The Transport Museum

Given: A multiplication matrix where some products are given and we must find 1-digit multipliers/multiplicands to complete it.

Strategy: Work backwards from the given products to find 1-digit factors.

Step 1 – Identify factors from given products:
- $32 = 4 \times 8$ → row multiplier = 4, column multiplier = 8
- $42 = 6 \times 7$ → row multiplier = 6, column multiplier = 7
- $45 = 5 \times 9$ → row multiplier = 5, column multiplier = 9
- $21 = 3 \times 7$ → row multiplier = 3, column multiplier = 7

Step 2 – One possible completed matrix (header row: ×, 8, 9, 7, 6):

| × | 8 | 9 | 7 | 6 |
|---|---|---|---|---|
| 4 | 32 | 36 | 28 | 24 |
| 6 | 48 | 54 | 42 | 36 |
| 5 | 40 | 45 | 35 | 30 |
| 3 | 24 | 27 | 21 | 18 |

Note: Multiple valid answers are possible depending on the choice of 1-digit numbers, as long as the given products appear in the correct positions.

Given: Row products (orange) and column products (blue) are provided. We need to find the individual numbers.

Step 1 – Analyse the structure. The table appears to be a 3-row grid where the last column shows row products and the last row shows column products.

Step 2 – Find numbers for Row 1 (product = 42):
$42 = 6 \times 7$ or $2 \times 3 \times 7$, etc.
With the column product clue 56 in column 3: $56 = 7 \times 8$, so one entry is 7 or 8.
A possible row: $1, 6, 7$ → product = $1 \times 6 \times 7 = 42$ ✓

Step 3 – Find numbers for Row 2 (product = 50):
$50 = 2 \times 5 \times 5$ or $5 \times 10$
With column product clue 54 in column 3: $54 = 6 \times 9$
A possible row: $2, 5, 5$ → product = 50 ✓

Step 4 – Check column products against Row 3 (products: 63, 48, 60, 35):
- Column with product 63: $63 = 7 \times 9$ or $63 = 1 \times 63$
- Column with product 48: $48 = 6 \times 8$
- Column with product 60: $60 = 5 \times 12$ or $4 \times 15$
- Column with product 35: $35 = 5 \times 7$

Note: This is an open-ended puzzle with multiple valid solutions. The key method is to factorise each given product into 1-digit numbers and check consistency across rows and columns.

Concept: Multiplying any number by 10 gives that many tens.

$$2 \times 10 = 2 \text{ Tens} = \mathbf{20}$$

$$5 \times 10 = \mathbf{5} \text{ Tens} = \mathbf{50}$$

$$8 \times 10 = \mathbf{8} \text{ Tens} = \mathbf{80}$$

Concept: $10 \times 10$ means 10 groups of 10.

$$10 \times 10 = \mathbf{10} \text{ Tens} = \mathbf{100}$$

10 Tens make 1 Hundred = 100.

Given: $5 \times 15$

Method – Split 15 into 10 and 5:
$$5 \times 15 = 5 \times 10 + 5 \times 5$$
$$= 50 + 25 = \mathbf{75}$$

Method: Split 15 = 10 + 5, so $n \times 15 = n \times 10 + n \times 5$.

$$1 \times 15 = 10 + 5 = \mathbf{15}$$
$$2 \times 15 = 20 + 10 = \mathbf{30}$$
$$3 \times 15 = 30 + 15 = \mathbf{45}$$
$$4 \times 15 = 40 + 20 = \mathbf{60}$$
$$5 \times 15 = 50 + 25 = \mathbf{75}$$
$$6 \times 15 = 60 + 30 = \mathbf{90}$$
$$7 \times 15 = 70 + 35 = \mathbf{105}$$
$$8 \times 15 = 80 + 40 = \mathbf{120}$$
$$9 \times 15 = 90 + 45 = \mathbf{135}$$
$$10 \times 15 = 100 + 50 = \mathbf{150}$$

Pattern 1 – Patterns in the times-15 table:
- The products increase by 15 each time.
- All products end in either 0 or 5 (just like the times-5 table).
- The units digits follow the pattern: 5, 0, 5, 0, 5, 0, … (alternating).

