Chapter 2 โ Operations with Integers
Complete StepโbyโStep Solutions
Every in-text question and every "Figure it Out" exercise from the chapter, solved in full detail โ number games, the carrom-coin model, token (bag) models, multiplication & division rules, and the big 16-question exercise set at the end.
In-Text Questions
These are the guided "?" questions, Math Talk prompts, and Try This activities woven through the chapter's explanations โ solved in the order they appear.
2.1 Rakesh's Puzzle โ A Number Game
Rakesh thinks of two numbers whose sum is 25 and difference is 11. Can you find the two numbers?
We use the guess-and-check method, trying pairs until both conditions are satisfied (remember: difference = first number โ second number).
| First Number | Second Number | Sum | Difference |
|---|---|---|---|
| 10 | 15 | 25 | โ5 |
| 20 | 5 | 25 | 15 |
| 19 | 6 | 25 | 13 |
| 18 | 7 | 25 | 11 |
We can also solve it directly: if $a+b=25$ and $a-b=11$, adding the two equations gives $2a = 36 \Rightarrow a = 18$, and then $b = 25-18=7$.
Now find two numbers whose sum is 25 but difference is โ11.
Swapping the two numbers from the first puzzle reverses the sign of the difference: $18 - 7 = 11$, so $7 - 18 = -11$. Solving directly: $a+b=25,\ a-b=-11 \Rightarrow 2a=14 \Rightarrow a=7,\ b=18$.
Find a pair of numbers for each given sum and difference. Use $\text{first} = \dfrac{\text{sum}+\text{difference}}{2}$, $\ \text{second} = \dfrac{\text{sum}-\text{difference}}{2}$.
(a) Sum = 27, Difference = 9
first $=\frac{27+9}{2}=18$, second $=\frac{27-9}{2}=9$
Check: $18+9=27$ โ, $18-9=9$ โ
(b) Sum = 4, Difference = 12
first $=\frac{4+12}{2}=8$, second $=\frac{4-12}{2}=-4$
Check: $8+(-4)=4$ โ, $8-(-4)=12$ โ
(c) Sum = 0, Difference = 10
first $=\frac{0+10}{2}=5$, second $=\frac{0-10}{2}=-5$
(d) Sum = 0, Difference = โ10
first $=\frac{0-10}{2}=-5$, second $=\frac{0+10}{2}=5$
(e) Sum = โ7, Difference = โ1
first $=\frac{-7-1}{2}=-4$, second $=\frac{-7+1}{2}=-3$
(f) Sum = โ7, Difference = โ13
first $=\frac{-7-13}{2}=-10$, second $=\frac{-7+13}{2}=3$
2.1 Carrom Coin Integers
The coin is struck twice: first by 4 units, then by 3 units (both rightward). What is its final position?
Both strikes move the coin the same way (rightward), so we simply add the distances: $4+3=7$ units from 0.
1. If the first movement is $-4$ and the final position is $5$, what is the second movement?
Using $P = a+b$: $-4 + b = 5 \Rightarrow b = 5-(-4) = 9$.
2. Strikes occur in the order $1, -2, 3, -4, \dots, -10$. What is the final position of the coin?
Add all the movements. Pair consecutive terms:
$$(1-2)+(3-4)+(5-6)+(7-8)+(9-10) = (-1)+(-1)+(-1)+(-1)+(-1) = -5$$
From the three figures (arrows $a$ and $b$ starting at 0), compare the magnitudes and directions of $a$ and $b$.
In each figure, both arrows start at 0. The length of an arrow shows the magnitude (how far the coin travels) and the side it lands on (left of 0 = negative, right of 0 = positive) shows the direction.
- Figure 1 โ Position P lies to the left of 0. Both $a$ and $b$ point left, so both are negative movements. Since arrow $b$ reaches further left than $a$, we get $|b| > |a|$.
- Figure 2 โ Position P lies to the right of 0. Both $a$ and $b$ point right (positive movements), and since $a$ reaches further than $b$, we get $|a| > |b|$.
- Figure 3 โ Position P coincides with 0. Both $a$ and $b$ point left of 0 initially, but $a$'s arc is longer, showing $|a| > |b|$; the key idea is that the landing point (not the arc's starting side) tells you the true final position.
The general takeaway: the arrow's endpoint side tells you the sign, and its length tells you the magnitude โ exactly how the formula $P=a+b$ works for any two strikes.
2.1 Token (Bag) Model for Addition & Subtraction
Find $(+7) - (+18)$ using tokens.
