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Sunday, June 17, 2012

A Little Mathematical Riddle

I sure seem to be in the habit of taking sabbaticals! :)

Imagine you are Chief of Police Montgomery, and you want to arrest the Rogerson brothers.
You want to know how many brothers there are, in order to fit them all in your prison cell. You can't see how many brothers because they have taken sanctuary.
You do, however, know 6 facts:
  1. The Rogerson brothers' father, Peter, was a TzuTzuist. The TzuTzuists believe that you should only have a single digit number of kids.
  2. Peter was extremely rich. His fortune was worth an eight-figure odd number.
  3. The last digit of the fortune is 9.
  4. The digit sum of the fortune is 17.
  5. Peter divided the fortune evenly among his sons.
  6. The fortune each brother received is a whole number.
How many Rogerson brothers are there? 

5 comments:

  1. Hi Dash, this is Alex and Geo. The answer is 7 brothers. It's not 1 because "brothers" is plural. It's not 2, 4, 6, 8 because the fortune is not divisible by 2. It's not 5 because nothing ending in 9 is divisible by 5. It's not 9 because nothing times 9 will produce a product that ends in 9. That leaves 3 and 7. There are only 7 possible values of fortune, and 11111219 is divisible by 7. Actually, we did not rule out 3, yet, but we certainly know how. Anyhow, 7 jail spaces will be needed. Thank you for a fun problem.

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    Replies
    1. George: You can rule out 3 because all numbers that are a multiple of 3 have a digit sum that is a multiple of 3. But the digit sum is 17.

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    2. Thank you very much! I had forgotten about that divisible-by-3 trick! Now I get to teach it to Alex. -- George

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    3. [2nd try to reply] Thank you! I had forgotten about that divid-by-3 trick! Now I get to teach it to Alex. -- George

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