Mental Addition and Subtraction

Partition both numbers: 76 + 45 = 70 + 40 + 6 + 5 = 110 + 11 = 121. The crucial step is recognising that 6 + 5 = 11 (not just 1), and then adding the full 11 to 110. This addition bridges through 100 — two two-digit numbers combine to make a three-digit answer.

Alternatively, add in steps: 76 + 40 = 116, then 116 + 5 = 121. This sequencing approach avoids having to recombine two partial sums. We can check: 121 − 45 = 76 ✔. This is a good question for highlighting how different mental strategies handle the hundred-boundary in different ways.

Use compensation: 83 − 47 = 83 − 50 + 3 = 33 + 3 = 36. We subtracted 3 too many, so we add 3 back.

Alternatively, count on from 47: from 47 to 50 is 3, from 50 to 83 is 33, so the total gap is 3 + 33 = 36. We can check: 36 + 47 = 83 ✔. This uses the relationship between addition and subtraction.

Adding 10 gives us 1 more than we need. Subtracting 1 corrects the overshoot. The net effect is +10 − 1 = +9. For example: 56 + 9 = 65, and 56 + 10 − 1 = 66 − 1 = 65 ✔.

Think of it like money: Imagine you owe a shopkeeper £9, but you only have a £10 note. You hand over the £10 (adding 10), but you’ve given too much! The shopkeeper must hand you £1 back (subtracting 1). The net result to your wallet is exactly a £9 difference.

This is the compensation strategy: round the number you’re adding (or subtracting) to something friendlier, then adjust. It works because 9 = 10 − 1, so adding 9 is identical to adding 10 and subtracting 1. The same idea extends: adding 19 = adding 20 − 1, adding 29 = adding 30 − 1, and so on.

Count back from 1000: one less is 999, two less is 998, three less is 997. Alternatively, we know 997 + 3 = 1000 (since 7 + 3 = 10, which takes us neatly to the next thousand), so 1000 − 3 must be 997.

This feels tricky because 1000 has zeros in the tens, hundreds, and ones columns — it looks like there’s “nothing to subtract from.” But thinking of 1000 as 999 + 1, we get 999 + 1 − 3 = 999 − 2 = 997. Subtracting across multiple place-value boundaries is a common source of error in written methods, but mental strategies like counting back or using known number facts can bypass those difficulties entirely.

99 is the same as 100 − 1. So 99 + 99 = (100 − 1) + (100 − 1) = 200 − 2 = 198. This is a compensation strategy — round up to a friendly number, then adjust.

Alternatively, think of it as a near-double: double 100 is 200, but we’ve added 1 too many twice, so 200 − 2 = 198. Or add in steps: 99 + 99 = 99 + 100 − 1 = 199 − 1 = 198. All three approaches exploit the closeness of 99 to 100, showing that choosing a nearby friendly number can make mental calculation much easier.

If you add 2 to both 83 and 48, the gap between the two numbers remains exactly the same. So 83 − 48 is mathematically identical to (83 + 2) − (48 + 2), which is 85 − 50.

This is the constant difference strategy. By shifting both numbers by the same amount (+2), we keep the difference identical but create a much friendlier calculation, as subtracting 50 is significantly easier mentally than subtracting 48.

Example: 37 + 63 = 100

Another: 82 + 18 = 100

Creative: 50 + 50 = 100 — using the same number twice! Or 91 + 09, which sparks debate about whether 09 counts as a two-digit number.

Trap: 46 + 64 = 110, not 100. A student might spot that 37 and 63 have “reversed digits” and leap to the rule “reverse the digits to find the partner to 100.” But reversing the digits of any two-digit number and adding always gives a multiple of 11 (here, 46 + 64 = 110 = 11 × 10), so it can never give exactly 100. The actual requirement is that the ones digits add to 10 and the tens digits add to 9.

Example: 103 − 97. The numbers are close together, so counting on from 97 is quick: 97 → 100 → 103, giving a difference of 6.

Another: 501 − 496. Count on: 496 → 500 → 501, difference = 5.

Creative: 10,001 − 9,998 = 3. The written subtraction looks daunting (crossing many boundaries) but counting on from 9,998 gives the answer in seconds.

Trap: 85 − 12 = 73. The numbers are far apart, so counting on from 12 all the way to 85 would be slow and error-prone. Partitioning (85 − 10 − 2 = 73) is much more efficient here. Counting on works best when the two numbers are close together.

Example: 64 + 52 = 116. Two 2-digit numbers giving a 3-digit answer.

Another: 7 + 8 = 15. Two 1-digit numbers giving a 2-digit answer.

Creative: 999 + 1 = 1000. A 3-digit and a 1-digit number giving a 4-digit answer — the tiniest addition that tips over into a new place value.

Trap: 50 + 30 = 80. Both are 2-digit and the answer is also 2-digit. A student might think “the numbers are big, so the answer will have more digits,” but size alone doesn’t determine digit count — what matters is whether the sum crosses the next power of 10.