We start with 7 green (positive) tokens, but need to remove 18 positives โ there aren't enough. So we insert enough zero pairs (one green + one red = 0) to have 18 positives available to remove.
We need $18 - 7 = 11$ more positives, so we add 11 zero pairs (11 green + 11 red). Now removing all 18 greens leaves the 11 reds behind.
Using tokens, argue that: (a) $7-18 = 7+(-18)$ (b) $4-(-12) = 4+12$
(a) Removing 18 positive tokens from the bag (subtracting $+18$) leaves the exact same result as directly adding 18 red/negative tokens (adding $-18$) โ in both cases the bag ends up 18 "more negative". So $7-18=7+(-18)=-11$.
(b) Removing 12 negative (red) tokens from the bag (subtracting $-12$) has the same net effect as adding 12 positive (green) tokens โ removing a debt is the same as gaining that amount. So $4-(-12)=4+12=16$.
In general, subtracting a number is the same as adding its additive inverse: $a-b = a+(-b)$.
2.2 Multiplication of Integers โ Bag Model
Find $4 \times (-6)$ and $9 \times (-7)$ using the bag model. How do we interpret $(-4)\times 2$?
$4\times(-6)$: place 6 negatives into the bag, 4 times $\Rightarrow -24$.
$9\times(-7)$: place 7 negatives into the bag, 9 times $\Rightarrow -63$.
$(-4)\times 2$: here the multiplier is negative, so instead of placing tokens, we remove tokens from the bag. We remove 2 positive (green) tokens, 4 times. Since the bag starts empty, we first insert 4 zero-pairs, then remove the 4ร2 = 8 greens, leaving 8 negatives.
How do we model $(-4)\times(-2)$ with tokens?
Negative multiplier means remove tokens; negative multiplicand means we are removing negative (red) tokens. We remove 2 reds, 4 times. Since the bag is empty, we insert 4 zero-pairs first (4 green + 4 red for each of the 4 removals = total 8 zero pairs), then remove the 8 reds, leaving 8 greens behind.
1. Using the token interpretation, find: (a) $3\times(-2)$ (b) $(-5)\times(-2)$ (c) $(-4)\times(-1)$ (d) $(-7)\times 3$
(a) Place 2 negatives, 3 times $\Rightarrow -6$
(b) Negative ร negative โ remove 2 negatives, 5 times $\Rightarrow +10$
(c) Remove 1 negative, 4 times $\Rightarrow +4$
(d) Place 3 negatives, 7 times $\Rightarrow -21$
2. Given $123\times456=56088$, without calculating find: (a) $(-123)\times456$ (b) $(-123)\times(-456)$ (c) $(123)\times(-456)$
Only the sign changes โ the magnitude 56088 stays the same. One negative factor โ negative product; two negative factors โ positive product.
3. Frame a simple rule to multiply two integers, using the token sets shown (all representing $-2$) and the $5\times4$ check.
Whether $-2$ is shown as 2 red tokens, or as 2 reds + 2 zero pairs, or as 4 reds + 4 greens โ it always nets out to $-2$. So multiplying any of these representations by 4 gives the same answer, $-8$: the final product only depends on the actual value being multiplied, not how it's built from tokens. Similarly $5\times4=20$ regardless of which token arrangement represents 4.
2.2 Patterns in Integer Multiplication โ Practice
Find the following products: (a) $4\times(-3)$ (b) $(-6)\times(-3)$ (c) $(-5)\times(-1)$ (d) $(-8)\times4$ (e) $(-9)\times10$ (f) $10\times(-17)$
(a) $4\times(-3)=-12$
(b) $(-6)\times(-3)=18$
(c) $(-5)\times(-1)=5$
(d) $(-8)\times4=-32$
(e) $(-9)\times10=-90$
(f) $10\times(-17)=-170$
2.2 Is Multiplication Commutative for Integers?
Fill in the blanks and check whether swapping the multiplier and multiplicand changes the product.
| Statement 1 | Statement 2 |
|---|---|
| $3\times(-4)=-12$ | $(-4)\times3=-12$ |
| $-30\times12=\mathbf{-360}$ | $12\times(-30)=\mathbf{-360}$ |
| $-15\times(-8)=120$ | $-8\times(-15)=120$ |
| $14\times(-5)=-70$ | $-5\times\mathbf{14}=-70$ |
The product is unchanged when the two numbers are swapped โ multiplication is commutative for integers: $a\times b = b\times a$.
