Operations with Integers

first = (27+(9))/2 = 18

second = 27−(18) = 9

first = (4+(12))/2 = 8

second = 4−(8) = -4

first = (0+(10))/2 = 5

second = 0−(5) = -5

first = (0+(-10))/2 = -5

second = 0−(-5) = 5

first = (-7+(-1))/2 = -4

second = -7−(-4) = -3

first = (-7+(-13))/2 = -10

second = -7−(-10) = 3

−4+b = 5

b = 5+4 = 9

(1−2)+(3−4)+(5−6)+(7−8)+(9−10)

= −1−1−1−1−1 = −5

(3)×(-2) = -6

(-5)×(-2) = 10

(-4)×(-1) = 4

(-7)×(3) = -21

(-123)×(456) = -56088

(-123)×(-456) = 56088

(123)×(-456) = -56088

Multiply their magnitudes.

Same signs → positive; different signs → negative.

If either integer is 0, the product is 0.

|product| = 4×3 = 12

(4)×(-3) = -12

|product| = 6×3 = 18

(-6)×(-3) = 18

|product| = 5×1 = 5

(-5)×(-1) = 5

|product| = 8×4 = 32

(-8)×(4) = -32

|product| = 9×10 = 90

(-9)×(10) = -90

|product| = 10×17 = 170

(10)×(-17) = -170

14×(−15) = −210

(−16)×(−5) = 80

36÷(−18) = −2

(−46)÷(−23) = 2

32+10(−5) = 32−50 = −18

3000×8+5000×(−5) = 24000−25000 = −1000

grey loss = 6400×5 = 32000

white bags = 32000÷8 = 4000

□ = 27÷(−3) = −9

□ = −35÷5 = −7

□ = −56÷(−8) = 7

□ = 132÷(−12) = −11

□ = 7×(−8) = −56

□ = (−11)×12 = −132

(−5)×15 = −75

(−28)×(−1) = 28

2×15 = 30

−27÷9 = −3

84÷(−4) = −21

(−56)÷(−2) = 28

x = −(27) = -27

x = −(-31) = 31

x = −(-1) = 1

x = −(1) = -1

x = −(0) = 0

new expression = −(old expression) = −(−4) = 4

-21 → 64 → 32 → 16 → 8 → 4 → 2 → 1 → -2 → -1 → 4

-6 → -3 → 10 → 5 → -14 → -7 → 22 → 11 → -32 → -16 → -8 → -4 → -2 → -1 → 4 → 2 → 1 → -2

Both eventually repeat the cycle 4 → 2 → 1 → −2 → −1 → 4.

correct marks = 15×4 = 60

incorrect count = (60−40)/2 = 10

total = 15+10 = 25

correct marks = 5×4 = 20

incorrect count = (20−(−10))/2 = 15

attempted = 5+15 = 20

unanswered = 25−20 = 5

rule: a−bc

4−8(−3) = 28; 6−9(6) = −48

−16−(−6)(−9) = −16−54 = −70

8+4(−5) = 8−20 = −12

(-3)×(-2)×(-1) = -6

(4)×(5)×(6) = 120

10×13 + 5×(−9) = 130−45 = 85

5×13 + 5×(−9) = 65−45 = 20

10×13 + 10×(−9) = 130−90 = 40

1×13 + 7×(−9) = 13−63 = -50

2×13 + 2×(−9) = 26−18 = 8

7×13 + 9×(−9) = 91−81 = 10

4×13 + 6×(−9) = 52−54 = -2

7×13 + 10×(−9) = 91−90 = 1

122×13 + 2×(−9) = 1586−18 = 1568

(−576)÷(−36) = 16

32÷648 = 4/81

(−300)÷(−1215) = 20/81

(−1960)÷280 = −7

a = −1412; b = 716; c = 1412

d = 370272; e = −370272; f = 335472

e < a < b < c < f < d

(−548+1)×972 = −532656+972 = −531684

(−548)(972−1) = −532656−(−548) = −532108

(−548+1)(972−1)

= −532656+548+972−1 = −531137

−207(33−7) = 207(−33+7) = −5382

(−6)×((−2)−(5+3)) = 60

((3+5)−(−2))×(−6) = -60

0+6×(−6) = −36

4+5×(−8) = −36

−6+5×(−6) = −36

−12+4×(−6) = −36

12+6×(−8) = −36

(4−1)×4 = 12

(8−2)×2 = 12

(13−1)×1 = 12

(1−4)×(−4) = 12

(0−6)×(−2) = 12

0−(1−0) = −1

1−(2−0) = −1

2−(4−1) = −1

−2−(0−1) = −1

3−(5−1) = −1

rule: a+b−c

−10+(−12)−(−9) = −13

rule: −ab−c

−4×8−(−3) = −29

−6×(−11)−12 = 54

−(−10)(−12)−(−9) = −111

The first pair must total 25 and be 11 apart. 18+7=25 and 18−7=11.

