Times Tables

Seven groups of 8 can be split into five groups of 8 and two groups of 8, because \( 5 + 2 = 7 \). Working it out: \( 5 \times 8 = 40 \) and \( 2 \times 8 = 16 \), so \( 40 + 16 = 56 \). And \( 7 \times 8 = 56 \). Both sides give 56.

This is the distributive property — you can break one factor into parts and multiply each part separately. It’s a powerful strategy for working out tricky times tables from ones you already know.

Imagine 9 rows of 4 counters arranged in a rectangle. That’s \( 9 \times 4 = 36 \) counters. Now rotate the rectangle 90° — you see 4 rows of 9 counters, which is \( 4 \times 9 = 36 \). The total number of counters hasn’t changed.

This is the commutative property of multiplication: swapping the order of the factors doesn’t change the product. It means that if you know \( 9 \times 4 \), you automatically know \( 4 \times 9 \) — halving the number of facts you need to learn!

\( 99 \times 5 = 100 \times 5 – 1 \times 5 = 500 – 5 = 495 \). Since 99 is one less than 100, we can use the easy fact \( 100 \times 5 = 500 \) and subtract one group of 5.

This shows how known times table facts can be used as building blocks for harder calculations. The compensation strategy — rounding to a nearby friendly number and adjusting — is one of the most powerful mental multiplication techniques.

Every number in the 6 times table can be written as \( 6 \times n \) for some whole number \( n \). But \( 6 = 3 \times 2 \), so \( 6 \times n = 3 \times 2 \times n = 3 \times (2n) \). This is a multiple of 3, so it’s in the 3 times table.

We can check: 6, 12, 18, 24, 30… all appear in the 3 times table. The reverse isn’t true though — 9 and 15 are in the 3 times table but not the 6 times table. This is because 6 is a multiple of 3, so the 6 times table is contained within the 3 times table.

If we follow the order of operations, the first calculation is \( 2 \times 12 \), which equals 24. The second calculation is \( 6 \times 4 \), which also equals 24. They both yield the exact same total.

This demonstrates the associative property of multiplication: when multiplying three or more numbers, the grouping of the factors does not change the product. This is incredibly useful for mental maths (e.g., solving \( 5 \times 14 \) by thinking of it as \( 5 \times (2 \times 7) = (5 \times 2) \times 7 = 10 \times 7 = 70 \)).

Example: \( 7 \times 8 = 56 \)

Another: \( 9 \times 7 = 63 \)

Creative: \( 9 \times 9 = 81 \) — the largest product of two single-digit numbers. There are actually only six combinations that work (ignoring order): 7×8, 7×9, 8×8, 8×9, 9×9, and 6×9 = 54.

Trap: \( 6 \times 8 = 48 \) — a student might feel confident this is over 50 because 6 and 8 are “big” single-digit numbers, but the product falls just short.

Follow-up prompt: Ask students: What is the highest product we can make with two different single-digit numbers? (Ans: \( 8 \times 9 = 72 \)).

Example: \( 3 \times 8 = 24 \) and \( 4 \times 6 = 24 \)

Another: \( 2 \times 9 = 18 \) and \( 3 \times 6 = 18 \)

Creative: \( 1 \times 36 = 36 \) and \( 6 \times 6 = 36 \) — one uses a number outside the standard tables. Or \( 2 \times 2 \times 6 = 24 \) and \( 3 \times 8 = 24 \) — using three factors against two!

Trap: \( 4 \times 8 = 32 \) and \( 5 \times 7 = 35 \) — a student might think “I added 1 to the 4 and subtracted 1 from the 8, so the answer stays the same.” This “balance” misconception feels logical but doesn’t work for multiplication (it does work for addition: \( 4 + 8 = 5 + 7 \), which is why students get confused).

Example: 12 (since \( 3 \times 4 = 12 \) and \( 4 \times 3 = 12 \))

Another: 24 (since \( 3 \times 8 = 24 \) and \( 4 \times 6 = 24 \))

Creative: 0 — it’s in every times table! Or 120, 360, or any multiple of 12. These numbers are called common multiples of 3 and 4, and 12 is the lowest common multiple.

Trap: 15 — a student might think this works because it’s in the 3 times table (\( 3 \times 5 = 15 \)) and it “feels” like a multiple of 4. But \( 15 \div 4 = 3.75 \), so it’s not in the 4 times table.

Example: \( 0.5 \times 0.4 = 0.2 \) — the answer (0.2) is smaller than both 0.5 and 0.4.

Another: \( 0.3 \times 0.7 = 0.21 \)

Creative: \( 0.1 \times 0.1 = 0.01 \) — the answer is one hundred times smaller than either factor. Or \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \) using fractions.

