Corrigé

7.1 Positive Integers

Pack A — Réponses

#	Question	Réponse
1	Fill in the missing digit: 3_7 + 285 = 632. Which digit is missing and in which place? Justify your answer.	The missing digit is 4 (tens place of 347)
2	Round 2364 to the nearest hundred and name the digit that decided the rounding. Justify.	2 400; the tens digit (6) decided — it is 5 or more, so round up
3	If 523 − ?? = 355, what is the missing number? Show how you would check your answer.	168; check: 355 + 168 = 523 ✓
4	Without fully calculating, decide: is 34 × 6 closer to 180 or 210? Justify.	Closer to 210; 30 × 6 = 180 and 40 × 6 = 240, so 34 × 6 is between, nearer 210
5	Is 4728 divisible by 4? Use short division to check. Show your working and state whether there is a remainder.	Yes — 4728 ÷ 4 = 1182 with no remainder
6	Is 84 divisible by 4? Use short division to find out. Show enough working to be sure.	Yes — 84 ÷ 4 = 21 exactly
7	Find ONE factor of 36 (other than 1 and 36) and justify that it is a factor.	E.g. 4 — because 36 ÷ 4 = 9 (a whole number) ✓
8	Which is the smallest multiple of 7 that is greater than 40? Justify.	42; because 42 = 7 × 6 and 35 = 7 × 5 which is not greater than 40
9	Is 91 a prime number? Justify with at least one trial division.	No — 91 = 7 × 13
10	Without working out the exact value of $9^2$, decide: is $9^2$ greater or less than 100? Justify.	Less than 100; because 10² = 100 and 9 < 10, so 9² < 10²
11	Round 3467 to the nearest ten.	3 470
12	Calculate 2847 + 1965.	4 812
13	Calculate 5023 − 2648.	2 375
14	Calculate 1848 ÷ 7.	264
15	Work out 5 + 3 × 8 using the correct order of operations.	29
16	Find the HCF of 24 and 36.	12
17	Find the LCM of 6 and 8.	24
18	What is 4 cubed? Write ${4}^3$.	64
19	Write 28 as a product of prime factors.	$2^2 \times 7$
20	Each notebook costs £3. How much do 14 notebooks cost?	£42
21	Work out $(5 + 7) \times 4 - 18 \div 3$.	42
22	Find the HCF of 12, 18 and 30.	6
23	Express 84 as a product of prime factors in index form.	$2^2 \times 3 \times 7$
24	Work out $\sqrt[3]{{216}}$.	6
25	Without calculating exactly, decide which is larger: 23 × 24 or 22 × 25. Explain your reasoning.	23 × 24 is larger
26	Calculate ${4}^2 \times 3$.	48
27	Find a number that is both a multiple of 6 and a factor of 60.	12 (or 6, 60)
28	Estimate 47 × 18 by rounding each number to 1 significant figure.	1 000 (estimate)
29	Apples cost 35p each. Oranges cost 50p each. Sam buys 4 apples and 3 oranges and pays with a £5 note. How much change does he receive?	£2.10
30	Work out $(7^2 - 4) \div 5 + 6$.	15
31	Given $p = 2^3 \times 3 \times 5$ and $q = 2^2 \times 3^2 \times 7$, find the HCF and LCM of $p$ and $q$.	HCF = 12, LCM = 2 520
32	Find the largest prime factor of 180.	5
33	In the multiplication below, each letter stands for a different digit. Find A and B. $$\overline{4A} \times 3 = \overline{14B}$$	A = 8, B = 4 (since 48 × 3 = 144)
34	Two whole numbers have HCF = 6 and LCM = 60. Find all possible pairs of numbers.	(6, 60) and (12, 30)
35	Is 91 a prime number? Show your reasoning.	No — 91 = 7 × 13
36	A number satisfies all of these: it is two-digit, it is prime, and the sum of its digits is 7. Find all such numbers.	16 (no — not prime), 25 (no), 34 (no), 43 ✓, 61 ✓, 70 (no). Answer: 43 and 61.
37	Two whole numbers $a$ and $b$ satisfy $a \times b = 56$ and $a + b = 15$, with $a > b$. Find $a$ and $b$.	a = 8, b = 7
38	Explain why the product of any two consecutive even numbers is always divisible by 8.	Any two consecutive even numbers are 2k and 2(k+1). Their product is 4k(k+1). Since k and k+1 are consecutive integers, one of them is even, so k(k+1) is even. Therefore 4k(k+1) = 4 × (even) = 8 × (integer). Always divisible by 8.
39	Work out $(9^2 - (2 + 3)^2) \div ((9 - 2 - 3) \times 4)$.	3.5
40	A school raises £1800 for a trip. Transport costs £480. The remaining money is split equally among 24 students. Each student pays £15 on top of their share. How much does each student have in total to spend?	£70