History Brahmagupta's Rules
Brฤhmasphuแนญasiddhฤnta (628 CE)
"The product or quotient of two fortunes is a fortune. The product or quotient of two debts is a fortune. The product or quotient of a debt and a fortune is a debt. The product or quotient of a fortune and a debt is a debt."
Brahmagupta used dhana (fortune) for positive numbers and แนแนa (debt) for negative numbers โ the very first recorded rules for multiplying and dividing signed numbers, nearly 1,400 years ago.
Example 1 Exam Marks Problem
An exam has 50 MCQs. +5 marks for every correct answer, โ2 for every wrong answer. Mala got 30 correct and 20 wrong. Find her total marks.
Marks from correct answers $= 30 \times 5 = 150$
Marks from wrong answers $= 20 \times (-2) = -40$
$$\text{Total} = 150 + (-40) = 110$$
What are the maximum and minimum possible marks in this exam?
Maximum โ every question correct: $50\times5=250$.
Minimum โ every question wrong: $50\times(-2)=-100$.
Example 2 The Mining-Shaft Elevator
(a) The elevator descends from ground level (0) at 3 m/min. Position after 1 hour? (b) It descends from 15 m above ground for 45 minutes. Final position?
Method 1 (subtraction): Distance in 60 min $= 60\times3=180$ m. Starting at 0 and descending: $0-180=-180$.
Method 2 (signed speed): Downward speed $=-3$ m/min for 60 min: $60\times(-3)=-180$.
Solve part (b) using Method 1 (subtraction), and verify with Method 2.
Method 1: Distance travelled in 45 min $=45\times3=135$ m descent. Starting 15 m above ground: $15-135=-120$.
Method 2 (check): Ending position $=15+(45\times(-3))=15+(-135)=-120$.
Activity A Magic Grid of Integers
Circle any number, strike out its row & column, circle any unstruck number, repeat. Multiply all circled numbers. Try again with different choices โ is the product always the same?
| 8 | โ4 | 12 | โ6 |
| โ28 | 14 | โ42 | 21 |
| 12 | โ6 | 18 | โ9 |
| 20 | โ10 | 30 | โ15 |
For the example round shown: circled numbers are $20,\ 14,\ 18,\ -6$.
$$20 \times 14 \times 18 \times (-6) = 280\times18\times(-6) = 5040 \times (-6) = -30240$$
This grid is specially constructed as a multiplication table (row-header ร column-header), so no matter which numbers you circle following the rule, you always end up picking exactly one number from every row and every column โ and the product always works out to the same value.
2.2 Division of Integers
1. Find the values: (a) $14\times(-15)$ (b) $-16\times(-5)$ (c) $36\div(-18)$ (d) $(-46)\div(-23)$
(a) $14\times(-15)=-210$
(b) $-16\times(-5)=80$
(c) $36\div(-18)=-2$
(d) $(-46)\div(-23)=2$
2. Room temperature drops from 32ยฐC at 5ยฐC/hour. Find the temperature 10 hours later.
Rate $=-5ยฐC$ per hour. After 10 hours: $32 + 10\times(-5) = 32-50=-18$.
3. A cement company earns โน8/bag profit on white cement and โน5/bag loss on grey cement.
(a) Sells 3,000 white & 5,000 grey bags in a month โ profit or loss?
(b) If 6,400 grey bags are sold, how many white bags are needed for no profit, no loss?
(a) Profit/loss $= 3000\times(8) + 5000\times(-5) = 24000 - 25000 = -1000$.
(b) Loss from grey cement $=6400\times(-5)=-32000$. Let $w$ = white bags needed: $8w - 32000 = 0 \Rightarrow w = 4000$.
4. Replace the blank with an integer to make each statement true.
(a) $(-3)\times\underline{\ \ }=27 \Rightarrow -9$
(b) $5\times\underline{\ \ }=(-35) \Rightarrow -7$
(c) $\underline{\ \ }\times(-8)=(-56) \Rightarrow 7$
(d) $\underline{\ \ }\times(-12)=132 \Rightarrow -11$
(e) $\underline{\ \ }\div(-8)=7 \Rightarrow -56$
(f) $\underline{\ \ }\div12=-11 \Rightarrow -132$
2.3 Expressions Using Integers โ Associative & Distributive Properties
Evaluate $5\times(-3)\times4$ by grouping differently. Does the grouping matter?
