Chapter 2: Operations With Integers

Let the two numbers be the first number $$a$$ and the second number $$b$$. We need $$a+b=25$$ and $$a-b=11$$.

Try a few pairs adding to $$25$$ and check the difference:

The pair $$18$$ and $$7$$ satisfies both conditions.

Check: $$18+7=25$$ and $$18-7=11$$. Both are correct.

We need $$a+b=25$$ and $$a-b=-11$$. Since $$a-b$$ is negative, the first number is smaller than the second.

The pair $$a=7$$ and $$b=18$$ works: $$7+18=25$$ and $$7-18=-11$$.

Notice that this is the same pair as before with the two numbers swapped — swapping changes only the sign of the difference.

Both strikes move the coin to the right, so the total rightward distance is the sum.

Starting at $$0$$: after the first strike, the coin is at $$0+4=4$$.

After the second strike: $$4+3=7$$.

Hence the coin ends at position $$7$$ units to the right of $$0$$.

Let the first strike move the coin by $$a$$ units and the second strike move it by $$b$$ units. Since both are rightward, the total distance moved from $$0$$ is $$a+b$$.

So the final position is

Adopt the convention that rightward is positive and leftward is negative. Then a rightward strike of $$5$$ units is $$+5$$ and a leftward strike of $$5$$ units is $$-5$$.

Let $$a$$ be the first movement and $$b$$ the second (each a positive or negative integer). Adding the two signed movements gives the final position:

The same formula covers all four cases (both right, both left, or one of each).

Using the formula $$\mathrm{P}=a+b$$ with $$a=5$$ and $$b=-7$$:

The coin ends up $$2$$ units to the left of $$0$$.

Start with $$7$$ green tokens (representing $$+7$$). We must remove $$18$$ green tokens, but there are only $$7$$ to begin with.

Add $$11$$ zero-pairs (each pair is one green + one red, worth $$0$$). The bag now has $$7+11=18$$ green tokens and $$11$$ red tokens. The total value is unchanged.

Now remove $$18$$ green tokens. What is left: $$11$$ red tokens.

Therefore

Two green tokens added $$4$$ times gives

In multiplication form, $$4\times 2=8$$. So there are $$8$$ positive tokens in the bag, representing $$+8$$.

$$4\times(-6)$$: Put $$6$$ red tokens ($$-6$$) into the bag $$4$$ times. Total red tokens $$=6\times 4=24$$. So $$4\times(-6)=-24$$.

$$9\times(-7)$$: Put $$7$$ red tokens into the bag $$9$$ times. Total red tokens $$=7\times 9=63$$. So $$9\times(-7)=-63$$.

Interpreting $$(-4)\times 2$$: A positive multiplier means "put in", so a negative multiplier $$-4$$ can be interpreted as "remove" $$4$$ times. Thus $$(-4)\times 2$$ means remove $$2$$ (green tokens) from the bag $$4$$ times. Starting from an empty bag, we first add $$4$$ zero-pairs ($$8$$ green and $$8$$ red) so we can remove $$2$$ greens on each of the $$4$$ occasions; after all removals only $$8$$ red tokens remain, giving $$(-4)\times 2=-8$$.

In the expression $$(-4)\times 2$$, the multiplicand is $$2$$, which stands for $$2$$ green tokens (green tokens represent positive units).

The multiplier $$-4$$ is a negative integer, and we interpret a negative multiplier as "remove" that many times. So we must remove what the multiplicand represents — i.e. remove $$2$$ green tokens, $$4$$ times.

If we removed red tokens instead, we would be modelling $$(-4)\times(-2)$$, a completely different expression.

The multiplicand $$-2$$ represents $$2$$ red tokens. The multiplier $$-4$$ means "remove" $$4$$ times. So $$(-4)\times(-2)$$ means: remove $$2$$ red tokens from the bag, $$4$$ times.