Swap the two numbers to reverse the difference: 7−18=−11.

Add S and D: (first+second)+(first−second)=2×first, so first=(S+D)÷2.

Subtract D from S: (first+second)−(first−second)=2×second, so second=(S−D)÷2.

Check that S and D have the same parity so both answers are integers.

Starting position is 0. The first movement puts it at −4.

To reach +5 from −4, move 9 units right.

Pair terms: (1−2)+(3−4)+…+(9−10).

Each pair equals −1; there are five pairs.

To take 18 positive tokens from 7 positive tokens, add 11 zero pairs; after removing the 18 positives, 11 negatives remain.

Removing 12 negative tokens has the same effect as adding 12 positive tokens.

Four groups of six negative tokens make 24 negative tokens.

Nine groups of seven negative tokens make 63 negative tokens.

A negative multiplier means “remove this group” four times.

The multiplicand is +2, so each group to be removed contains two positive tokens.

Start by adding four zero pairs; after removing eight positives, eight negatives remain.

The negative multiplier means remove a group of two negative tokens four times.

Add four zero pairs first so there are eight negative tokens to remove.

After removing them, eight positive tokens remain.

Each set has net value −2, even if it contains extra zero pairs.

Four copies have net value 4×(−2).

The value of every starting arrangement is 4.

Five copies therefore have value 5×4.

Repeated addition shows positive×negative is negative.

Removing negative groups shows negative×negative is positive.

Multiplication by zero gives zero.

Inside the brackets, 4+(−3)=1, so the left side is (−2)×1=−2.

On the right, (−2)×4=−8 and (−2)×(−3)=+6.

Add: −8+6=−2.

First simplify inside: 2+(−3)=−1, so −4×(−1)=+4.

Distribute instead: (−4)×2=−8 and (−4)×(−3)=+12.

Combine: −8+12=+4.

The first arrow a moves right from 0, so a is positive.

The next arrow b moves left and ends at point P, which lies left of 0, so b is negative and its leftward length is greater than a’s rightward length.

Arrow a moves right from 0 to the far point, so a is positive.

Arrow b then moves left to P, which is still right of 0. Therefore the rightward movement a is longer than the leftward movement b.

Arrow a moves left from 0, so a is negative.

Arrow b moves right from the left point back to 0, so b is positive.

The final position is exactly 0, so the two movements cancel.

Removing two positive tokens four times has the same effect as adding two negative tokens four times.

Add four groups of −2: (−2)+(−2)+(−2)+(−2)=−8.

One copy of a is still a: 1×a=a.

Multiplying by −1 gives the additive inverse of a: (−1)×a=−a.

For example, 3×(−4)=−12 and (−4)×3=−12.

The other rows also give the same product after the factors are swapped.

Calculate each product in order: 9, 6, 3, 0, −3, −6.

Each time the second factor drops by 1, the product drops by 3.

Calculate: −9, −6, −3, 0, 3, 6.

When the second factor decreases by 1, the product increases by 3.

Multiply the absolute values to find the product size.

The product is positive for two matching signs and negative for opposite signs.

Treat a fortune as positive and a debt as negative.

Same signs give a positive product or quotient; opposite signs give a negative one.

Score = 30×5 + 20×(−2) = 150−40 = 110.

For the greatest score, get all 50 correct: 50×5.

For the least score, get all 50 wrong: 50×(−2).

A downward movement is negative: 60×(−3)=−180.

Starting at 0 gives 0+(−180).

Movement = 45×(−3)=−135 m.

Final position = 15+(−135)=−120 m.

The sample circles −6, 14, 20 and 18, one in each row and column.

(−6)×14×20×18=−30,240.

Each pick uses one row factor and one column factor.

Choose row factors 1, 2, −3, 4 and column factors 2, −1, 3, −2.

Multiply each row factor by each column factor to fill the grid.

A legal selection multiplies every row factor and every column factor once.

Divide the absolute values first.

Use a positive answer for matching signs and a negative answer for opposite signs.

Division by zero is not defined.

(5×(−3))×4 = (−15)×4 = −60.

5×((−3)×4) = 5×(−12) = −60.

(25×12)×(−6)=300×(−6)=−1,800.

25×((−6)×12)=25×(−72)=−1,800.

Count the negative factors.

Pair the negative factors.

Inside first: 4+(−2)=2, so 5×2=10.

Distribute: 5×4+5×(−2)=20−10=10.

Left side: 2+(−3)=−1, then (−4)×(−1)=4.

Right side: (−4)×2=−8 and (−4)×(−3)=12; −8+12=4.

For example, choose the rule first number + second number − third number.

Substitute: −10+(−12)−(−9)=−22+9.

The aim is to capture all opposing coins or leave the opponent with no legal move.

On a turn, move one coin to a neighbouring empty point, or jump an adjacent opposing coin into an empty point beyond it and capture it.