Chapter 2: Operations with Integers (Integers)

Quick way: first number $=(\text{sum}+\text{difference})÷2$; second number $=\text{sum}-\text{first}$.

(a) 18, 9 (b) 8, −4 (c) 5, −5 (d) −5, 5 (e) −4, −3 (f) −10, 3

(1) $-4+b=5$, so $b=9$: 9 units to the right.

(2) $(1-2)+(3-4)+\cdots +(9-10)=5\times (-1)=$ −5: 5 units to the left of 0.

(1) +9 (2) −5

1. a goes right (positive), b goes left (negative), and b is longer: $|b|>|a|$. So P is to the left of 0 (negative).
2. a goes right, b goes left, and b is shorter: $|b|<|a|$. So P is to the right of 0 (positive).
3. a goes left (negative), b goes right (positive), and they are equal in magnitude: P = 0, i.e. $b=-a$.

(1) a > 0, b < 0, |b| > |a|, P < 0 (2) a > 0, b < 0, |b| < |a|, P > 0 (3) a < 0, b > 0, |a| = |b|, P = 0

(a) To remove 18 positives from 7 positives, add 11 zero pairs and remove 18 positives: 11 negatives remain. Adding 18 negatives to 7 positives instead makes 7 zero pairs and also leaves 11 negatives. Both give −11.

(b) To remove 12 negatives from 4 positives, add 12 zero pairs and remove the 12 negatives: 4 + 12 = 16 positives remain. Adding 12 positives to 4 positives also gives 16. Both give 16.

(a) both leave 11 red tokens: −11 (b) both leave 16 green tokens: 16

$4\times (-6)$: put 6 negatives into the bag 4 times: −24. $9\times (-7)=$ −63.

$(-4)\times 2$ means removing 2 positives 4 times (a negative multiplier means removal). We remove green tokens because the multiplicand, 2, is positive; the multiplier only tells us how many times and whether to put in or take out. (To remove them from an empty bag, we first put in zero pairs.) The result is −8.

−24, −63; (−4) × 2 means removing 2 positives 4 times (green, because the multiplicand 2 is positive), giving −8.

(a) Put 2 red tokens in 3 times: −6

(b) Remove 2 red tokens 5 times (after adding 10 zero pairs): 10 green remain: 10

(c) Remove 1 red token 4 times: 4

(d) Remove 3 green tokens 7 times: 21 red remain: −21

(a) −6 (b) 10 (c) 4 (d) −21

(a) −56088 (b) 56088 (c) −56088

(a) −56088 (b) 56088 (c) −56088

Multiply the magnitudes (ignoring signs). Then:

• same signs (both positive or both negative) → the product is positive;
• different signs → the product is negative.

Multiply the sizes; the product is positive if the signs are the same and negative if they differ.

(a) 8 red: −8. (b) 16 red + 8 green: 8 zero pairs cancel, leaving 8 red: −8. (c) 24 red + 16 green: 16 pairs cancel, leaving 8 red: −8.

The same answer, −8, because each set represents the same number −2; only extra zero pairs differ, and zero pairs add nothing.

For $5\times 4$: 4 green → 20 green (20); 6 green + 2 red → 30 green + 10 red = 20. Always 20.

−8 in every case (zero pairs cancel); similarly 5 × 4 = 20 with any token set for 4.

Yes. Removing 2 positives is the same as adding 2 negatives (adding the inverse). So $(-4)\times 2$ = adding 2 red tokens to the empty bag 4 times = −8.

Yes: add 2 negatives 4 times, giving −8.

(a) −12 (b) 18 (c) 5 (d) −32 (e) −90 (f) −170

(a) −12 (b) 18 (c) 5 (d) −32 (e) −90 (f) −170

$-30\times 12=$ −360 and $12\times (-30)=$ −360; $-5\times$ 14 $=-70$.

The product stays the same when the multiplier and multiplicand are swapped: multiplication of integers is commutative.

−360, −360, 14; swapping does not change the product.

Maximum: all 50 correct: $50\times 5=$ 250. Minimum: all 50 wrong: $50\times (-2)=$ −100.

Elevator: it moves $45\times 3=135$ m downwards: $15-135=$ −120, i.e. 120 m below the ground.

Maximum 250, minimum −100; the elevator is at −120 m.