Starting from an empty bag, we add $$4$$ zero-pairs (each pair is $$1$$ green + $$1$$ red, worth $$0$$). Actually, to be able to remove $$2$$ red tokens on each of $$4$$ occasions, we need at least $$8$$ reds — so add $$8$$ zero-pairs (i.e. $$8$$ greens + $$8$$ reds), value still $$0$$.

Now remove $$2$$ red tokens $$4$$ times ($$= 8$$ red tokens removed). We are left with $$8$$ green tokens.

Hence $$(-4)\times(-2)=+8$$. The product of two negative integers is positive.

(a) $$2$$ red $$\times\, 4 = 8$$ red $$=-8$$.

(b) $$(4\,\text{red}+2\,\text{green})\times 4 = 16$$ red $$+\;8$$ green. Cancelling $$8$$ zero-pairs leaves $$8$$ red $$=-8$$.

(c) $$(6\,\text{red}+4\,\text{green})\times 4 = 24$$ red $$+\;16$$ green. Cancelling $$16$$ zero-pairs leaves $$8$$ red $$=-8$$.

All three give the same answer, $$-8$$. This is not a coincidence: since each set has value $$-2$$, multiplying by $$4$$ must produce $$-8$$ regardless of how the value $$-2$$ is arranged in tokens.

The number $$4$$ can be represented by many token sets: e.g. (i) $$4$$ green tokens; (ii) $$5$$ green + $$1$$ red; (iii) $$7$$ green + $$3$$ red. Each set has value $$4$$.

(i) $$4$$ green $$\times 5 = 20$$ green $$=20$$.

(ii) $$(5\text{ green}+1\text{ red})\times 5 = 25$$ green $$+5$$ red. Cancel $$5$$ zero-pairs → $$20$$ green $$=20$$.

(iii) $$(7\text{ green}+3\text{ red})\times 5 = 35$$ green $$+15$$ red. Cancel $$15$$ zero-pairs → $$20$$ green $$=20$$.

All three give $$5\times 4=20$$. The answer does not depend on which token set is chosen.

Yes. Instead of "remove $$2$$ green tokens $$4$$ times", we can equivalently add the additive inverse of $$2$$, which is $$-2$$ (i.e. $$2$$ red tokens), $$4$$ times.

Adding $$2$$ red tokens $$4$$ times gives $$8$$ red tokens in the bag, which represents $$-8$$.

So $$(-4)\times 2$$ can be defined as "add $$-2$$ four times", giving

Both interpretations (removal or addition of inverses) yield the same result.

Reading down the list, the multiplier decreases by $$1$$ at each step: $$4,3,2,1,0,\ldots$$

Correspondingly, the product decreases by $$3$$ at each step: $$12,9,6,3,0,\ldots$$

So the pattern is: every time the multiplier goes down by $$1$$, the product goes down by $$3$$ (which is the multiplicand).

Yes, the pattern continues. Each step the product decreases by $$3$$:

This is consistent with our token interpretation: $$(-1)\times 3$$ means "remove $$3$$ greens once", giving $$-3$$; similarly for the others.

Reading down the list, the multiplier decreases by $$1$$ each step: $$4,3,2,1,0,\ldots$$

The product now increases by $$3$$ each step: $$-12,-9,-6,-3,0,\ldots$$

So when the multiplicand is $$-3$$: every time the multiplier goes down by $$1$$, the product goes up by $$3$$ (i.e. the product changes by $$-(-3)=+3$$).

Continuing the same rule (the product goes up by $$3$$ each time the multiplier drops by $$1$$):

All three products are positive, consistent with the rule: negative times negative equals positive.

Yes. If $$a$$ is a negative integer, then $$1\times a$$ means "put $$a$$ into the bag once", which of course gives $$a$$.

For example, $$1\times(-5)=-5$$ and $$1\times(-100)=-100$$.

So $$1\times a=a$$ holds for every integer $$a$$, positive, negative, or zero. The number $$1$$ is the multiplicative identity.

Compute each blank using the rules for signed multiplication.

