11.1 Mental Addition, Subtraction, Multiplication, and Division Strategies

Purpose and Design of the Non-Calculator Section

Section 2 of the LANTITE Numeracy test consists of 13 questions completed without access to an on-screen or handheld calculator. Candidates generally allocate the final 25 to 30 minutes of the two-hour exam window to this section, allowing roughly two minutes per question. Because ACER designs Section 2 to assess functional numerical fluency for classroom teachers, the numbers in every problem are carefully chosen to reward structural mental strategies rather than tedious, multi-line paper algorithms.

If you find yourself writing out four rows of long multiplication or performing long division with messy two-digit divisors on your scratch paper, pause immediately: there is almost always a cleaner decomposition, factor split, or compensation shortcut built directly into the numbers.

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Compensation and Equal Additions for Addition and Subtraction

Rounding and Adjusting (Compensation)

Compensation replaces a number that lies close to a multiple of $10$, $100$, or $1.00 (such as numbers ending in $7, 8, 9$ or $1, 2, 3$) with the nearest round benchmark, and then applies an opposite adjustment at the end.

• Addition Compensation: To calculate $468 + 197$, round $197$ up by $3$ to get $200$. Add the benchmark first, then subtract the extra $3$ you introduced:

$$468 + 197 = 468 + 200 - 3 = 668 - 3 = 665$$

• Subtraction Compensation and the Sign-Reversal Trap: To calculate $832 - 296$, round $296$ up to $300$ and subtract: $832 - 300 = 532$. Because you subtracted $300$ instead of $296$, you took away $4$ too many. To restore the correct balance, you must add back $4$:

$$832 - 296 = 832 - 300 + 4 = 532 + 4 = 536$$

A very common non-calculator error under exam pressure is subtracting the $4$ instead of adding it, yielding the distractor $528$. Always ask yourself: "Did I subtract too much or too little?" Subtracting too much requires adding back.

Equal Additions (Constant Difference) Across Zeros

Traditional column subtraction across multiple zeros—such as $2{,}000 - 647$ or $50.00 minus $18.65—is notorious for producing borrowing errors in the tens or hundreds column. The constant difference rule states that shifting both numbers in a subtraction by the exact same amount leaves the gap between them unchanged:

$$a - b = (a - c) - (b - c)$$

By subtracting $1$ from both numbers in $2{,}000 - 647$, the minuend becomes all nines after the leading digit, eliminating every borrow:

$$2{,}000 - 647 = 1{,}999 - 646 = 1{,}353$$

Each place value now subtracts independently from left to right: $1 - 0 = 1$, $9 - 6 = 3$, $9 - 4 = 5$, and $9 - 6 = 3$. The same technique works for currency: $50.00 - 18.65 = 49.99 - 18.64 = 31.35$ dollars.

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Mental Multiplication: Partitioning, Doubling-and-Halving, and Scaling

Distributive Partitioning (Split and Area Strategy)

The distributive law allows you to split one factor into friendly place-value components—or a round benchmark minus a small unit adjustment—and multiply each part separately:

• Additive Split: For $28 \times 15$, split $15$ into $10 + 5$. Since multiplying by $5$ is always exactly half of multiplying by $10$, you obtain:

$$28 \times 15 = (28 \times 10) + (28 \times 5) = 280 + 140 = 420$$

• Subtractive Split: For $39 \times 14$, rewrite $39$ as $40 - 1$:

$$39 \times 14 = (40 \times 14) - (1 \times 14) = 560 - 14 = 546$$

Doubling and Halving

Whenever you multiply an even number by a number ending in $5$ (or $.5$), double the multiple of $5$ and halve the even number. Because $(2a) \times \left(\frac{b}{2}\right) = a \times b$, the product remains identical while the numbers collapse into a single-digit table fact:

• $35 \times 16 = 70 \times 8 = 560$
• $45 \times 18 = 90 \times 9 = 810$
• $2.5 \times 28 = 5 \times 14 = 10 \times 7 = 70$

Multiplying and Dividing by $5$, $25$, and $50$

Because $5 = \frac{10}{2}$, $25 = \frac{100}{4}$, and $50 = \frac{100}{2}$, you should never perform long multiplication or long division by $5$, $25$, or $50$. Replace them with powers of $10$ and halving or doubling:

• Multiply by $5$: Multiply by $10$, then halve ($64 \times 5 = 640 \div 2 = 320$).
• Multiply by $25$: Divide by $4$ (halve twice), then multiply by $100$ ($36 \times 25 = (36 \div 4) \times 100 = 9 \times 100 = 900$).
• Multiply by $50$: Halve the number, then multiply by $100$ ($48 \times 50 = 24 \times 100 = 2{,}400$).
• Divide by $5$: Double the number, then divide by $10$ ($435 \div 5 = 870 \div 10 = 87$).
• Divide by $25$: Multiply by $4$ (double twice), then divide by $100$ ($725 \div 25 = 2{,}900 \div 100 = 29$).

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Mental Division: Factor-Splitting, Chunking, and Divisibility Rules

Factor-Splitting and Chunking (Partial Quotients)

When dividing by a composite two-digit number, combine factor-splitting with chunking (splitting the dividend into convenient multiples of the divisor):

$$756 \div 14 = 756 \div 2 \div 7 = 378 \div 7$$

Next, split $378$ into two obvious multiples of $7$, namely $350 + 28$:

$$378 \div 7 = (350 + 28) \div 7 = (350 \div 7) + (28 \div 7) = 50 + 4 = 54$$

Rapid Divisibility Rules for Mental Factoring

Before choosing a division path or simplifying a fraction, scan the digits using these divisibility tests:

• Divisible by $2$: The last digit is even ($0, 2, 4, 6, 8$).
• Divisible by $3$: The sum of the digits is a multiple of $3$ (e.g., for $756$, $7 + 5 + 6 = 18$, which is divisible by $3$).
• Divisible by $4$: The last two digits form a number divisible by $4$ (e.g., $536$ ends in $36$, and $36 \div 4 = 9$).
• Divisible by $5$: The last digit is $0$ or $5$.
• Divisible by $6$: The number passes the tests for both $2$ and $3$ (an even number whose digits sum to a multiple of $3$).
• Divisible by $8$: The last three digits are divisible by $8$ (or the number can be halved three times cleanly).
• Divisible by $9$: The sum of the digits is a multiple of $9$ (e.g., $756$ has digit sum $18$, so $756 \div 9 = 84$).
• Divisible by $10$: The last digit is $0$.

Worked Section 2 Classroom Example

Scenario: A primary school orders $25$ numeracy resource kits at $14.80 each, plus a flat $15.00 courier fee, paid from a $500.00 department budget. How much budget remains?

1. Kit cost via $\times 25$ shortcut: Divide $14.80 by $4$ (halve to $7.40, halve again to $3.70) and multiply by $100$: $3.70 \times 100 = 370.00$ dollars.
2. Add courier fee: $370.00 + 15.00 = 385.00$ dollars.
3. Subtract from $500.00: Count up from $385 to $500 ($385 + 15 = 400$, plus $100 = 500$), leaving $115.00.

Summary of Mental Arithmetic Strategies