Picking parity

Kishor has some number cards and a puzzle: there are 5 boxes, each to hold exactly 1 card, and the numbers in the boxes should sum to 30. Every card in his collection shows an ODD number. Can he do it?

There are a great many ways of choosing 5 cards. Is there a way to find out without checking all the possibilities? There is, and it is worth more than the answer itself.

Start with even numbers. Any even number can be arranged in PAIRS with nothing left over — that is what being even means. So if you put several even numbers together, the whole collection still arranges into pairs with nothing left over. Adding any number of even numbers always gives an even number.

Now odd numbers. An odd number CANNOT be arranged in pairs: it is one more than a collection of pairs, with a single counter left over.

Put TWO odd numbers together. Each contributes a collection of pairs plus one leftover — and the two leftovers pair up with each other. So the whole thing arranges into pairs: the sum of two odd numbers is always EVEN. The picture of the counters is itself a proof.

What about three odd numbers? Two of them pair up neatly, and the third leaves one counter over. So the sum of 3 odd numbers is odd. Explore 4, 5 and 6 odd numbers the same way and you will find the pattern: an even COUNT of odd numbers gives an even sum, an odd count gives an odd sum.

Three rules to remember, each proved by the counters picture:

• even + even = even
• odd + odd = even (the two leftovers pair up)
• even + odd = odd (one leftover stays alone)

Back to Kishor. He has 5 boxes — an ODD number of boxes — and every card is odd. Five odd numbers must sum to an odd number. But 30 is even. So Kishor cannot arrange these cards to add up to 30, and we know it without trying a single combination.