Row 2: $$-30\times 12$$: magnitudes multiply to $$30\times 12=360$$; signs are $$-$$ and $$+$$ so the product is negative. $$-30\times 12=-360$$. Similarly $$12\times -30=-360$$.

Row 4: We need $$-5\times \underline{\phantom{x}}=-70$$. Divide: $$-70\div(-5)=14$$. So the missing number is $$14$$, matching $$14\times -5=-70$$.

Completed table:

In every pair, swapping the two integers being multiplied does not change the product. For instance, $$3\times(-4)=-12$$ and $$(-4)\times 3=-12$$; $$(-15)\times(-8)=120$$ and $$(-8)\times(-15)=120$$.

The magnitudes of the two products are the same (because $$a\times b=b\times a$$ for whole numbers) and the signs are the same (because the sign of the product is determined only by the parity of negative signs, not their order).

Yes, this always happens. For any two integers $$a$$ and $$b$$,

This is called the commutative property of multiplication. It holds for positive integers, negative integers, and zero.

No. Since $$a\times b=b\times a$$, both the magnitude and the sign of the product remain the same after swapping.

For example, $$3\times(-4)=-12$$ and $$(-4)\times 3=-12$$; both are negative. Similarly $$(-15)\times(-8)=(-8)\times(-15)=120$$; both are positive.

The exam has $$50$$ questions, $$+5$$ for each correct answer and $$-2$$ for each wrong answer.

Maximum: all $$50$$ answers correct, none wrong.

Minimum: all $$50$$ answers wrong, none correct.

In Example 2(b), the elevator starts at $$+15$$ m (above ground) and descends at $$3$$ m per minute for $$45$$ minutes.

Method 1 (subtraction): Compute the total descent and subtract it from the starting position.

Total descent $$=3\times 45=135$$ m.

So the elevator is $$120$$ m below ground level.

Each entry in the grid is a product of a row-value and a column-value. Writing the row-values on the side and column-values on the top:

The game forces you to pick exactly one entry from each row and each column. So the four picked entries are $$r_1\,c_{\sigma(1)},\;r_2\,c_{\sigma(2)},\;r_3\,c_{\sigma(3)},\;r_4\,c_{\sigma(4)}$$ for some rearrangement $$\sigma$$ of the columns.

Their product is

Computing: rows give $$2\times(-7)\times 3\times 5=-210$$; columns give $$4\times(-2)\times 6\times(-3)=144$$. So every product equals

The two grids in the problem are identical, so the answer for both grids is $$-30240$$, regardless of the starting number.

What is special: The grid is a full multiplication table. The magic is in the arrangement — any grid that is a product table (entry $$(i,j)=r_i\times c_j$$) will have this property. You can build your own by choosing any row-values and column-values, filling in the products, and playing the game.

Rules for the sign of a quotient are the same as for multiplication:

• positive $$\div$$ positive $$=$$ positive, e.g. $$100\div 25=4$$.
• negative $$\div$$ positive $$=$$ negative, e.g. $$(-100)\div 25=-4$$.
• positive $$\div$$ negative $$=$$ negative, e.g. $$50\div(-25)=-2$$.
• negative $$\div$$ negative $$=$$ positive, e.g. $$(-100)\div(-4)=25$$.

In short: same signs give a positive quotient, different signs give a negative quotient. The magnitude is obtained by dividing the absolute values in the usual way.

First way $$((5\times -3)\times 4)$$:

Second way $$(5\times(-3\times 4))$$:

Both give $$-60$$. So it does not matter how we group the numbers — this is the associative property of multiplication.

Example 1: $$2\times(-5)\times 3$$.
$$(2\times -5)\times 3=-10\times 3=-30$$.
$$2\times(-5\times 3)=2\times -15=-30$$. Same.

Example 2: $$(-4)\times(-2)\times 5$$.
$$((-4)\times(-2))\times 5=8\times 5=40$$.
$$(-4)\times((-2)\times 5)=(-4)\times(-10)=40$$. Same.

