Multiples
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.
8 × 9 = 72, so 72 appears in the 8 times table. Equally, 9 × 8 = 72, so 72 appears in the 9 times table too. A number can sit in more than one times table — it doesn’t “belong” to just one.
Alternatively, check by division: 72 ÷ 8 = 9 (exact) and 72 ÷ 9 = 8 (exact). Since both divisions leave no remainder, 72 is a multiple of both. In fact, 72 is also a multiple of 1, 2, 3, 4, 6, 12, 18, 24, 36, and 72 itself.
Since 6 = 2 × 3, any multiple of 6 can be written as 6 × n = 2 × 3 × n = 3 × (2n). This means every multiple of 6 is automatically 3 times some whole number, making it a multiple of 3.
Check the 6 times table: 6, 12, 18, 24, 30, 36… Every one of these appears in the 3 times table too. This isn’t a coincidence — because 3 is a factor of 6, every multiple of 6 must also be a multiple of 3. The same logic works for any factor: every multiple of 6 is also a multiple of 2.
A multiple of 7 is any number that can be written as 7 × (a whole number). Since 7 × 0 = 0, zero fits the definition perfectly — it is 7 multiplied by 0.
This surprises many students because they think multiples “start at” the number itself (7, 14, 21…). But zero is actually a multiple of every whole number, because any number times 0 is 0. We could also check: 0 ÷ 7 = 0 with no remainder, confirming 0 is in the 7 times table.
Take any whole number n. Its multiples are: n, 2n, 3n, 4n, 5n… Since we can keep multiplying by larger and larger whole numbers, the list never ends. For example, the multiples of 3 are 3, 6, 9, 12… and there is no “last one.”
Suppose someone claimed the list stopped — say k × n was the largest multiple. But then (k + 1) × n would be an even larger multiple, contradicting the claim. This is very different from factors, which are always a finite list.
Give an Example Of…
Think carefully — the fourth box is a trap! Give a non-example that looks right but isn’t.
Example: 12
Another: 24
Creative: 0 — it’s a multiple of every number, so it’s a multiple of both 4 and 6. Or 120 — a much larger common multiple students rarely consider.
Trap: 46 — a student might combine the digits of 4 and 6 to make 46. But 46 ÷ 4 = 11.5 and 46 ÷ 6 = 7.67, so it’s a multiple of neither. You can’t find common multiples by sticking the numbers together.
Example: 14
Another: 49
Creative: 91 — many students won’t realise that 7 × 13 = 91. Or 98 — the largest two-digit multiple of 7.
Trap: 17 — a student might think “1 + 7 = 8, which is close to 7, so 17 is a multiple of 7.” They are misapplying the digit-sum divisibility rule that works for 3 and 9, but no such rule exists for 7. In fact, 17 ÷ 7 = 2 remainder 3.
Example: 6
Another: 15
Creative: 111 — the digit sum is 3, confirming it’s a multiple of 3 (since 3 × 37 = 111), but 111 ÷ 9 = 12.33… so it’s not a multiple of 9.
Trap: 36 — a student might think “36 is 3 × 12, so it’s a multiple of 3 but not 9.” But 36 = 9 × 4, so it is also a multiple of 9. Students often stop checking as soon as they find a number in the 3 times table, forgetting that they must actively test the second condition to rule out the 9 times table.
Example: −8
Another: −20
Creative: −400 — a very large negative multiple. Or −4 itself — since 4 × (−1) = −4.
Trap: −6 — a student might think any negative even number is a multiple of 4. But −6 ÷ 4 = −1.5, which is not a whole number. Being negative and even is not enough — the number must be exactly divisible by 4.
Always, Sometimes, Never
Is the statement always true, sometimes true, or never true? Students should justify their decision with examples.
The 5 times table goes: 5, 10, 15, 20, 25, 30, 35, 40… The last digit alternates between 5 and 0, and this pattern repeats forever. No multiple of 5 can ever end in any other digit.
This is because 5 × any even number ends in 0, and 5 × any odd number ends in 5. This is one of the most reliable “last digit” rules in mathematics, and it works in the opposite direction too: any whole number ending in 0 or 5 is a multiple of 5.
True case: 3 × 1 = 3 (odd), 3 × 3 = 9 (odd), 7 × 5 = 35 (odd). False case: 3 × 2 = 6 (even), 5 × 4 = 20 (even), 7 × 6 = 42 (even).
Odd × odd = odd, but odd × even = even. So the multiples of any odd number alternate between odd and even: 3, 6, 9, 12, 15, 18… Students who only check the first multiple (e.g. 3 × 1 = 3) may wrongly conclude the answer is “always.”
