Chapter 2 — Operations with Integers
Complete Step‑by‑Step Solutions
Every in-text question and every “Figure it Out” exercise from the chapter, solved in full detail — 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 — solved in the order they appear.
2.1 Rakesh’s Puzzle — 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 − second number).
| First Number | Second Number | Sum | Difference |
|---|---|---|---|
| 10 | 15 | 25 | −5 |
| 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 −11.
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: $18+9=27$ ✓, $18-9=9$ ✓
(b) Sum = 4, Difference = 12
first $=\frac{4+12}{2}=8$, second $=\frac{4-12}{2}=-4$
Check: $8+(-4)=4$ ✓, $8-(-4)=12$ ✓
(c) Sum = 0, Difference = 10
first $=\frac{0+10}{2}=5$, second $=\frac{0-10}{2}=-5$
(d) Sum = 0, Difference = −10
first $=\frac{0-10}{2}=-5$, second $=\frac{0+10}{2}=5$
(e) Sum = −7, Difference = −1
first $=\frac{-7-1}{2}=-4$, second $=\frac{-7+1}{2}=-3$
(f) Sum = −7, Difference = −13
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 — 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 — 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 — 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 — 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 — 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 $-18$) — 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 — 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 — 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 4×2 = 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 × negative → 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 — the magnitude 56088 stays the same. One negative factor → negative product; two negative factors → 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 — 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 — 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 — multiplication is commutative for integers: $a\times b = b\times a$.
History Brahmagupta’s Rules
Brāhmasphuṭasiddhānta (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 ṛṇa (debt) for negative numbers — 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, −2 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 — every question correct: $50\times5=250$.
Minimum — 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 — is the product always the same?
| 8 | −4 | 12 | −6 |
| −28 | 14 | −42 | 21 |
| 12 | −6 | 18 | −9 |
| 20 | −10 | 30 | −15 |
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 × 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 — 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 32°C at 5°C/hour. Find the temperature 10 hours later.
Rate $=-5°C$ per hour. After 10 hours: $32 + 10\times(-5) = 32-50=-18$.
3. A cement company earns ₹8/bag profit on white cement and ₹5/bag loss on grey cement.
(a) Sells 3,000 white & 5,000 grey bags in a month — 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 — 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.
$(-2)\times(4+(-3)) = (-2)\times1=-2$
$(-2)\times4 + (-2)\times(-3) = -8+6=-2$ ✓ — matches!
Another check: $(-5)\times(3+(-7)) = (-5)\times(-4)=20$, and $(-5)\times3+(-5)\times(-7)=-15+35=20$ ✓
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: $-4\times2 + (-4)\times(-3) = -8+12=4$ — 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 — 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 | −3 | 16 |
| −3 | 10 | 2 | 5 |
| −4 | −1 | −6 | 1 |
| −10 | −12 | −9 | ? |
The operation is $a+b-c$: check $5+8-3=10$ ✓, and $(-4)+(-1)-(-6)=1$ ✓.
Last row: $(-10)+(-12)-(-9) = -10-12+9 = -13$
Find the operation used by Machine 2.
| a | b | c | Result |
|---|---|---|---|
| 4 | 8 | −3 | −29 |
| 6 | −11 | 12 | 54 |
| 5 | 3 | 7 | −22 |
| −3 | 9 | −8 | 35 |
| −7 | 4 | 6 | 22 |
| −10 | −12 | −9 | ? |
Testing $-(a\times b + c)$: row 1: $-(4\times8+(-3))=-(32-3)=-29$ ✓. Row 4: $-(-3\times9+(-8))=-(-27-8)=35$ ✓.
Last row: $-((-10)\times(-12)+(-9)) = -(120-9)=-111$
Figure It Out — 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 → negative)
(b) $84\div(-4)=-21$ (different signs → negative)
(c) $(-56)\div(-2)=28$ (same signs → positive)
Find the integer whose product with $(-1)$ is: (a) 27 (b) −31 (c) −1 (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 −21 (odd → ×(−3)+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 −6 (even → 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, −2 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 −10 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 — find the operation done by the machine:
| a | b | c | Result |
|---|---|---|---|
| 4 | 8 | −3 | 28 |
| 6 | 9 | 6 | −48 |
| 2 | 3 | −2 | 8 |
| −9 | 5 | −8 | 31 |
| 7 | −4 | −6 | −17 |
| −16 | −6 | −9 | ? |
Testing $a-b\times c$: row 1: $4-8\times(-3)=4+24=28$ ✓. Row 4: $-9-5\times(-8)=-9+40=31$ ✓. Row 5: $7-(-4)\times(-6)=7-24=-17$ ✓.
Last row: $-16 – (-6)\times(-9) = -16-54=-70$
Temperature drops 5°C each hour, currently at 8°C. 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) −6 (b) 120
(a) Try $-3,-2,-1$: $(-3)\times(-2)\times(-1) = 6\times(-1)=-6$ ✓
(b) Try $4,5,6$: $4\times5\times6=120$ ✓
An alien currency uses only $+13$ pibs and $-9$ pibs coins. Show combinations that total: (a) +20 (b) +40 (c) −50 (d) +8 (e) +10 (f) −2 (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) $+20$: $13(5)-9(5)=65-45=20$ → 5 coins of +13, 5 coins of −9
(b) $+40$: $13(10)-9(10)=130-90=40$ → 10 coins of +13, 10 coins of −9
(c) $-50$: $13(1)-9(7)=13-63=-50$ → 1 coin of +13, 7 coins of −9
(d) $+8$: $13(2)-9(2)=26-18=8$ → 2 coins of +13, 2 coins of −9
(e) $+10$: $13(7)-9(9)=91-81=10$ → 7 coins of +13, 9 coins of −9
(f) $-2$: $13(4)-9(6)=52-54=-2$ → 4 coins of +13, 6 coins of −9
(g) $+1$: $13(7)-9(10)=91-90=1$ → 7 coins of +13, 10 coins of −9
(h) Since $\gcd(13,9)=1$, every integer target can be expressed as $13x-9y$ for some non-negative integers $x,y$ (Bézout’s identity guarantees a solution, and using enough extra $+13,-9$ coin pairs — which net to $13-9\times$something close to zero when scaled — we can always reach a non-negative solution). For 1568: $13(122) – 9(2) = 1586-18=1568$ ✓
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 “−” and “×” 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 | −6 | $0+6\times(-6)=-36$ |
| −36 | 0 | 5 | $-36+0\times5=-36$ |
| 4 | −8 | 5 | $4+(-8)\times5=-36$ |
| −6 | 6 | −5 | $-6+6\times(-5)=-36$ |
| 12 | −4 | 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 | $(10-4)\times2=12$ |
| −1 | −4 | 4 | $(-1-(-4))\times4=12$ |
| 0 | −2 | 6 | $(0-(-2))\times6=12$ |
| −2 | −14 | 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 | $5-(6-0)=-1$ |
| −1 | 0 | 0 | $-1-(0-0)=-1$ |
| 2 | 4 | 1 | $2-(4-1)=-1$ |
| 10 | 12 | 1 | $10-(12-1)=-1$ |