You always get −30240, whichever numbers you choose (e.g. $(-6)\times 14\times 18\times 20=-30240$).

Why: the grid is a multiplication table. Every entry is (a row number) × (a column number), with row numbers 2, −7, 3, 5 and column numbers 4, −2, 6, −3. For example $-42=(-7)\times 6$ and $30=5\times 6$.

Choosing one number from each row and each column uses every row number and every column number exactly once, so the product is always $(2\times (-7)\times 3\times 5)\times (4\times (-2)\times 6\times (-3))=(-210)\times 144=-30240$.

The magic is in both the numbers and their arrangement. To make your own grid, choose any four row numbers and four column numbers and fill in their products.

Always −30240: the grid is a multiplication table (rows 2, −7, 3, 5; columns 4, −2, 6, −3), so every choice uses each row and column number once.

Divide the magnitudes; the quotient is positive if the dividend and divisor have the same sign, and negative if their signs differ (the same rule as for multiplication). E.g. $(-100)÷25=-4$, $(-100)÷(-4)=25$, $50÷(-25)=-2$.

Same signs → positive quotient; different signs → negative quotient.

(a) −210 (b) 80 (c) −2 (d) 2

(a) −210 (b) 80 (c) −2 (d) 2

$32+10\times (-5)=32-50=$ −18°C

−18°C

(a) $3000\times 8+5000\times (-5)=24,000-25,000=-1,000$: a loss of ₹1,000.

(b) Loss on grey cement = $6400\times 5$ = ₹32,000. White bags needed $=32,000÷8=$ 4,000 bags.

(a) Loss of ₹1,000 (b) 4,000 bags

(a) −9 (b) −7 (c) 7 (d) −11 (e) −56 (f) −132

(a) −9 (b) −7 (c) 7 (d) −11 (e) −56 (f) −132

$(25\times -6)\times 12=-150\times 12=-1800$; $25\times (-6\times 12)=25\times -72=-1800$; $(25\times 12)\times -6=300\times -6=-1800$. Every order gives −1800.

Always −1800.

$(-1)$ multiplied an even number of times gives 1, and an odd number of times gives −1. (The book's line "−1 × −1 × −1 × −1 × −1 = 1" should read −1: five is odd.)

Rule: a product of non-zero integers is positive if the number of negative factors is even and negative if it is odd. (If any factor is 0, the product is 0.)

Even number of negative factors → positive; odd number → negative.

$(-2)\times (4+(-3))=(-2)\times 1=-2$, and $(-2)\times 4+(-2)\times (-3)=-8+6=-2$. ✓

Tokens: multiplying by −4 means adding the inverse 4 times. The inverse of $2+(-3)$ is 2 red and 3 green tokens. Placing this row 4 times gives 8 red (that is $-4\times 2=-8$) and 12 green (that is $-4\times (-3)=12$). In all: $-8+12=4$, and also $-4\times (-1)=4$. ✓

Both sides give −2; with tokens, 4 rows of (2 red, 3 green) show −4 × (2 + (−3)) = (−8) + 12 = 4.

Machine 1: $(-10)+(-12)-(-9)=$ −13.

Machine 2: $-(a\times b)-c$, i.e. multiply the first two numbers, change the sign, then subtract the third:

$-(4\times 8)-(-3)=-32+3=-29$ ✓; $-(6\times -11)-12=66-12=54$ ✓; $-(15)-7=-22$ ✓.

For $(-10,-12,-9)$: $-(120)-(-9)=$ −111.

Machine 1: −13. Machine 2: −(a × b) − c, giving −111.

(a) $(-5)\times 15=$ −75 (b) $-28\times -1=$ 28 (c) four negative factors (even): 30

(a) −75 (b) 28 (c) 30

(a) −3 (b) −21 (c) 28

(a) −3 (b) −21 (c) 28

(a) −27 (b) 31 (c) 1 (d) −1 (e) 0

(a) −27 (b) 31 (c) 1 (d) −1 (e) 0

The second expression has every term with the opposite sign, i.e. it is $(-1)\times$ the first. So its value is $(-1)\times (-4)=$ 4.