Trap: \( 2 \times 3 = 6 \) — a student might pick any two small numbers, thinking “small times small gives something smaller.” But with whole numbers greater than 1, the product is always larger. Both factors must be between 0 and 1 for this to work. This challenges the “multiplication makes bigger” misconception.

True case: \( 3 \times 4 = 12 \) is bigger than \( 3 + 4 = 7 \). For most pairs of whole numbers greater than 2, the product beats the sum.

False case: \( 1 \times 2 = 2 \) is less than \( 1 + 2 = 3 \). Also \( 2 \times 2 = 4 \) equals \( 2 + 2 = 4 \). Students often assume multiplication always “wins,” but small numbers and numbers less than 1 break this rule.

If you know \( 7 \times 8 = 56 \), you automatically know \( 56 \div 8 = 7 \) and \( 56 \div 7 = 8 \). Every times table fact gives you two division facts (or one, if both factors are the same, like \( 6 \times 6 = 36 \) gives \( 36 \div 6 = 6 \)).

This is because multiplication and division are inverse operations. Understanding this connection is one of the most powerful reasons to learn times tables — they unlock division too. Students who see times tables and division as completely separate skills are missing this fundamental relationship.

Odd × odd always gives odd. Check: \( 3 \times 5 = 15 \), \( 7 \times 9 = 63 \), \( 11 \times 3 = 33 \). Every product is odd.

Why? An odd number can be written as \( 2k + 1 \). Multiplying two odd numbers: \( (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1 \), which is odd. For a product to be even, at least one factor must be even. Students often mix this up with the rule for adding two odd numbers, which does give an even result.

True cases: the first nine multiples of 9 all have digits that add to exactly 9. Check: 9, 18 (1+8), 27 (2+7), 36 (3+6), 45 (4+5), 54 (5+4), 63 (6+3), 72 (7+2), 81 (8+1) — every digit sum is 9.

False case: \( 9 \times 11 = 99 \), and \( 9 + 9 = 18 \), not 9. Also \( 9 \times 21 = 189 \), and \( 1 + 8 + 9 = 18 \). The correct rule is that the digits of any multiple of 9 always add up to a multiple of 9 (not necessarily 9 itself). Students who only practise the 9 times table up to \( 9 \times 9 \) often over-generalise this beautiful pattern.

True case: For 4 and 6, the HCF is 2 and the LCM is 12. In almost all instances, this statement holds true, as factors are generally smaller than the numbers themselves, and multiples are generally larger.

False case: For 5 and 5, the HCF is 5 and the LCM is also 5. They are equal! This forces students to really clarify the difference between factors and multiples, and remember to test edge cases (like identical numbers).

The student is ordering by the first factor only without actually calculating the products. This ignores the fact that both factors contribute to the size of the answer.

The actual products are: \( 3 \times 9 = 27 \), \( 4 \times 7 = 28 \), \( 5 \times 5 = 25 \). So the correct order from smallest to largest is \( 5 \times 5 \), \( 3 \times 9 \), \( 4 \times 7 \). A larger first factor doesn’t guarantee a larger product — you must work out the answers.

The answer is correct — 56 is in the 8 times table (\( 8 \times 7 = 56 \)) — but the reasoning is completely wrong. The student has invented a false divisibility test: “if the digit sum eventually gives an even number, the number is in the 8 times table.” This has no mathematical basis.

A counterexample disproves it instantly: 38 has digits summing to 11, then 1 + 1 = 2 (even), but \( 38 \div 8 = 4.75 \), so 38 is not in the 8 times table. To check divisibility by 8, you need to actually divide or count through the 8 times table. Digit-sum tests only work reliably for 3 and 9.

The student is confusing multiplying by 0 with multiplying by 1. Multiplying by 1 leaves a number unchanged (the identity property), but multiplying by 0 always gives 0. The correct answer is 0.

Think of it this way: \( 3 \times 4 \times 0 \) means “zero groups of twelve,” which is nothing. Any number multiplied by 0 equals 0 — no matter how large the other factors are. This is a fundamental property that students sometimes resist because it feels like the 12 should “survive.”

The student has over-extended the “add a zero” shortcut for multiplying by 10. They reasoned that since 5 is half of 10, you should “add a 5” instead of “add a zero.” This is a creative but completely false generalisation — multiplication doesn’t work by appending digits.

The correct answer is \( 12 \times 5 = 60 \). A good strategy: \( 12 \times 10 = 120 \), then halve it to get \( 120 \div 2 = 60 \). This does correctly use the relationship between ×5 and ×10, but through halving the result, not halving the “add a zero” rule.