Parity of numbers and expressions

Every whole number falls into exactly one of two camps. An even number is a multiple of 2, written $2k$ for some whole number $k$ (like $0, 2, 4, 6,\dots$). An odd number is one more than a multiple of 2, written $2k+1$ (like $1, 3, 5, 7,\dots$). Nothing is both, and nothing is neither — this is what makes parity such a reliable tool. From this, three combining rules follow, and they are worth memorising because they appear constantly in Number Play puzzles:

• even $\pm$ even $=$ even, and odd $\pm$ odd $=$ even
• even $\pm$ odd $=$ odd
• even $\times$ (anything) $=$ even, and odd $\times$ odd $=$ odd

These are not rules to memorise blindly — they follow directly from the $2k$ and $2k+1$ definitions. Take an even number $2a$ and an odd number $2b+1$, where $a$ and $b$ are whole numbers, and add them:

where: $a$ and $b$ are whole numbers labelling which even and odd number was chosen, and $2(a+b)+1$ is exactly the odd-number pattern — twice something, plus one. So even plus odd is odd, always, regardless of which even and odd numbers you picked. The same idea proves odd $\times$ odd is odd:

where: the right-hand side is again $2(\text{whole number}) + 1$, so the product is odd no matter which two odd numbers were multiplied. This is the real engine behind every parity proof: an expression that always simplifies to $2(\text{something}) + 1$ is odd for every valid choice of the letters inside it, and one that simplifies to $2(\text{something})$ is even for every valid choice — the parity is fixed by the algebraic form, not by which numbers you happen to try.

Two more facts round out the toolkit. First, a count matters as much as the numbers themselves: adding an even count of odd numbers gives an even total, while adding an odd count of odd numbers gives an odd total — this is the pattern behind the derivation below. Second, parity arguments only bite when the problem's structure forces a fixed number of odd/even terms to combine in a fixed way; if a quantity can range freely (like how many dots in a grid happen to be coloured red), parity alone cannot pin down its outcome.

Can five odd numbers add up to an even total? Settle it with a parity argument, not by testing examples.

Step 1. From the derivation above, the sum of any five odd numbers always simplifies to $2(a_1+a_2+a_3+a_4+a_5+2)+1$, which is exactly the algebraic pattern for an odd number.

Step 2. Compare this to the claim. If someone insists five odd numbers could sum to an even total, that total would need to fit the form $2(\text{whole number})$ instead — but the algebra shows the sum is always $2(\text{whole number})+1$, never the even form, regardless of which five odd numbers were chosen.

Step 3. Conclude. The sum of five odd numbers can never be even — it is always odd, and this is proven for every possible choice of five odd numbers at once, not just checked for a few. A quick sanity check with real numbers confirms the pattern without replacing the proof: $1+3+5+7+9 = 25$ (odd), and $11+13+15+17+19 = 75$ (odd) — consistent with, but not a substitute for, the general argument.

Competitive stretch

The same reasoning instantly answers a follow-up: can four odd numbers sum to an odd total? Following Step 2 of the derivation with $n=4$ terms gives $2(a_1+a_2+a_3+a_4) + 4 = 2(a_1+a_2+a_3+a_4+2)$, which is the even-number pattern — so four odd numbers always sum to an even total, never odd. In general, an odd count of odd numbers gives an odd sum, and an even count gives an even sum; counting the terms correctly is the first move in any such problem.

• Assuming any sum of odd numbers is automatically odd — the count of terms decides this: an even count of odd numbers (like $3+5=8$) gives an even total, while an odd count gives an odd total.
• Testing a handful of specific numbers and calling it "proof" — a few examples can suggest a pattern but can never rule out every possible case; only a general argument using $2k$ or $2k+1$ can guarantee it for all numbers.
• Forgetting that a parity argument only applies when the problem structure fixes how many odd/even terms are being combined — if a quantity (like a count of coloured dots) can genuinely vary, parity alone cannot force it to be one particular value.
• Losing track of which letters in an expression are "fixed by the problem" versus "can be any whole number" — the strength of a parity proof comes from showing the result holds for every value the free letters could take.

Ask the tutor: Want to try a few more "prove it's impossible" parity puzzles, or check whether your own odd/even argument actually holds for every case?