4

• −21 → 64 → 32 → 16 → 8 → 4 → 2 → 1 → −2 → −1 → 4 → 2 → 1 → …
• −6 → −3 → 10 → 5 → −14 → −7 → 22 → 11 → −32 → −16 → −8 → −4 → −2 → −1 → 4 → 2 → 1 → …
• 7 → −20 → −10 → −5 → 16 → 8 → 4 → 2 → 1 → …

Patterns: every starting number tried ends in the same loop 1 → −2 → −1 → 4 → 2 → 1 (except 0, which stays at 0). Odd numbers flip sign when multiplied by −3, so the signs keep switching; once a power of 2 (like 64 or −32) appears, the numbers simply halve down to ±1.

All sequences tried (except 0) end in the loop 1, −2, −1, 4, 2, 1.

(a) Correct answers give $15\times 4=60$. She lost $60-40=20$ marks, i.e. $20÷2=$ 10 incorrect. Total questions $=15+10=$ 25.

(b) Correct answers give $5\times 4=20$. To end at −10 he lost 30 marks: $30÷2=$ 15 incorrect. He answered $5+15=20$ questions, so in the same 25-question test he left 5 unanswered.

(a) 10 incorrect; 25 questions (b) 15 incorrect; yes, 5 unanswered (of 25).

The rule is $a-b\times c$ (first number minus the product of the other two):

$4-8\times (-3)=4+24=28$ ✓; $6-9\times 6=-48$ ✓; $2-3\times (-2)=8$ ✓; $-9-5\times (-8)=31$ ✓; $7-(-4)(-6)=-17$ ✓.

For $(-16,-6,-9)$: $-16-(-6)(-9)=-16-54=$ −70.

a − b × c; the last result is −70.

$8+4\times (-5)=8-20=$ −12°C

8 + 4 × (−5) = −12°C

(a) $(-3)\times (-2)\times (-1)=-6$ (b) $4\times 5\times 6=120$

(a) −3, −2, −1 (b) 4, 5, 6

(h) Yes. In fact, since 7 coins of +13 and 10 coins of −9 make +1, any whole-number total can be made by repeating that combination; using 122 coins of +13 and 2 of −9 is one short way.

(a) 5, 5 (b) 10, 10 (c) 1, 7 (d) 2, 2 (e) 7, 9 (f) 4, 6 (g) 7, 10 (h) yes: 122 × 13 − 2 × 9 = 1568

(a) $(-576)÷(-36)=$ 16

(b) $32÷648=\frac{32}{648}=$ $\frac{4}{81}$ (positive; the answer is a fraction, not an integer)

(c) $(-300)÷(-1215)=\frac{300}{1215}=$ $\frac{20}{81}$ (positive)

(d) $(-1960)÷280=$ −7

(a) 16 (b) 4/81 (c) 20/81 (d) −7

Values: (a) −1412 (b) 716 (c) 1412 (d) 3,70,272 (e) −3,70,272 (f) 3,35,472

Increasing order: (e) < (a) < (b) < (c) < (f) < (d)

(e), (a), (b), (c), (f), (d)

(a) One −548 replaced by −547 adds 972: $-532656+972=$ −531684

(b) One fewer 972: $-532656-(-548)=-532656+548=$ −532108

(c) From (a), with one fewer: $-531684+547=$ −531137

(a) −531684 (b) −532108 (c) −531137

$-207\times (33-7)=-207\times 26$ and $207\times (-33+7)=207\times (-26)$. Both equal $-(207\times 26)$, so the value is −5382.

−5382

Multiplying by −6 a large negative or large positive number gives the extremes. (A computer check of all arrangements confirms these.)

(a) Maximum: $(-2-(3+5))\times (-6)=(-10)\times (-6)=$ 60

(b) Minimum: $(3-(-2)+5)\times (-6)=10\times (-6)=$ −60

Maximum 60: (−2 − (3 + 5)) × (−6); minimum −60: (3 − (−2) + 5) × (−6)

(a) $0+6\times (-6)$; $4+(-8)\times 5$; $-6+(-6)\times 5$; $2+(-2)\times 19$; $-36+7\times 0$

(b) $(5-2)\times 4$; $(2-5)\times (-4)$; $(10-4)\times 2$; $(0-6)\times (-2)$; $(-1-2)\times (-4)$

(c) $1-(3-1)$; $0-(3-2)$; $2-(5-2)$; $-4-(-1-2)$; $5-(10-4)$

E.g. (a) 4 + (−8) × 5 (b) (5 − 2) × 4 (c) 2 − (5 − 2), and others as listed.