Sandy Wood
Brain Game: Zero Hour
by Sandy Wood - June 12, 2009 - 7:30 AM

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I couldn’t determine the original source of today’s math puzzle chestnut; I’ve seen it in a few places, so suffice to say that it’s several years old. It’s still a great puzzle and a fun challenge. Good luck:

Group the digits 1-9 into a set of three 3-digit numbers,
using each digit only once in a set.

The second number should be exactly DOUBLE the first number;

and the third number should be exactly TRIPLE the first number.

There are three FOUR* possible sets of numbers that correctly
solve this puzzle; can you find one (or all of them)?

* see in comments!
Here are the SOLUTIONS.
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Comments (5)

  1. I think there is a fourth answer.

    327…654…981

    327*2=654 and 327*3=981

    posted by MikeTrig on 6-12-2009 at 8:17 am

  2. Looks like you’re correct, MikeTrig; I’ve added yours to the list of solutions.

    One thing I noticed when adding yours is that the top two answers use the same digit-trios in each group… same with the bottom two answers. Interesting!

    posted by Sandy Wood on 6-12-2009 at 8:52 am

  3. Sandy, it looks like those two trio are both there because they preserve the order and the digit that moves from the ones digit to the hundreds digit doesn’t carry when multiplied by 3. E.g. 273 -> 327 and 192 -> 219.

    posted by Eric Y. on 6-12-2009 at 11:25 am

  4. I notice all the answers sum up to 12…15…18

    Is that some mathmatical rule? Like how the two digits of multiples of 9 add up to 9?

    posted by Michelle on 6-12-2009 at 2:11 pm

  5. Yes, Michelle, it is a rule of number theory for both 3 and 9. Any multi-digit number whose digits add up to a multiple of 3 (or 9) is divisible by 3 (or 9). I was taught this as a technique called “casting out 3s”, which can be used for checking math calculations.

    posted by Fred on 6-16-2009 at 5:06 pm

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