at 08:14:35, FiBsTeR wrote:

Although the question did not ask for proof, here's my attempt at doing it anyway

Nice Proof.
Well Done!