Odd and Even Numbers
Questions designed to stretch thinking, reveal misconceptions, and spark mathematical reasoning.
Convince Me That…
Students must construct a mathematical argument for why each statement is true.
Even numbers can be split into exact pairs, while odd numbers always have one left over. When you add them, the even part pairs up perfectly, but the odd number’s “leftover 1” stays unpaired — so the total always has a remainder of 1 when divided by 2. For example: 4 + 7 = 11, 10 + 3 = 13, 2 + 1 = 3.
Visually, imagine arranging the even number as complete pairs of counters and the odd number as pairs with one spare. Combining them gives a bigger set of pairs, but that one spare counter is still left over, making the total odd. This also works with 0 + 1 = 1 (odd), confirming that 0 behaves as an even number here.
Every odd number is one more than an even number. So if we add two odd numbers, we’re adding two even numbers plus two extra ones: even + even + 1 + 1 = even + 2, which is still even. For example: 7 + 5 = (6 + 1) + (4 + 1) = 6 + 4 + 2 = 12.
Visually, imagine each odd number as pairs of counters with one left over. Two odd numbers each have one “lonely” counter — but those two lonely counters pair up together, leaving no remainder. The result is always even.
We just need one counterexample: half of 6 is 3, and 3 is odd. Similarly, half of 2 is 1 (odd), half of 10 is 5 (odd), and half of 14 is 7 (odd). So halving an even number does not guarantee an even result.
The pattern: even numbers that are multiples of 4 (like 4, 8, 12, 16…) do halve to give even numbers. But even numbers that are not multiples of 4 (like 2, 6, 10, 14…) halve to give odd numbers. Students who test only 4, 8, and 12 might wrongly conclude the result is always even.
A number is even if it can be written as 2 × (a whole number), and odd if it cannot. −7 ÷ 2 = −3.5, which is not a whole number, so −7 is not a multiple of 2. Therefore −7 is odd.
On a number line, odd and even numbers alternate: … −4 (even), −3 (odd), −2 (even), −1 (odd), 0 (even), 1 (odd), 2 (even) … Following this pattern, −7 sits between −8 (even) and −6 (even), confirming it is odd. The concept of odd and even applies to all integers, not just positive ones.
We just need a counterexample to prove it is not always even. Consider 6 ÷ 2 = 3. Both 6 and 2 are even, but the answer 3 is odd!
We can also have results that aren’t whole numbers at all. For example, 2 ÷ 4 = 0.5. Since parity only applies to integers, 0.5 is neither odd nor even. While an even divided by an even can be even (like 8 ÷ 2 = 4), it is definitely not guaranteed. Students often falsely assume division will share the same predictable parity patterns as multiplication.
Give an Example Of…
Think carefully — the fourth box is a trap! Give a non-example that looks right but isn’t.
Example: 3
Another: 5
Creative: 15 — a composite odd factor that students often overlook because they jump to prime factors only.
Trap: 6 — it is a factor of 30 (since 30 ÷ 6 = 5), but 6 is even, not odd. A student might grab the first factor they think of without checking whether it’s odd.
Example: 13 (digit sum: 1 + 3 = 4, which is even; 13 is odd)
Another: 37 (digit sum: 3 + 7 = 10, which is even; 37 is odd)
Creative: 1001 (digit sum: 1 + 0 + 0 + 1 = 2, which is even; 1001 is odd). Or any single odd digit wouldn’t work since a single odd digit has an odd digit sum.
Trap: 24 (digit sum: 2 + 4 = 6, which is even — but 24 itself is even, not odd). This exposes the misconception that the digit sum tells you whether the number is odd or even.
Example: 3 × 8 = 24 (3 is odd, 8 is even)
Another: 1 × 24 = 24 (1 is odd, 24 is even)
Creative: (−3) × (−8) = 24 — using negative numbers! −3 is still odd and −8 is still even, and two negatives multiply to give a positive.
Trap: 4 × 6 = 24 — the product is correct, but both 4 and 6 are even. Students often latch onto the first times table fact they remember for 24, completely forgetting to check the secondary constraint (the odd/even condition) once their brain has successfully solved the multiplication part.
Example: 3 + 5 = 8 (both numbers are odd, and the answer is even)
Another: 7 + 1 = 8
Creative: 3 × 5 − 7 = 8. Or 1 + 1 + 1 + 1 = 4 — adding an even quantity of odd numbers always gives an even result.
Trap: 3 × 5 = 15 — all the numbers used are odd, but the answer is also odd! The product of two odd numbers is always odd, so multiplication alone won’t produce an even result. Students need to combine odd numbers using addition or subtraction to get an even answer.
Always, Sometimes, Never
Is the statement always true, sometimes true, or never true? Students should justify their decision with examples.
If two numbers are both multiples of 2, then their sum is also a multiple of 2. For example: 4 + 6 = 10, 12 + 8 = 20, 0 + 2 = 2. Every case works.
