4 June 2013

Too early for a puzzle?

I know it is only Tuesday but I am on my flight back home and there are two of my friends who have sent me puzzles that I thought can be posted here…

This one comes courtesy of Amrith Kumar. I sent him an answer – but I have not received confirmation whether I am right or not. So, we might have to wait till he publishes the answer.

Think of all the 4 digit numbers ( leading 0s count). Essentially 0000 to 9999. Now pick only those numbers where the first two digits add up to the same number as the last two digits. So 1533 qualifies as does 0312 and so on.

You have to prove that the total of all such numbers (if you add them up) is divisible by 3 as well as 11.

Posted June 4, 2013 by Rajib Roy in category "Puzzles


  1. By Rajib Roy on

    Amrith, I posted your problem. After a day or so, you might have to acknowledge the right answers or post it. I am tagging you so you can see all the answers. Did you get a chance to see my response this morning?

  2. By Rajib Roy on

    Answer to this puzzle: (Claire, you can go to bed now ๐Ÿ™‚ )
    For every number wxyz from the set of numbers where w+x=y+z e.g. 3470, there is a corresponding number 9999-wxyz in that set which has exactly the same characteristics. In our case 9999-3470=6529 and 6+5=2+9. Therefore, if you add all those numbers in the set, you will get a number that is a multiple of 9999. And therefore is divisible by 3, 9, 11, 99 etc etc


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