Pack B — Réponses

#	Question	Réponse
1	Fill in the missing digit: 4_3 + 378 = 841. Which digit is missing and in which place? Justify your answer.	The missing digit is 6 (tens place of 463)
2	Round 7819 to the nearest thousand and name the digit that decided the rounding. Justify.	8 000; the hundreds digit (8) decided — it is 5 or more, so round up
3	If 741 − ?? = 448, what is the missing number? Show how you would check your answer.	293; check: 448 + 293 = 741 ✓
4	Without fully calculating, decide: is 47 × 8 closer to 350 or 400? Justify.	Closer to 400; 50 × 8 = 400 and 40 × 8 = 320, so 47 × 8 is between, nearer 400
5	Is 5913 divisible by 3? Use short division to check. Show your working and state whether there is a remainder.	Yes — 5913 ÷ 3 = 1971 with no remainder
6	Is 135 divisible by 9? Use short division to find out. Show enough working to be sure.	Yes — 135 ÷ 9 = 15 exactly
7	Find ONE factor of 48 (other than 1 and 48) and justify that it is a factor.	E.g. 6 — because 48 ÷ 6 = 8 (a whole number) ✓
8	Which is the smallest multiple of 9 that is greater than 50? Justify.	54; because 54 = 9 × 6 and 45 = 9 × 5 which is not greater than 50
9	Is 43 a prime number? Justify with at least one trial division.	Yes — 43 has no factors other than 1 and 43
10	Without working out the exact value of $7^2$, decide: is $7^2$ greater or less than 50? Justify.	Less than 50; because 7² = 49 and 8² = 64, so 7² is between 49 and 64, below 50
11	Round 8352 to the nearest hundred.	8 400
12	Calculate 5384 + 2769.	8 153
13	Calculate 7104 − 3857.	3 247
14	Calculate 2484 ÷ 9.	276
15	Work out 12 + 4 × 7 using the correct order of operations.	40
16	Find the HCF of 30 and 42.	6
17	Find the LCM of 4 and 10.	20
18	What is 5 cubed? Write ${5}^3$.	125
19	Write 30 as a product of prime factors.	$2 \times 3 \times 5$
20	Each notebook costs £7. How much do 12 notebooks cost?	£84
21	Work out $(8 + 6) \times 3 - 20 \div 4$.	37
22	Find the HCF of 20, 28 and 36.	4
23	Express 120 as a product of prime factors in index form.	$2^3 \times 3 \times 5$
24	Work out $\sqrt[3]{{343}}$.	7
25	Without calculating exactly, decide which is larger: 31 × 34 or 32 × 33. Explain your reasoning.	32 × 33 is larger
26	Calculate ${6}^2 \times 5$.	180
27	Find a number that is both a multiple of 4 and a factor of 48.	8 (or 4, 12, 16, 24, 48)
28	Estimate 63 × 29 by rounding each number to 1 significant figure.	1 800 (estimate)
29	Apples cost 45p each. Oranges cost 60p each. Sam buys 3 apples and 4 oranges and pays with a £5 note. How much change does he receive?	£1.25
30	Work out $(8^2 - 4) \div 6 + 3$.	13
31	Given $p = 2^3 \times 3 \times 5$ and $q = 2^2 \times 3^2 \times 7$, find the HCF and LCM of $p$ and $q$.	HCF = 6, LCM = 1 260
32	Find the largest prime factor of 156.	13
33	In the multiplication below, each letter stands for a different digit. Find A and B. $$\overline{4A} \times 3 = \overline{14B}$$	A = 8, B = 4 (same reasoning)
34	Two whole numbers have HCF = 4 and LCM = 48. Find all possible pairs of numbers.	(4, 48) and (12, 16)
35	Is 97 a prime number? Show your reasoning.	Yes — 97 is prime
36	A number satisfies all of these: it is two-digit, it is prime, and the sum of its digits is 8. Find all such numbers.	17 (1+7=8, prime ✓), 53 (5+3=8, prime ✓), 71 (7+1=8, prime ✓). Answer: 17, 53, 71.
37	Two whole numbers $a$ and $b$ satisfy $a \times b = 60$ and $a + b = 17$, with $a > b$. Find $a$ and $b$.	a = 12, b = 5
38	Explain why the product of any two consecutive even numbers is always divisible by 8.	Examples: 2×4=8 ✓; 4×6=24=8×3 ✓; 6×8=48=8×6 ✓. General: 2k × 2(k+1) = 4k(k+1). Consecutive integers k, k+1 → one is even → k(k+1) = 2m for some integer m → product = 8m. QED.
39	Work out $(10^2 - (3 + 2)^2) \div ((10 - 3 - 2) \times 5)$.	3
40	A school raises £2200 for a trip. Transport costs £640. The remaining money is split equally among 40 students. Each student pays £12 on top of their share. How much does each student have in total to spend?	£51.50