$(5\times-3)\times4 = -15\times4=-60$... (the textbook uses $-60$ but note the worked value shown is $-60$; grouping the other way:)
$5\times(-3\times4)=5\times(-12)=-60$
Also grouping 5 and 4 first: $(5\times4)\times(-3)=20\times(-3)=-60$.
Multiply $25\times(-6)\times12$ in every possible order and check the product stays the same.
$(25\times-6)\times12 = -150\times12=-1800$
$25\times(-6\times12)=25\times(-72)=-1800$
$(-6\times12)\times25=-72\times25=-1800$
Using $-1\times-1=1,\ -1\times-1\times-1=-1,\ldots$, give a rule for the sign of the product of many integers.
Each extra factor of $-1$ flips the sign. So:
Does the distributive property hold for $(-2)\times(4+(-3))$? Check a few other examples.
$(-2)\times(4+(-3)) = (-2)\times1=-2$
$(-2)\times4 + (-2)\times(-3) = -8+6=-2$ โ โ matches!
Another check: $(-5)\times(3+(-7)) = (-5)\times(-4)=20$, and $(-5)\times3+(-5)\times(-7)=-15+35=20$ โ
Visually show the distributive property for $-4\times(2+(-3))$.
Multiplying by $-4$ means adding the additive inverse of the quantity, 4 times. The quantity is $2+(-3)=-1$, so its inverse is $1$, added 4 times: $4\times1=4$. Directly: $-4\times(-1)=4$.
Split form: $-4\times2 + (-4)\times(-3) = -8+12=4$ โ same answer, confirming the property visually (splitting the "4 copies of the inverse of $-1$" into "4 copies of the inverse of 2" plus "4 copies of the inverse of $-3$").
Activity Pick the Pattern โ Machines
Find the operation used by Machine 1 and fill in the last result.
| a | b | c | Result |
|---|---|---|---|
| 5 | 8 | 3 | 10 |
| 10 | 11 | 12 | 9 |
| 5 | 8 | โ3 | 16 |
| โ3 | 10 | 2 | 5 |
| โ4 | โ1 | โ6 | 1 |
| โ10 | โ12 | โ9 | ? |
The operation is $a+b-c$: check $5+8-3=10$ โ, and $(-4)+(-1)-(-6)=1$ โ.
Last row: $(-10)+(-12)-(-9) = -10-12+9 = -13$
Find the operation used by Machine 2.
| a | b | c | Result |
|---|---|---|---|
| 4 | 8 | โ3 | โ29 |
| 6 | โ11 | 12 | 54 |
| 5 | 3 | 7 | โ22 |
| โ3 | 9 | โ8 | 35 |
| โ7 | 4 | 6 | 22 |
| โ10 | โ12 | โ9 | ? |
Testing $-(a\times b + c)$: row 1: $-(4\times8+(-3))=-(32-3)=-29$ โ. Row 4: $-(-3\times9+(-8))=-(-27-8)=35$ โ.
Last row: $-((-10)\times(-12)+(-9)) = -(120-9)=-111$
Figure It Out โ Exercise
The full 16-question end-of-chapter exercise, solved step by step.
Find the values of the following expressions:
(a) $(-5)\times(18+(-3))$ (b) $(-7)\times4\times(-1)$ (c) $(-2)\times(-1)\times(-5)\times(-3)$
(a) $(-5)\times(18+(-3)) = (-5)\times15 = -75$
(b) $(-7)\times4\times(-1) = (-28)\times(-1) = 28$
(c) $(-2)\times(-1)\times(-5)\times(-3)$: step by step $(-2)\times(-1)=2$; $2\times(-5)=-10$; $-10\times(-3)=30$.
Find the values: (a) $(-27)\div9$ (b) $84\div(-4)$ (c) $(-56)\div(-2)$
(a) $(-27)\div9=-3$ (different signs โ negative)
(b) $84\div(-4)=-21$ (different signs โ negative)
(c) $(-56)\div(-2)=28$ (same signs โ positive)
Find the integer whose product with $(-1)$ is: (a) 27 (b) โ31 (c) โ1 (d) 1 (e) 0
We need $x$ such that $-1\times x = $ given value, so $x = -(\text{given value})$.
Given $47-56+14-8+2-8+5=-4$, find the value of $-47+56-14+8-2+8-5$ without calculating the full expression.
