Time for some math problems! I was initially going to share two interesting problems featuring the number 2, but then it turned out I’d done one of them wrong and the answer wasn’t 2 at all. But then I turned to the digits 1 and 3, and they saved the day. Hooray!

Problem #1 is about amicable pairs. These are numbers each of whose proper divisors add to the other number. For example, 220 and 284 are amicable, because the proper divisors of 220 (1,2,4,5,10,11,20,22,44,55,110) add to 284, and the proper divisors of 284 (1,2,4,71,142) add to 220. And that’s the example you’re going to find if you do **any **sort of search for amicable pairs, because the next pair is 1184 and 1210. Not that they are so uncommon (there are about 12 million known pairs), but they do tend to be pretty big.

So anyway, take any amicable pair. Start with the first number, add the reciprocals of all of its divisors, and then take the reciprocal of the sum [i.e. (1/1+1/2+1/4+…+1/220)^{-1}]. Do the same thing to the other number [i.e. (1/1+1/2+1/4+…+1/284)^{-1}]. Then add those two numbers together, and you’ll get 1.

Pretty neat, huh? And the proof isn’t too bad (it involves putting each sum over a common denominator and falls into place from that, where “falls into place” can mean anything from you see it right away to six hours of staring).

Problem #2: I learned of this problem via Ted’s comment here. (Thanks Ted!) He linked to Ron Knott’s site, which has all sorts of stuff about generating Pythagorean Triples, including this algorithm:

- Start with two fractions that multiply to 2: for example, 3/4 and 16/6.
- Add 2 to each number: 11/4 and 28/6
- Put the numbers over a common denominator: 33/12 and 56/12
- The numerators will be the first two numbers of a Pythagorean Triple: 33
^{2}+56^{2}=65^{2}

Isn’t that cool?? Furthermore, you’ll get primitive triples exactly when you start with reduced fractions (and use the least common denominator in the third step).

Problem #3 is about base 3. Find all positive integers *n* such that neither *n* nor *n*^{2} have any 1s when written in Base 3. And since I told you the answers to the first two problems and just left off the proofs, I think I’ll leave off the answer to this and let you see if you can find them all. Happy Problem Solving!

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March 8, 2009 at 12:09 am |

Re #2 , you’re welcome, and thank you for your very enjoyable and educational site!

Re #3, I’ve found them all; I wrote a crude proof that I can add tomorrow night so as not to spoil this for anyone else (unless someone writes a better one before I get online again.)

Here’s my list, in base 10.

March 8, 2009 at 12:17 am |

I figured the one without an answer would be the easiest.

I won’t spoil #3 either.

March 14, 2009 at 12:17 am |

Hm, no other answerers? Then I will hint: Square various values of n base 3 and see what happens.

March 23, 2009 at 8:17 am |

I would love to see your proof of this, Ted. This problem has been driving me nuts for days. I went out to dinner with my mom and sister and was working on it on the back of a placemat and was so close, but I can’t put the last piece together. I noticed the last digit of any number in base three is the remainder of that number when divided by three, so n^2 in base three will always have a final digit of one or zero. The only time it is zero and not one is obviously when the number is divisible by three. My theory was that there aren’t any numbers where neither n nor n^2 contain any ones when written in base three, but I couldn’t figure out the complete reason why the multiples of three always contained a one when squared. Or I could be completely off base on this piece too…