Problèmes — Solutions détaillées

**Locker puzzle.** 100 lockers are numbered 1 to 100, and they are all closed. 100 students walk past them in order. - Student 1 opens every locker. - Student 2 closes every **2nd** locker (lockers 2, 4, 6, …). - Student 3 changes the state of every **3rd** locker. - This continues until student 100. How many lockers are open at the end? Which lockers are they?

Réponse

10 lockers are open: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

A locker ends up open if and only if it is toggled an **odd** number of times. Locker $n$ is toggled once for each of its factors. Most numbers have an even number of factors (they pair up: if $d$ divides $n$, so does $n \div d$). The exception is **perfect squares**, where one factor equals its own pair ($\sqrt{n}$), giving an odd total. The perfect squares from 1 to 100 are $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$ — that is $1^2$ through $10^2$. So exactly **10 lockers** are open.

**Coin combinations.** Using only 5¢, 10¢, and 20¢ coins, in how many different ways can you make exactly 50¢? (Order does not matter — {3 × 10¢, 4 × 5¢} is one way.)

Réponse

12 ways

Systematically vary the number of 20¢ coins (0, 1, or 2): - **Two 20¢ coins** (40¢ used, 10¢ remaining): (1×10¢, 0×5¢) or (0×10¢, 2×5¢). **2 ways.** - **One 20¢ coin** (20¢ used, 30¢ remaining from 5¢ and 10¢): (0×10, 6×5), (1×10, 4×5), (2×10, 2×5), (3×10, 0×5). **4 ways.** - **No 20¢ coins** (50¢ from 5¢ and 10¢ only): (0×10, 10×5), (1×10, 8×5), (2×10, 6×5), (3×10, 4×5), (4×10, 2×5), (5×10, 0×5). **6 ways.** Total: 2 + 4 + 6 = **12 ways.** Since the number of 20¢ coins is fixed in each group, the groups are mutually exclusive and exhaustive.

**The 1089 trick.** Take any 3-digit number where the first digit is at least 2 more than the last digit (e.g. 731). Step 1: Reverse the digits (137). Step 2: Subtract the smaller from the larger (731 − 137 = 594). Step 3: Reverse your result (495). Step 4: Add the result from Step 2 to its reverse (594 + 495). (a) Try the trick with 731. What do you get? (b) Try it with 852. What do you get? (c) Does it always give 1089? Explain why by using a general 3-digit number with hundreds digit $a$, tens digit $b$, and units digit $c$ where $a > c$.

Réponse

(a) 1089 (b) 1089 (c) Always 1089 — see working.

(a) 731 reversed = 137. 731 − 137 = 594. 594 reversed = 495. 594 + 495 = **1089** ✓. (b) 852 reversed = 258. 852 − 258 = 594. 594 reversed = 495. 594 + 495 = **1089** ✓. (c) Write the number as $100a + 10b + c$. Its reverse is $100c + 10b + a$. Difference (assuming $a > c$): $(100a + 10b + c) − (100c + 10b + a) = 99a − 99c = 99(a − c)$. Since $a$ and $c$ are single digits with $a > c$, we have $a − c \in \{2, 3, 4, 5, 6, 7, 8, 9\}$, giving $99(a−c) \in \{198, 297, 396, 495, 594, 693, 792, 891\}$. These are all 3-digit multiples of 99. For each: its digit-reversal is also in the list, and the sum of any of these with its reversal equals 1089. For example $198 + 891 = 1089$, $297 + 792 = 1089$, $396 + 693 = 1089$, $495 + 495 = 990$ — wait, for $a − c = 5$ we get 495; its reversal 594; $495 + 594 = 1089$ ✓. The result is always **1089**.