Students might try to find a counterexample but won’t succeed. The key reasoning: adding two groups that can each be split into pairs always gives a result that can be split into pairs.
Some even numbers are multiples of 4: for example, 8, 12, and 20 are all even and multiples of 4. But 6, 10, and 14 are even yet not multiples of 4.
Every multiple of 4 is even (since 4 = 2 × 2), but not every even number is a multiple of 4. Even numbers alternate between being multiples of 4 and not: 2 (no), 4 (yes), 6 (no), 8 (yes), 10 (no), 12 (yes)…
Odd × odd always gives odd. For example: 3 × 5 = 15 (odd), 7 × 9 = 63 (odd), 1 × 1 = 1 (odd). You will never get an even product from two odd numbers.
This is because for a product to be even, at least one factor must be a multiple of 2. If neither number has 2 as a factor, the product can’t have 2 as a factor either. Students who confuse the addition rule (odd + odd = even) with the multiplication rule often get this wrong.
Starting from an odd number: 1 + 2 + 3 = 6 (even), 3 + 4 + 5 = 12 (even), 5 + 6 + 7 = 18 (even). Starting from an even number: 2 + 3 + 4 = 9 (odd), 4 + 5 + 6 = 15 (odd).
Three consecutive numbers always contain one odd, one even, one odd (OEO) or one even, one odd, one even (EOE). In the OEO case, odd + odd = even, then even + even = even. In the EOE case, even + odd = odd, then odd + even = odd. So the sum is even when the first number is odd, and odd when the first number is even.
Almost all prime numbers are indeed odd (e.g., 3, 5, 7, 11, 13…). However, 2 is the only even prime number, so the statement isn’t always true.
This perfectly tests the intersection of two key definitions. Because every even number greater than 2 is divisible by 2 (meaning it has more than two factors), 2 stands entirely alone as the solitary even prime.
Squaring an odd number simply means multiplying an odd number by itself (Odd × Odd). Since the product of two odd numbers is always odd, an odd number squared is always odd.
For example: 3 × 3 = 9, 5 × 5 = 25, 7 × 7 = 49. Students will sometimes confuse squaring (multiplying by itself) with multiplying by 2 (which would result in an even number).
Odd One Out
Which is the odd one out? Can you make a case for each one? There’s no single right answer!
Explain the Mistake
Each example contains a deliberate error targeting a common misconception. Can you find where and why the reasoning goes wrong?
Answer: Neither
Reasoning: “0 is nothing. It’s not really a number, so it can’t be odd or even.”
The student has the “zero is nothing” misconception — confusing the concept of zero the number with the idea of “nothing at all.” Zero is absolutely a number: it sits on the number line, you can add it, subtract it, and use it in calculations.
0 is even. It satisfies every definition of “even”: 0 = 2 × 0 (so it’s a multiple of 2), 0 ÷ 2 = 0 with no remainder, and it sits between two odd numbers (−1 and 1). The pattern … −2, −1, 0, 1, 2 … shows even and odd alternating, with 0 in an even position.
Answer: Odd ✓
Reasoning: “7 × 3 = 21, which is odd. I know it’s odd because the answer takes on the type of the bigger number. 7 is bigger and 7 is odd, so the answer is odd.”
The answer is correct — 21 is odd — but the reasoning is dangerously wrong. The student has invented a “bigger number decides” rule for multiplication. This is a made-up rule that happens to give the right answer here by coincidence.
A counterexample breaks it immediately: 5 × 4 = 20. The bigger number is 5 (odd), but the answer 20 is even. The correct rule is: if either factor is even, the product is even. The product of two odd numbers is always odd — it has nothing to do with which number is larger.
This mistake often happens because students conflate the rules of parity with the rules for adding negative numbers (where the sign of the ‘bigger’ absolute value determines the sign of the answer, e.g., −7 + 3 = −4).
Answer: Odd
Reasoning: “136 has three digits: 1, 3, and 6. Two of those digits (1 and 3) are odd and only one (6) is even. Since the majority are odd, 136 is odd.”
The student has applied a “majority of digits” misconception — looking at all the digits and “voting” on whether the number is odd or even based on how many odd digits it contains. This is completely wrong.
Whether a number is odd or even depends only on the last digit (the ones digit). 136 ends in 6, which is even, so 136 is even. The other digits (1 and 3) are irrelevant to this decision. A quick check: 136 ÷ 2 = 68 with no remainder.
Answer: Even
Reasoning: “20 is even. Subtracting doesn’t change whether a number is odd or even — you start with an even number so the answer is still even.”
The student believes the “starting number keeps its parity” misconception — that subtracting from an even number always leaves it even. This ignores the effect of what you’re subtracting. 20 − 7 = 13, which is odd.
Subtraction follows the same parity rules as addition: even − odd = odd (just as even + odd = odd). Subtracting an odd number always flips the parity. More counterexamples: 10 − 3 = 7 (odd), 8 − 5 = 3 (odd). Only subtracting an even number preserves parity: 20 − 6 = 14 (even).