Any even number can be written as 2k. Any multiple of 2k is 2k × n = 2(kn), which is always even. So the result of multiplying an even number by any whole number is always even — you can never get an odd answer.
Students might confuse this with “an odd number is a multiple of an odd number” (which is sometimes true, e.g. 15 = 3 × 5). The key insight is that even numbers “pass on” their evenness to all of their multiples, so an odd number can never appear in an even number’s times table.
True case: 12 is a multiple of 3, and 12 > 3. Most multiples students encounter in times tables (6, 12, 18, 24…) are indeed larger than the number. False case: 5 × 1 = 5, so 5 is a multiple of 5 but not greater than 5 — it’s equal to it. And 5 × 0 = 0, so 0 is a multiple of 5 but is less than 5. Also, negative multiples apply here: −10 is a multiple of 5, and −10 < 5.
Students often picture multiples as a list that “starts at the number and goes up” (5, 10, 15, 20…), forgetting that the number itself is a multiple (equal, not greater), that zero is a multiple of every number (less, not greater), and that negative multiples exist. This connects to the common belief that multiples are “bigger versions” of a number.
Because 6 is a multiple of 3, any multiple of 6 is automatically a multiple of 3. We can show this algebraically by factorising out a 3: 6n = 3(2n).
Since n is a whole number, 2n is also a whole number. Therefore, the expression is always 3 multiplied by some whole number, which is the exact definition of a multiple of 3. This is a crucial step in bridging arithmetic multiples with algebraic proof!
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: 1, 2, 3, 6, 12
Reasoning: “I started with the numbers that go into 6, then I carried on to the next one.”
The student has confused multiples with factors. The numbers 1, 2, 3, and 6 are factors of 6 (numbers that divide exactly into 6). The student then added 12, perhaps thinking “keep going” means doubling.
The first five multiples of 6 are found by multiplying 6 by 1, 2, 3, 4, and 5: 6, 12, 18, 24, 30. Multiples are in the 6 times table; factors are numbers that divide into 6. Only 6 and 12 from the student’s list are actual multiples of 6.
Answer: Yes, 48 is a multiple of 8 ✔
Reasoning: “48 ends in 8, and multiples of 8 always end in 8.”
The answer is correct — 48 is a multiple of 8 (since 8 × 6 = 48) — but the reasoning is dangerously wrong. The student has invented a “last digit” rule: “multiples of 8 always end in 8.” This rule is false.
Counterexamples break the rule in both directions: 18 ends in 8 but is not a multiple of 8 (18 ÷ 8 = 2.25). And 16, 24, 32, 40 are all multiples of 8 that don’t end in 8. Unlike multiples of 5, the 8 times table has no simple last-digit rule. The correct check is to see whether the number divides exactly by 8.
Answer: Yes
Reasoning: “53 has a 3 in it, so it must be in the 3 times table.”
The student believes that containing the digit 3 makes a number a multiple of 3. This is a common misconception — students confuse “having a digit” with “being divisible by.”
53 ÷ 3 = 17 remainder 2, so 53 is not a multiple of 3. The correct divisibility test for 3 is to add the digits: 5 + 3 = 8, and since 8 is not a multiple of 3, neither is 53. Meanwhile, 42 has no digit 3 at all but is a multiple of 3 (since 4 + 2 = 6). Having a particular digit and being a multiple of that digit are completely different things.
Answer: All four are multiples of 6
Reasoning: “They are all in the 3 times table, and 3 is half of 6, so multiples of 3 are the same as multiples of 6.”
The student assumes that multiples of a factor are the same as multiples of the number. Because 3 is a factor of 6, the student concludes that the 3 times table and the 6 times table contain the same numbers. This logic is backwards.
In fact, while every multiple of 6 is a multiple of 3, the reverse is not true. To be a multiple of 6, a number must be divisible by both 2 and 3. Checking: 12 ÷ 6 = 2 ✔, 15 ÷ 6 = 2.5 ✘, 18 ÷ 6 = 3 ✔, 21 ÷ 6 = 3.5 ✘. Only 12 and 18 are multiples of 6. The numbers 15 and 21 are multiples of 3 but are odd, so they cannot be multiples of 6.
Answer: 48
Reasoning: “6 × 8 = 48, so the LCM is 48.”
48 is a common multiple of 6 and 8, but not the lowest. The student has just multiplied the two numbers together without checking for smaller common multiples.
The correct Lowest Common Multiple (LCM) is 24 (since 24 ÷ 6 = 4 and 24 ÷ 8 = 3). Multiplying the two numbers always produces a common multiple, but if the numbers share any common factors (in this case, they share a factor of 2), a smaller common multiple will always exist.