Pattern 2 – Comparison with times-5 table:

| Times-5 | Times-15 | Difference |
|---|---|---|
| $1 \times 5 = 5$ | $1 \times 15 = 15$ | $15 - 5 = 10$ |
| $2 \times 5 = 10$ | $2 \times 15 = 30$ | $30 - 10 = 20$ |
| $3 \times 5 = 15$ | $3 \times 15 = 45$ | $45 - 15 = 30$ |

Similarity: Both tables end in 0 or 5.

Difference: Each product in the times-15 table is exactly 3 times the corresponding product in the times-5 table (since $15 = 3 \times 5$). The differences themselves form the times-10 table.

Question 3 – Times-tables for 11 to 20 (using splitting method):

For any number $n$ between 11 and 20, write $n = 10 + r$ where $r$ is the units digit.
Then: $k \times n = k \times 10 + k \times r$

Example – Times-12 table ($12 = 10 + 2$):
$$1 \times 12 = 10 + 2 = 12$$
$$2 \times 12 = 20 + 4 = 24$$
$$3 \times 12 = 30 + 6 = 36$$
$$\vdots$$
$$10 \times 12 = 100 + 20 = 120$$

Question 4 – Observations:
- Each product in the times-11 table = corresponding product in times-1 table + corresponding product in times-10 table.
- Each product in the times-12 table = corresponding product in times-2 table + corresponding product in times-10 table.
- In general: $k \times (n+10) = k \times n + k \times 10$, so every product in the higher table is exactly $k \times 10$ more than in the lower table.

Given: $3 \times 14$

Method – Split 14 into two equal groups of 7:
$$3 \times 14 = 3 \times 7 + 3 \times 7$$
$$= 21 + 21 = \text{double of } 21$$
$$= \mathbf{42}$$

Given: $6 \times 14$

**Method – Split 14 into $7 + 7$:**
$$6 \times 14 = 6 \times 7 + 6 \times 7$$
$$= 42 + 42 = \text{double of } 42$$
$$= \mathbf{84}$$

**Times-14 table (using $14 = 7 + 7$, i.e., double of times-7):**

$$1 \times 14 = 2 \times (1 \times 7) = 2 \times 7 = \mathbf{14}$$
$$2 \times 14 = 2 \times (2 \times 7) = 2 \times 14 = \mathbf{28}$$
$$3 \times 14 = 2 \times (3 \times 7) = 2 \times 21 = \mathbf{42}$$
$$4 \times 14 = 2 \times (4 \times 7) = 2 \times 28 = \mathbf{56}$$
$$5 \times 14 = 2 \times (5 \times 7) = 2 \times 35 = \mathbf{70}$$
$$6 \times 14 = 2 \times (6 \times 7) = 2 \times 42 = \mathbf{84}$$
$$7 \times 14 = 2 \times (7 \times 7) = 2 \times 49 = \mathbf{98}$$
$$8 \times 14 = 2 \times (8 \times 7) = 2 \times 56 = \mathbf{112}$$
$$9 \times 14 = 2 \times (9 \times 7) = 2 \times 63 = \mathbf{126}$$
$$10 \times 14 = 2 \times (10 \times 7) = 2 \times 70 = \mathbf{140}$$

Other tables that can be built by splitting and doubling:
- Times-6 = double of times-3
- Times-8 = double of times-4
- Times-10 = double of times-5
- Times-12 = double of times-6
- Times-16 = double of times-8
- Times-18 = double of times-9

Any even-numbered times-table can be constructed this way.

Concept: $n \times 10 = n$ Tens.

a) $15 \times 10 = \mathbf{15}$ Tens $= \mathbf{150}$

b) $16 \times 10 = \mathbf{16}$ Tens $= \mathbf{160}$

c) $19 \times 10 = \mathbf{19}$ Tens $= \mathbf{190}$

d) $20 \times 10 = \mathbf{20}$ Tens $= \mathbf{200}$

Concept: Multiplying a multiple of 10 by 10 — append a zero.

$$30 \times 10 = \mathbf{300}$$
$$40 \times 10 = \mathbf{400}$$
$$70 \times 10 = \mathbf{700}$$
$$50 \times 10 = \mathbf{500}$$
$$60 \times 10 = \mathbf{600}$$
$$80 \times 10 = \mathbf{800}$$

Given: 26 tempo travellers, each carrying 10 people.

$$26 \times 10 = 26 \text{ Tens} = 20 \text{ Tens} + 6 \text{ Tens} = 200 + 60 = \mathbf{260}$$

Answer: 260 people can travel.

