KA-0049Parity of numbers (odd/even behavior)
5 points, difficulty 3 of 3Level 5–6about 135sThe numbers 1 to 9 are placed in a row in some order. Can every neighbouring pair add up to an odd number?
Hints
Take them one at a time. The first gives nothing away.
1A nudge
Odd plus even is odd. What does that force?
2The strategy
A pair adds to an odd number only when one is odd and one is even, so the whole row has to alternate. Then count how many odd and even numbers you actually have.
3The full solution
Alternating nine places needs five of one parity and four of the other. The numbers 1 to 9 give exactly five odd and four even, so an alternating row exists - and it must start and end odd.
Solution
The reliable way
A pair adds to an odd number exactly when one member is odd and the other even, so the row must alternate parity all the way along. Nine alternating places need five of one parity and four of the other, and 1 to 9 gives exactly five odd numbers and four even ones. So it can be done, with the odd numbers in the first, third, fifth, seventh and ninth places - for example 1, 2, 3, 4, 5, 6, 7, 8, 9. Many orders work: the five odd numbers can be arranged among their places in any order, and so can the four even ones. The transferable idea: turn a question about sums into a question about parity, then count the parities you have.
The elegant way
Alternation is forced, the parity counts match exactly, so the natural order 1 to 9 already works and every reshuffle within each parity works too.
Why this is on the test: It looks like a search and is really a counting-by-parity argument, which is the jump a 5-point question asks a student to make.