Sandy Wood
Brain Game: Pick Three
by Sandy Wood - April 16, 2009 - 7:30 AM

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Herbert served as the neighborhood’s local runner; he’d collect money and ”Pick Three” lottery numbers from his neighbors and would then buy the tickets all at once. When players won, he delivered the cash and received a percentage for his trouble.

Josephine was a regular, and she seemed to win more than most of Herbert’s other customers. “What’s your secret?” he finally asked her. “There’s no secret,” winked Josephine. “I’ve just been lucky. But all the three-digit numbers I choose do have something in common. You can figure it out.”

And she was right. That night, Herbert did some playing around with the six numbers Josephine chose that day and learned her secret. Can you?

Here were Josephine’s numbers for that day:

297   429   583   616   704   836

Here is the SOLUTION.

 

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Comments (5)

  1. OK it took a little while and an extra cup of coffee but it finally came into sight. Good one.

    posted by Kev in GA on 4-16-2009 at 8:20 am

  2. How the heck would that help her win?

    posted by dave tsunami on 4-16-2009 at 11:24 am

  3. Huh, how interesting, this is what I got:

    Add the numbers in the one’s and hundred’s place and, from the resulting sum, subtract the number in the ten’s place from the number in the one’s place. The difference is equal to the original number’s digit in the ten’s spot.

    For example:

    297 : 2 + 7 = 9 (09) -> 9 – 0 = 9
    429 : 4 + 9 = 13 -> 3 – 1 = 2

    Without taking any effort whatsoever to research or prove this, this looks like it might be a property of all 3 digit multiples of 11.

    posted by jso on 4-16-2009 at 2:47 pm

  4. It doesn’t (and wouldn’t) help her at all, Dave. She admitted herself she she was “just lucky.”

    posted by Sandy Wood on 4-16-2009 at 3:16 pm

  5. jso,

    You are right. It’s an old, pre-computer method of testing for divisibility by 11.

    Take an integer and pull out every other digit. Those digits pulled out form one group and the digits remaining form another group. Add up the digits of each group. Subtract one sum from the other. If the difference is 0 or is divisible by 11 (recursion!), then the original integer is divisible by 11.

    For any base, this works as a test for divisibility by base+1.

    There are all sorts of tricks like that for doing mental math quickly. It’s a testament to what we’ve forgotten since we started relying on computers.

    posted by Bryan on 4-17-2009 at 2:42 pm

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