# Chocolate Math

From my half-sister by way of my dad (yay for e-mail forwards):

YOUR AGE BY CHOCOLATE MATH This is pretty neat. DON’T CHEAT BY SCROLLING DOWN FIRST! It takes less than a minute . Work this out as you read … Be sure you don’t read the bottom until you’ve worked it out! This is not one of those waste of time things, it’s fun.

- First of all, pick the number of times a week that you would like to have chocolate (more than once but less than 10)
- Multiply this number by 2 (just to be bold)
- Add 5
- Multiply it by 50 – I’ll wait while you get the calculator
- If you have already had your birthday this year add 1755 …. If you haven’t, add 1754.
- Now subtract the four digit year that you were born. You should have a three digit number The first digit of this was your original number (i.e., how many times you want to have chocolate each week). The next two numbers are YOUR AGE! (Oh YES, it is!!!!!) THIS IS THE ONLY YEAR (2005) IT WILL EVER WORK, SO SPREAD IT AROUND WHILE IT LASTS.</lj-cut>

My dad asked for an explanation of why this works. I didn’t exactly give what you’d call an

I’ll prove the “chocolate math” correct directly. A proof by contradiction would be kind of fun, but overkill. :-)

Let n be the number of chocolates per week, b be your birth year, a be your age, and k be a correction for if your birthday hasn’t come yet this year — specifically, let k be 0 if you’ve had your birthday already, or 1 otherwise.

Note that the equation

a = 2005 - b - k

defines the relationship between a (age) and b (birth year), for this year.

The claim is that as long as n is a one-digit number between 1 and 9 inclusive, and a is less than 100 (I’ve inferred this constraint though it wasn’t explicitly stated; the math is wrong otherwise), the number at the end will be a three digit number — at least when written in decimal :-) — where the first digit is n and the last two are a. In math terms, we can write that as

100n + a

It’s fairly straightforward to turn the rest of the instructions into this algebraic expression:

50(2n + 5) + 1755 - b - k

Then, the claim is that these two expressions are equal for all allowed values of the variables. It can be written and simplified this way:

100n + a | = 50(2n + 5) + 1755 - b - k |

= 100n + 250 + 1755 - b - k | |

= 100n + 2005 - b - k |

Cancelling 100n from both sides, we get this equation:

a = 2005 - b - k

Substituting the earlier equation relating a and b, we see that

a = a

Which is a tautology, and the claim is proven.

OK, so a full-fledged direct proof may have been overkill too. :-) </lj-cut>