**Calendar logic.** 1st January is a Wednesday. What day of the week is 1st March in the same (non-leap) year? Show your reasoning clearly.

Réponse

Saturday

January has 31 days; February has 28 days (non-leap year). From 1 January to 1 March is 31 + 28 = 59 days later. 59 ÷ 7 = 8 remainder 3. So 1 March is 3 days after Wednesday: +1 = Thursday, +2 = Friday, +3 = Saturday. **1st March is a Saturday.** (Note: teacher may wish to double-check with a real calendar for a specific year.)

**Handshake problem.** At a party, every person shakes hands exactly once with every other person. (a) If there are 5 people, how many handshakes are there in total? (b) If there are 10 people, how many handshakes are there? (c) Find a formula for the number of handshakes when there are $n$ people. (d) At a conference there were 190 handshakes in total. How many people attended?

Réponse

(a) 10 (b) 45 (c) $\dfrac{n(n-1)}{2}$ (d) 20 people

(a) Person 1 shakes hands with 4 others, person 2 with 3 remaining others (not counting person 1 again), and so on: $4 + 3 + 2 + 1 = \mathbf{10}$ handshakes. (b) Same pattern: $9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = \mathbf{45}$ handshakes. (c) Each of the $n$ people shakes hands with $(n-1)$ others. That gives $n(n-1)$ — but each handshake has been counted twice (once for each person). So the formula is $\dfrac{n(n-1)}{2}$. (d) Set $\dfrac{n(n-1)}{2} = 190$, so $n(n-1) = 380$. We need two consecutive integers with product 380. Try: $19 \times 20 = 380$ ✓. So $n = \mathbf{20}$ people attended.

**Three bells.** Three bells toll at the start of school assembly. After that: - Bell A tolls every 8 minutes. - Bell B tolls every 12 minutes. - Bell C tolls every 18 minutes. (a) After how many minutes will all three bells first toll together again? (b) How many times does Bell A toll in the first 2 hours (not counting the start)? (c) Between the start and the first time all three bells toll together, how many times does Bell B toll on its own (not at the same time as any other bell)?

Réponse

(a) 72 minutes (b) 15 times (c) 2 times

(a) We need the LCM of 8, 12, and 18. Prime factorisations: $8 = 2^3$, $12 = 2^2 \times 3$, $18 = 2 \times 3^2$. LCM = $2^3 \times 3^2 = 8 \times 9 = \mathbf{72}$ minutes. (b) In 120 minutes (2 hours), Bell A tolls at: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120. That is $120 \div 8 = \mathbf{15}$ times. (c) Bell B tolls at: 12, 24, 36, 48, 60, 72 minutes. We remove times when B coincides with A or C. LCM(8, 12) = 24: B and A both toll at 24, 48, 72. LCM(12, 18) = 36: B and C both toll at 36, 72. Check each: 12 — is it a multiple of 8? $12 ÷ 8 = 1.5$ ✗; of 18? $12 ÷ 18 < 1$ ✗ → **B alone** ✓. 24 — multiple of 8 ✓ → with A. 36 — multiple of 18 ✓ → with C. 48 — multiple of 8 ✓ → with A. 60 — $60 ÷ 8 = 7.5$ ✗, $60 ÷ 18 ≈ 3.33$ ✗ → **B alone** ✓. 72 — all three. B tolls alone at minutes 12 and 60: **2 times**.

**Unknown digits.** In the multiplication below, **A** and **B** represent single digits (0–9): $$A3 \times B = 161$$ Find the values of A and B. Show your working.

Réponse

A = 2, B = 7 (since 23 × 7 = 161)

For $A3 \times B = 161$, the two-digit number $A3$ must end in 3. Possibilities: 13, 23, 33, 43, 53, 63, 73, 83, 93. Divide 161 by each: $161 \div 13 \approx 12.4$ (no); $161 \div 23 = 7$ ✓ (7 is a single digit). Check: $23 \times 7 = 161$ ✓. So $A = 2$ and $B = 7$.