Example: 3 − 7 = −4

Another: 10 − 25 = −15

Creative: 0 − 1 = −1, the simplest possible example. Or think of real life: a temperature of 2°C drops by 5°C, giving −3°C.

Trap: 7 − 3 = 4. A student might write “3 − 7” but then swap the numbers and calculate 7 − 3 = 4 instead, believing “you always take the smaller from the larger.” This is the “you can’t subtract a bigger number from a smaller number” misconception — in fact you can, and the result is negative.

True case: 3 + 5 = 8, which is bigger than both 3 and 5. This works whenever both numbers are at least 1.

False case: 7 + 0 = 7, which is not bigger than 7 — it’s equal to it. Zero is a whole number, and adding it doesn’t increase the total. This challenges the common belief that “addition always makes things bigger.” If we extend this to integers, adding a negative number actually makes the result smaller (e.g. 7 + (−3) = 4), which is a powerful preview of a future topic.

If \( a – b = c \), then \( c + b = a \). Addition and subtraction are inverse operations, so this check always works. For example: 83 − 47 = 36. Check: 36 + 47 = 83 ✔.

This is one of the most powerful checking strategies students can learn. It works for any numbers — whole numbers, decimals, negatives — because the inverse relationship is a fundamental property of arithmetic, not a trick that only works in special cases.

True case: 5 − 5 = 0, and 5 − 5 = 0. When both numbers are equal, swapping makes no difference.

False case: 8 − 3 = 5, but 3 − 8 = −5. The answers are not the same. Subtraction is not commutative — the order matters. This is a crucial difference from addition, where swapping always works (3 + 8 = 8 + 3). Many students assume that because order doesn’t matter in addition, it shouldn’t matter in subtraction either.

5 + 0 = 5. 123 + 0 = 123. Zero added to any number leaves it unchanged. Zero is the identity element for addition.

Students sometimes confuse “adding zero” with “writing a zero on the end.” If adding 0 changed a number’s value, then eating 0 biscuits would make you gain weight! The confusion happens because students think of ‘putting a zero on the end’ (which is actually multiplying by 10). 5 + 0 = 5, but 5 × 10 = 50.

It is almost never a strict requirement to start with the first number, because addition is commutative. In fact, changing the order is often a core mental strategy.

For example, if the calculation is 4 + 78, starting at 4 and counting on 78 is incredibly tedious and error-prone. Swapping it to 78 + 4 makes it trivial. A flexible mathematician looks at the whole calculation before deciding the most efficient way to process it.

(Note: All three calculations have a difference of 6)

(Note: All three calculations sum to 85)

The student correctly partitioned and found 80 and 11, but then only added the ones digit of 11 to 80, getting 80 + 1 = 81. They lost the 10 from the 11. This is a partitioning recombination error — the student doesn’t know how to handle a two-digit partial sum.

The correct calculation is 80 + 11 = 91. The 11 is not “1” — it is 11, which means adding a full ten and a one: 80 + 10 + 1 = 91. This error is very common when students begin using partitioning as a mental strategy.

The answer is correct — 45 + 55 is 100 — but the reasoning is a dangerous overgeneralisation. The student has invented the rule: “if the ones digits add to 10, the answer is a round hundred.” This is wrong.

Counterexample: 25 + 35 = 60. The ones digits add to 10 (5 + 5), but the answer is 60, not a round hundred. The rule only worked for 45 + 55 because the tens digits also add to 9 (4 + 5 = 9). For two two-digit numbers to add to 100, you need the ones to add to 10 and the tens to add to 9. Getting the right answer with flawed reasoning is especially risky — it can go undetected until it causes problems later.

The student can’t subtract 7 from 4 in the ones column, so they reverse the subtraction and do 7 − 4 = 3 instead. This is sometimes called the “smaller from larger” error — the student always subtracts the smaller digit from the larger, regardless of which is on top.

The correct answer is 297. A good mental approach: 304 − 4 = 300, then 300 − 3 = 297 (splitting the 7 into 4 and 3). Or count back: 304, 303, 302, 301, 300, 299, 298, 297. The student’s answer of 303 should raise a red flag — subtracting 7 should move the answer much further from 304, yet 303 is only 1 away. That’s a sign something has gone wrong.

The student correctly partitioned 25 into 20 and 5, and correctly calculated 67 + 20 = 87. But then they subtracted 5 instead of adding it. This is a sign confusion in multi-step mental calculations — the student lost track of the operation during the second step of their strategy.

The correct answer is 87 + 5 = 92. This error is especially common when students mix up the compensation strategy (where you do subtract after adding too much, e.g. +30 then −5) with straightforward partitioning (where both parts must be added). The student may have confused the two approaches.

The student has made the classic “smaller from larger” error, but this time applied across the decimal point. Because they cannot subtract 5 tenths from 2 tenths, they simply reversed the subtraction in that column (5 − 2 = 3), ignoring place value and the direction of the operation.

The correct answer is 3.7. A robust mental strategy would be to partition the 0.5: first subtract 0.2 to get down to 4.0, then subtract the remaining 0.3 to get 3.7. This shows how crucial it is to extend whole-number boundary strategies to decimals.