Example 3: $$(-6)\times 7\times(-1)$$.
$$((-6)\times 7)\times(-1)=-42\times -1=42$$.
$$(-6)\times(7\times -1)=(-6)\times -7=42$$. Same.

Observation: Whichever way we group three integers in a multiplication, the product is the same. Multiplication of integers is associative:

Using commutativity, the three factors $$5,\;-3,\;4$$ can be arranged in $$3!=6$$ orders:

Each evaluates to

because multiplication of integers is both commutative and associative, so any order gives the same product.

All six orders of $$25,\,-6,\,12$$:

Every order gives $$-1800$$, confirming commutativity and associativity for these integers.

Yes. Every additional factor of $$-1$$ flips the sign of the product. Starting from $$1$$ and flipping repeatedly:

• If $$-1$$ appears an even number of times, the sign flips back to positive: the product is $$+1$$.
• If $$-1$$ appears an odd number of times, the sign lands on negative: the product is $$-1$$.

In symbols, for a positive integer $$n$$:

Count how many negative integers appear among the factors (ignore the positives):

• If the count of negative factors is even (including $$0$$), the product is positive.
• If the count of negative factors is odd, the product is negative.

The magnitude of the product is the product of the magnitudes.

Reason: Each negative factor can be split as $$(-1)\times(\text{its magnitude})$$. Collecting all the $$-1$$s gives $$(-1)^{k}$$ where $$k$$ is the number of negatives; this is $$+1$$ for even $$k$$ and $$-1$$ for odd $$k$$.

LHS: $$5\times(4+(-2))=5\times 2=10$$.

RHS: $$5\times 4+5\times(-2)=20+(-10)=10$$.

Both sides equal $$10$$, so yes, the expressions are equal. This is the distributive property of multiplication over addition — it holds for integers just as for positive integers:

Given expression:

LHS $$=(-2)\times(4+(-3))=(-2)\times 1=-2$$.

RHS $$=(-2)\times 4+(-2)\times(-3)=-8+6=-2$$. Equal. ✓

Example 1: $$3\times(5+(-8))$$.
LHS $$=3\times(-3)=-9$$. RHS $$=15+(-24)=-9$$. Equal. ✓

Example 2: $$(-4)\times((-6)+2)$$.
LHS $$=(-4)\times(-4)=16$$. RHS $$=24+(-8)=16$$. Equal. ✓

Example 3: $$(-5)\times((-3)+(-7))$$.
LHS $$=(-5)\times(-10)=50$$. RHS $$=15+35=50$$. Equal. ✓

The distributive property holds for all integers.

First simplify: $$2+(-3)=-1$$. So $$(-4)\times(2+(-3))=(-4)\times(-1)$$.

Using the hint, multiplying by $$-4$$ means "add the additive inverse $$4$$ times". The additive inverse of $$-1$$ is $$+1$$ (one green token). Adding one green token $$4$$ times gives $$4$$ green tokens $$=+4$$.

Right-hand side visually: $$(-4)\times 2+(-4)\times(-3)$$.

$$(-4)\times 2$$: add the inverse of $$+2$$ (i.e. $$2$$ red tokens) four times → $$8$$ red tokens $$=-8$$.

$$(-4)\times(-3)$$: add the inverse of $$-3$$ (i.e. $$3$$ green tokens) four times → $$12$$ green tokens $$=+12$$.

Sum: $$-8+12=+4$$.

Both sides visually give $$4$$ green tokens, confirming distributivity.

Guess the operation as $$a+b-c$$ and test it on the rows:

• $$5+8-3=10$$ ✓
• $$10+11-12=9$$ ✓
• $$5+8-(-3)=13+3=16$$ ✓
• $$-3+10-2=5$$ ✓
• $$-4+(-1)-(-6)=-5+6=1$$ ✓

The machine computes $$a+b-c$$. For the last row:

Try the rule $$-(a\times b+c)$$ (that is, take the negative of $$a\times b+c$$):

• $$-(4\times 8+(-3))=-(32-3)=-29$$ ✓
• $$-(6\times(-11)+12)=-(-66+12)=-(-54)=54$$ ✓
• $$-(5\times 3+7)=-(15+7)=-22$$ ✓
• $$-((-3)\times 9+(-8))=-(-27-8)=-(-35)=35$$ ✓
• $$-((-7)\times 4+6)=-(-28+6)=-(-22)=22$$ ✓

So Machine 2 computes $$-(a\times b+c)$$. For the last row with $$a=-10,\;b=-12,\;c=-9$$:

Marks from correct answers: $$30\times 5=150$$.

