Anodyne
Friday, September 26, 2014
 
This is the only useful thinking I've accomplished this week, & it fills me with weird giddy joy, like a chest full of helium.  This took two full days of non-work thinking to accomplish, btw, so I'm not planning on quitting the day job any time soon.  A MIT-bound high schooler could probably have solved it in 10 minutes, but that person isn't me.

Q:  Bob multiplies three different prime numbers together.  Is it possible for the digits of that product to sum to 18?  Why or why not? 

CJB:

1.  To make a number divisible by 9 (9, 18, 27 & etc.), 9 needs to factor in the initial multiplication process.  9 = 3 * 3, so you need 2 3s as factors.

2.  You can never get 2 3s using "3 different prime numbers" because 3 is prime and by the question's rules you can only use it once.

3.  Therefore you can never produce a 9.

4.  By the fundamental theorem of arithmetic, 18 uniquely factors as 2 * 3 *3.

4a. Bob is consequently SOL with his multiplication process, because it will never produce one of the three factors required for an 18.
- posted by cjb @ 8:42 AM


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