Every term in the second expression is exactly the additive inverse of the corresponding term in the first expression ($47\to-47$, $-56\to56$, and so on). So the whole second expression is the additive inverse of the first expression's sum.
$$-47+56-14+8-2+8-5 = -(47-56+14-8+2-8+5) = -(-4) = 4$$
Modified Collatz Conjecture with integers: if even, halve; if odd, multiply by $-3$ and add 1. Try starting numbers $-21$ and $-6$. Describe the pattern.
Starting at โ21 (odd โ ร(โ3)+1):
$-21 \to 64 \to 32 \to 16 \to 8 \to 4 \to 2 \to 1 \to -2 \to -1 \to 4 \to 2 \to 1 \to -2 \to -1 \to 4 \to \cdots$
Starting at โ6 (even โ halve):
$-6 \to -3 \to 10 \to 5 \to -14 \to -7 \to 22 \to 11 \to -32 \to -16 \to -8 \to -4 \to -2 \to -1 \to 4 \to 2 \to 1 \to -2 \to -1 \to 4 \to \cdots$
In a test, +4 marks for correct, โ2 for incorrect.
(a) Anita answered all questions and scored 40 with 15 correct answers. How many were incorrect? How many total questions?
(b) Anil scored โ10 with 5 correct answers. How many were incorrect? Did he leave any unanswered?
(a) Let incorrect answers $=x$. Marks: $15\times4 + x\times(-2) = 40$
$60 - 2x = 40 \Rightarrow 2x = 20 \Rightarrow x = 10$
Total questions $= 15+10 = 25$
(b) Let incorrect answers $=y$. Marks: $5\times4 + y\times(-2)=-10$
$20-2y=-10 \Rightarrow 2y=30 \Rightarrow y=15$
Anil answered $5+15=20$ questions. Since the test has 25 questions (from part a), he left $25-20=5$ questions unanswered.
Pick the pattern โ find the operation done by the machine:
| a | b | c | Result |
|---|---|---|---|
| 4 | 8 | โ3 | 28 |
| 6 | 9 | 6 | โ48 |
| 2 | 3 | โ2 | 8 |
| โ9 | 5 | โ8 | 31 |
| 7 | โ4 | โ6 | โ17 |
| โ16 | โ6 | โ9 | ? |
Testing $a-b\times c$: row 1: $4-8\times(-3)=4+24=28$ โ. Row 4: $-9-5\times(-8)=-9+40=31$ โ. Row 5: $7-(-4)\times(-6)=7-24=-17$ โ.
Last row: $-16 - (-6)\times(-9) = -16-54=-70$
Temperature drops 5ยฐC each hour, currently at 8ยฐC. Write an expression for the temperature after 4 hours.
$$8 + (4\times(-5)) = 8 - 20 = -12$$
Find 3 consecutive numbers with a product of (a) โ6 (b) 120
(a) Try $-3,-2,-1$: $(-3)\times(-2)\times(-1) = 6\times(-1)=-6$ โ
(b) Try $4,5,6$: $4\times5\times6=120$ โ
An alien currency uses only $+13$ pibs and $-9$ pibs coins. Show combinations that total: (a) +20 (b) +40 (c) โ50 (d) +8 (e) +10 (f) โ2 (g) +1, and (h) determine whether 1568 pibs can be purchased.
We need non-negative whole numbers of $+13$ coins ($x$) and $-9$ coins ($y$) so that $13x - 9y = \text{target}$.