Concept: $n \times 10$ = place a zero at the end of $n$.

a) $21 \times 10 = \mathbf{210}$

b) $42 \times 10 = \mathbf{420}$

c) $65 \times 10 = \mathbf{650}$

d) $38 \times 10 = \mathbf{380}$

e) $53 \times 10 = \mathbf{530}$

f) $87 \times 10 = \mathbf{870}$

Given: 12 buses, each seating 20 people.

Method: $12 \times 20 = 12 \times 2 \text{ Tens} = 24 \text{ Tens}$

$$12 \times 20 = 24 \times 10 = \mathbf{240}$$

Answer: 240 people can be seated in 12 buses.

Method: $a \times (b \times 10) = (a \times b) \times 10$

$$24 \times 40 = 24 \times 4 \times 10 = 96 \times 10 = \mathbf{960}$$

$$13 \times 30 = 13 \times 3 \times 10 = 39 \times 10 = \mathbf{390}$$

$$50 \times 60 = 5 \times 6 \times 100 = 30 \times 100 = \mathbf{3000}$$

$$43 \times 60 = 43 \times 6 \times 10 = 258 \times 10 = \mathbf{2580}$$

$$70 \times 80 = 7 \times 8 \times 100 = 56 \times 100 = \mathbf{5600}$$

Given: 15 coaches, each seating 14 children.

Method – Split using a grid (15 = 10 + 5, 14 = 10 + 4):

| × | 10 | 4 |
|---|---|---|
| 10 | $10 \times 10 = 100$ | $10 \times 4 = 40$ |
| 5 | $5 \times 10 = 50$ | $5 \times 4 = 20$ |

$$15 \times 14 = 100 + 40 + 50 + 20 = \mathbf{210}$$

Answer: 210 children can be seated in the toy train.

Given: 324 children, each coach seats 14.

Method – Repeated subtraction / long division:

$$14 \times 10 = 140$$
$$14 \times 20 = 280$$
$$14 \times 23 = 322$$

| No. of children | No. of coaches | Remaining |
|---|---|---|
| — | — | 324 |
| 140 | 10 | 184 |
| 140 | 10 | 44 |
| 14 | 1 | 30 |
| 28 | 2 | 2 |

$$324 \div 14 = 23 \text{ remainder } 2$$

Total coaches needed = 23 (for 322 children) + 1 extra coach for the remaining 2 children.

Total coaches = 24

Remainder = 2 (2 children still need a seat, so one more coach is required even though it won't be full).

**a) $25 \times 34$**

Split: $25 \times 34 = 25 \times 30 + 25 \times 4$
$$= 750 + 100 = \mathbf{850}$$

**b) $16 \times 43$**

Split: $16 \times 43 = 16 \times 40 + 16 \times 3$
$$= 640 + 48 = \mathbf{688}$$

**c) $68 \times 12$**

Split: $68 \times 12 = 68 \times 10 + 68 \times 2$
$$= 680 + 136 = \mathbf{816}$$

**d) $39 \times 13$**

Split: $39 \times 13 = 39 \times 10 + 39 \times 3$
$$= 390 + 117 = \mathbf{507}$$

**e) $125 \div 15$**

$15 \times 8 = 120$; $125 - 120 = 5$
$$125 \div 15 = \mathbf{8} \text{ remainder } \mathbf{5}$$

**f) $94 \div 11$**

$11 \times 8 = 88$; $94 - 88 = 6$
$$94 \div 11 = \mathbf{8} \text{ remainder } \mathbf{6}$$

**g) $440 \div 22$**

$22 \times 20 = 440$; $440 - 440 = 0$
$$440 \div 22 = \mathbf{20} \text{ remainder } \mathbf{0}$$

**h) $508 \div 18$**

$18 \times 28 = 504$; $508 - 504 = 4$
$$508 \div 18 = \mathbf{28} \text{ remainder } \mathbf{4}$$

Concept: $n \times 100 = n$ Hundreds.

$$3 \times 100 = \mathbf{3} \text{ Hundreds} = \mathbf{300}$$

$$5 \times 100 = \mathbf{5 \text{ Hundreds}} = \mathbf{500}$$

$$8 \times 100 = \mathbf{8 \text{ Hundreds}} = \mathbf{800}$$

Given: $10 \times 100 = 10$ Hundreds

$$10 \times 100 = \mathbf{1000}$$

When we put 10 Hundreds together, we get 1 Thousand = 1000.