Monday, March 2, 2015

1089 is all that and a bag of chips

Take any three-digit number in which the first and last digits are different.  Reverse the digits to get a new number, so now you have two numbers, and each one is the reverse of the other.  Subtract the smaller from the larger to get a third three-digit number.  (If subtracting gives you a two digit number, then please treat it as a three-digit number whose first digit is zero.)  Now add this number to its reverse.  The result will be 1089.
This is the first puzzle featured for this Saturday's Puzzle night at the Institute in Philadelphia. I love you guys though, and I feel I've been neglecting you as I focus more on the writing that I'm doing and this math group. I will give you a head start on working on the problem on yourself, then I'll give a walk through and some notes.

Something like 836 is a good example to try. So 836 - 638 = 198. 198 + 891 = 1098. Maybe you tried something like 655, in which case you get 655 - 556 = 99. As they point out in the puzzle, treat it like a 3 digit number, 099, in which case 099 + 990 = 1089. Lets now reduce this to the alphabet, because every smart ass in the world hates it when math uses letters. Our 3 digit number will simply be ABC. It COULD be AAC, where the first and second letter are the same, or it COULD be ACC where the second and last number are the same BUT never ABA, because the first and last number ARE different. And only old women like ABBA, but that's neither here nor there. Anyways the first rule of this puzzle is:
  • A > C
because the larger number ALWAYS subtracts the smaller number. Cool? Ice Cold.

Set this puzzle up, and you get the form ABC - CBA = DEF. If our first rule is true, then C - A is always going to be subtracting a larger number from a smaller number. Most of us don't even think about carrying the 1 from B anymore. That changes C from a 1 digit number to a 2 digit number. This tells us the algebraic answer though: C - A = (10 + C) - A. Why 10? Because we are using decimal, of course. If we take a 1 from B and add it to the front of C, it's the same as adding 10 to the number. Here's my second rule:
  • E = 9
The carry over rule applies here too. The thing is, this is not something like 7-8. Since it's in the 10's place, it would look something like 70 - 80. Or algebraically: 10(B-1) - 10B. When you carry the one, it becomes ((10B - 10) + 100) - 10B. As long as B is a number less than 9, than E is always 9. Which just leaves D as (A - C - 1). 100(A - C - 1) + 9 + (C + 10 - A). Or 100A -100C + C - A.

Here is next wild part: D + F = 9. D is (A - C - 1) and F is (C + 10 - A), so A and C cancel out and the whole thing becomes 10 -1, or 9. E is 9 and 9 + 9 is 18, so that extra 1 gets added to the front, making the whole thing 1089. Wow. That took me all afternoon, now I am going to bed. For anyone interested, does this formula work for Octal or Hexadecimal? Is there a way to make it work?

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