(a) $+20$: $13(5)-9(5)=65-45=20$ โ 5 coins of +13, 5 coins of โ9
(b) $+40$: $13(10)-9(10)=130-90=40$ โ 10 coins of +13, 10 coins of โ9
(c) $-50$: $13(1)-9(7)=13-63=-50$ โ 1 coin of +13, 7 coins of โ9
(d) $+8$: $13(2)-9(2)=26-18=8$ โ 2 coins of +13, 2 coins of โ9
(e) $+10$: $13(7)-9(9)=91-81=10$ โ 7 coins of +13, 9 coins of โ9
(f) $-2$: $13(4)-9(6)=52-54=-2$ โ 4 coins of +13, 6 coins of โ9
(g) $+1$: $13(7)-9(10)=91-90=1$ โ 7 coins of +13, 10 coins of โ9
(h) Since $\gcd(13,9)=1$, every integer target can be expressed as $13x-9y$ for some non-negative integers $x,y$ (Bรฉzout's identity guarantees a solution, and using enough extra $+13,-9$ coin pairs โ which net to $13-9\times$something close to zero when scaled โ we can always reach a non-negative solution). For 1568: $13(122) - 9(2) = 1586-18=1568$ โ
Find the values:
(a) $(32\times(-18))\div((-36))$
(b) $(32)\div((-36)\times(-18))$
(c) $(25\times(-12))\div((45)\times(-27))$
(d) $(280\times(-7))\div((-8)\times(-35))$
(a) $32\times(-18)=-576$; $-576\div(-36)=16$
(b) $(-36)\times(-18)=648$; $32\div648=\dfrac{4}{81}$
(c) $25\times(-12)=-300$; $45\times(-27)=-1215$; $-300\div(-1215)=\dfrac{300}{1215}=\dfrac{20}{81}$
(d) $280\times(-7)=-1960$; $(-8)\times(-35)=280$; $-1960\div280=-7$
Arrange in increasing order:
(a) $(-348)+(-1064)$ (b) $(-348)-(-1064)$ (c) $348-(-1064)$
(d) $(-348)\times(-1064)$ (e) $348\times(-1064)$ (f) $348\times964$
(a) $-348+(-1064)=-1412$
(b) $-348-(-1064)=-348+1064=716$
(c) $348-(-1064)=348+1064=1412$
(d) $(-348)\times(-1064)=370272$
(e) $348\times(-1064)=-370272$
(f) $348\times964=335472$
Given $(-548)\times972 = -532656$, write the values of:
(a) $(-547)\times972$ (b) $(-548)\times971$ (c) $(-547)\times971$
(a) $(-547)=(-548)+1$, so $(-547)\times972 = (-548)\times972 + 972 = -532656+972=-531684$
(b) $971=972-1$, so $(-548)\times971 = (-548)\times972 - (-548)\times1 = -532656+548=-532108$
(c) $(-547)\times971 = (-547)\times972 - (-547) = -531684+547=-531137$
Given $207\times(-33+7)=-5382$, find $-207\times(33-7)$.
Note $-33+7=-26$ and $33-7=26$, i.e. $33-7 = -(-33+7)$.
Since $207\times(-26)=-5382$, we know $207\times26=5382$.
$$-207\times(33-7) = -207\times26 = -5382$$
Using $3,-2,5,-6$ exactly once and $+,-,\times$ exactly once (with brackets), write an expression for: (a) maximum possible result (b) minimum possible result.
We systematically test how grouping the "โ" and "ร" operators changes the outcome, since multiplying by a large negative number can create large positive or negative swings.
(a) Maximum:
$$(-6)\times\big((-2)-5\big)+3 = (-6)\times(-7)+3 = 42+3 = 45$$
(b) Minimum:
$$\big(3+5-(-2)\big)\times(-6) = (3+5+2)\times(-6) = 10\times(-6) = -60$$
Fill in the blanks in at least 5 different ways with integers:
(a) $\square + \square\times\square = -36$
(b) $(\square-\square)\times\square = 12$
(c) $(\square-(\square-\square)) = -1$
(a) Need $a+b\times c=-36$. Five valid sets $(a,b,c)$:
| a | b | c | Check |
|---|---|---|---|
| 0 | 6 | โ6 | $0+6\times(-6)=-36$ |
| โ36 | 0 | 5 | $-36+0\times5=-36$ |
| 4 | โ8 | 5 | $4+(-8)\times5=-36$ |
| โ6 | 6 | โ5 | $-6+6\times(-5)=-36$ |
| 12 | โ4 | 12 | $12+(-4)\times12=-36$ |
(b) Need $(a-b)\times c=12$. Five valid sets $(a,b,c)$:
| a | b | c | Check |
|---|---|---|---|
| 5 | 3 | 6 | $(5-3)\times6=12$ |
| 10 | 4 | 2 | $(10-4)\times2=12$ |
| โ1 | โ4 | 4 | $(-1-(-4))\times4=12$ |
| 0 | โ2 | 6 | $(0-(-2))\times6=12$ |
| โ2 | โ14 | 1 | $(-2-(-14))\times1=12$ |
(c) Need $a-(b-c)=-1$, i.e. $a-b+c=-1$. Five valid sets $(a,b,c)$:
| a | b | c | Check |
|---|---|---|---|
| 0 | 1 | 0 | $0-(1-0)=-1$ |
| 5 | 6 | 0 | $5-(6-0)=-1$ |
| โ1 | 0 | 0 | $-1-(0-0)=-1$ |
| 2 | 4 | 1 | $2-(4-1)=-1$ |
| 10 | 12 | 1 | $10-(12-1)=-1$ |