### Solution to March's Puzzle: Free Tuna

You go to the grocery store and grab eight cans of tuna from the shelf. You go to the cash register to pay. In a good mood, the store owner hands you three plastic bags and says, "If you can arrange the eight cans in these three plastic bags such that each bag contains an odd number of cans, you can have them for free." Can you think of a way to get that free tuna?

Obviously, you can't divide the eight tuna cans into three separate plastic bags so that each holds an odd number of cans. However, nothing in the puzzle dictates the arrangement of the bags around the tuna cans. The sum of three odd numbers x+y+z, where each number is considered only once, naturally amounts to an odd number. However, taking one of the odd numbers into consideration twice allows for an arrangement in which one of the elements is even (say, y)—for example, (y+(x))+(z) = 8. The use of parentheses is intentional—each pair of parentheses represents a plastic bag. For example, let x equal 1, y equal 2, and z equal 5: You place 1 tuna can in plastic bag A, 2 tuna cans in plastic bag B, and 5 tuna cans in plastic bag C. Then, place plastic bag A in plastic bag B. You end up with 1 tuna can in bag A, 3 in B (x+y), and 5 in C.

As an aside, if you like trying to solve open puzzles, the tuna cans puzzle reminds me of a mathematical conjecture that so far hasn't been proven. The conjecture is named after its conjurer—"Goldbach's conjecture." The original conjecture says, Every odd number can be expressed as the sum of three prime numbers. Euler simplified the conjecture to the form Every even number can be expressed as the sum of two prime numbers. In case you're wondering, I'm not planning on publishing the proof to this conjecture next month.