Marks from wrong answers: $$20\times(-2)=-40$$.

Total marks:

Rule for multiplying two integers:

1. Multiply their absolute values (magnitudes) using the usual whole-number multiplication.
2. Decide the sign:
   • If both integers have the same sign (both positive or both negative), the product is positive.
   • If the two integers have different signs, the product is negative.
3. If either factor is $$0$$, the product is $$0$$.

For example, $$(-6)\times(-4)$$: magnitudes give $$24$$; both signs negative → product $$+24$$. And $$7\times(-9)$$: magnitudes give $$63$$; signs differ → product $$-63$$.

The temperature falls by $$5^{\circ}\mathrm{C}$$ each hour, so each hour changes the temperature by $$-5^{\circ}\mathrm{C}$$.

Total change in $$10$$ hours:

Starting temperature is $$32^{\circ}\mathrm{C}$$. Final temperature:

Compare the two expressions term by term:

Every sign is reversed, so the new sum is the additive inverse of the old sum:

Rule. If the current number $$n$$ is even, replace it with $$n\div 2$$; if it is odd, replace it with $$n\times(-3)+1$$.

Start $$-21$$: $$-21$$ is odd, so $$-21\times(-3)+1=63+1=64$$.
$$-21\to 64\to 32\to 16\to 8\to 4\to 2\to 1$$.

Start $$-6$$: $$-6$$ is even, so $$-6\div 2=-3$$. Then $$-3$$ is odd, $$-3\times(-3)+1=10$$. And so on:
$$-6\to -3\to 10\to 5\to -14\to -7\to 22\to 11\to 32\to -16\to -8\to -4\to -2\to -1\to 4\to 2\to 1$$.

Start $$5$$: $$5\to -14\to -7\to 22\to 11\to 32\to -16\to -8\to -4\to -2\to -1\to 4\to 2\to 1$$.

Start $$-1$$: $$-1\to 4\to 2\to 1$$.

Pattern observed: All starting numbers we try eventually reach $$1$$, after which the sequence enters the loop $$1\to -2\to -1\to 4\to 2\to 1\to\cdots$$ (odd $$1$$: $$1\times(-3)+1=-2$$; then $$-2\to -1\to 4\to 2\to 1$$). So the sequence always ends up cycling through $$\{1,-2,-1,4,2\}$$.

Guess the rule as $$a-b\times c$$ and verify:

• $$4-8\times(-3)=4-(-24)=28$$ ✓
• $$6-9\times 6=6-54=-48$$ ✓
• $$2-3\times(-2)=2-(-6)=8$$ ✓
• $$-9-5\times(-8)=-9-(-40)=31$$ ✓
• $$7-(-4)\times(-6)=7-24=-17$$ ✓

So the machine computes $$a-b\times c$$. For the last row with $$a=-16,\;b=-6,\;c=-9$$:

Each hour the temperature changes by $$-5^{\circ}\mathrm{C}$$. Over $$4$$ hours, total change $$=4\times(-5)^{\circ}\mathrm{C}$$.

Starting temperature is $$8^{\circ}\mathrm{C}$$. The expression for the temperature after $$4$$ hours is

Note that $$-33+7=-26$$ and $$33-7=26$$, which are additive inverses of each other. So

More directly, observe that

Alternatively: $$-207\times(33-7)=-(207\times(33-7))=-(207\times 26)$$. And $$207\times(-33+7)=207\times(-26)=-5382$$, so $$207\times 26=5382$$, giving $$-(207\times 26)=-5382$$.