**Sequence puzzle.** Here is a sequence: 2, 6, 12, 20, 30, … Find the 10th term, and write a rule for the $n$th term.

Réponse

110

Look at the differences: 6−2=4, 12−6=6, 20−12=8, 30−20=10 — second differences are constant (2), so it is quadratic. Notice: 2 = 1×2, 6 = 2×3, 12 = 3×4, 20 = 4×5, 30 = 5×6. The pattern is $n(n+1)$. For the 10th term: $10 \times 11 = \mathbf{110}$.

**Number theory.** Find the smallest three-digit number that is: - divisible by 7, **and** - has a digit sum of 9. Show how you checked your answer.

Réponse

126

Three-digit multiples of 7 start at 105. Check digit sums in order: 105 → 1+0+5=6 (no); 112 → 1+1+2=4 (no); 119 → 1+1+9=11 (no); 126 → 1+2+6=9 ✓. The smallest three-digit multiple of 7 with digit sum 9 is **126**. Check: 126 ÷ 7 = 18 ✓; digit sum = 1+2+6 = 9 ✓.

**Magic square.** In a 3 × 3 magic square, every row, every column, and both main diagonals have the same sum (the "magic sum"). The grid below has three numbers already placed. Find all nine entries. $$\begin{array}{|c|c|c|}\hline 2 & \square & \square \\\hline \square & 5 & \square \\\hline \square & \square & 8 \\\hline\end{array}$$ Hint: the magic sum can be found from the diagonal containing 2, 5, 8.

Réponse

Magic sum = 15. Grid (rows): [2, 7, 6], [9, 5, 1], [4, 3, 8].

The main diagonal (top-left to bottom-right) contains 2, 5, 8. These sum to 15, so the **magic sum = 15**. Now fill the grid. Let the entries be: $$\begin{array}{|c|c|c|}\hline 2 & a & b \\\hline c & 5 & d \\\hline e & f & 8 \\\hline\end{array}$$ Row 1: $2 + a + b = 15 \Rightarrow a + b = 13$. Row 3: $e + f + 8 = 15 \Rightarrow e + f = 7$. Col 1: $2 + c + e = 15 \Rightarrow c + e = 13$. Col 3: $b + d + 8 = 15 \Rightarrow b + d = 7$. Anti-diagonal (top-right to bottom-left): $b + 5 + e = 15 \Rightarrow b + e = 10$. From $a + b = 13$ and $b + d = 7$: subtract $\Rightarrow a - d = 6$. Col 2: $a + 5 + f = 15 \Rightarrow a + f = 10$. Row 2: $c + 5 + d = 15 \Rightarrow c + d = 10$. We know the standard 3×3 magic square using the integers 1–9 has magic sum 15. The only 3×3 magic square (up to rotation/reflection) containing 2, 5, 8 on the leading diagonal uses all of 1–9 exactly once. Testing: rows [2,7,6], [9,5,1], [4,3,8]. Verify: rows: 15,15,15 ✓; cols: 15,15,15 ✓; diagonals: 2+5+8=15, 6+5+4=15 ✓. Answer: **[2,7,6], [9,5,1], [4,3,8]**.

**Optimisation.** A farmer has exactly 120 m of fencing. He wants to enclose a rectangular field using all of it. What are the dimensions that give the **largest possible area**? What is that area?

Réponse

30 m × 30 m = 900 m²

Let the length be $l$ and the width be $w$. Perimeter: $2l + 2w = 120$, so $l + w = 60$. Area $A = l \times w$. Try values: if $l = 30, w = 30$: $A = 900$. If $l = 40, w = 20$: $A = 800$. If $l = 50, w = 10$: $A = 500$. The area is largest when the rectangle is a **square** (both sides equal). With $l + w = 60$, the maximum is at $l = w = 30$ giving $A = \mathbf{900}$ m².

**Modular reasoning.** A clock loses exactly 4 minutes every hour. It is set correctly at **6:00 am**. What time does the clock **show** at **6:00 pm** the same day?

Réponse

5:12 pm (the clock shows 11 hours 12 minutes have passed, so 5:12 pm)

From 6:00 am to 6:00 pm is 12 real hours. The clock loses 4 minutes every real hour, so in 12 hours it loses $12 \times 4 = 48$ minutes. The clock therefore shows $12 \text{ hours} - 48 \text{ minutes} = 11 \text{ hours and } 12 \text{ minutes}$ elapsed. Starting from 6:00 am + 11 h 12 min = **5:12 pm**. (The clock shows 5:12 pm even though the actual time is 6:00 pm.)