Number Play

Since the actual figure is not visible, the method is as follows:

Given: Each child counts how many children standing in front of them (i.e., between them and the front of the line) are taller than them.

Step 1: Start from the child at the front of the line — they have no one in front, so they always say 0.

Step 2: For each subsequent child, count only those children who are positioned in front of them AND are taller than them.

Step 3: Write that count as the child's number.

Apply this rule to the given figure to obtain the sequence.

Recall the rule: Each child says the number of children in front of them who are taller.

(a) 0, 1, 1, 2, 4, 1, 5

We need to construct a height arrangement of 7 children (positions 1 to 7 from front to back) such that:
- Child 1 (front): 0 taller in front → always 0. ✓
- Child 2: 1 taller in front (Child 1 is taller than Child 2).
- Child 3: 1 taller in front.
- Child 4: 2 taller in front.
- Child 5: 4 taller in front (all 4 children in front are taller).
- Child 6: 1 taller in front.
- Child 7: 5 taller in front.

One valid height arrangement (tallest height = 7 units for reference):
Positions (front to back): heights could be arranged as 6, 3, 5, 4, 1, 7, 2.

Verification:
- Pos 1 (h=6): 0 in front → says 0 ✓
- Pos 2 (h=3): h=6 > 3, count=1 → says 1 ✓
- Pos 3 (h=5): h=6 > 5, h=3 < 5, count=1 → says 1 ✓
- Pos 4 (h=4): h=6>4, h=3<4, h=5>4, count=2 → says 2 ✓
- Pos 5 (h=1): h=6,3,5,4 all >1, count=4 → says 4 ✓
- Pos 6 (h=7): h=6<7, h=3<7, h=5<7, h=4<7, h=1<7, count=0...

Let us try another arrangement. The key insight: the number a child says equals the number of inversions at that position.

A valid arrangement for (a): Place children with heights such that:
Front→Back heights: 5, 3, 4, 2, 1, 6, 0 (using relative ranks 0–6).
- Pos 1 (h=5): 0 → 0 ✓
- Pos 2 (h=3): 5>3 → 1 ✓
- Pos 3 (h=4): 5>4, 3<4 → 1 ✓
- Pos 4 (h=2): 5>2, 3>2, 4>2 → but need 2, so only 2 taller. Try h=2 with only 2 taller in front: 5>2, 3>2 but 4>2 gives 3. Adjust.

Let heights (front to back) be: 4, 2, 3, 1, 0, 5, 6 (values 0–6).
- Pos 1 (4): 0 ✓
- Pos 2 (2): 4>2 → 1 ✓
- Pos 3 (3): 4>3, 2<3 → 1 ✓
- Pos 4 (1): 4>1, 2>1, 3>1 → 3 (need 2) ✗

Try: 3, 1, 2, 0, 4, 5, 6 — but last child (6) would have 5 taller in front only if all 5 before are taller, but 4,5 are taller and 3,1,2,0 are not all taller than 6...

Best approach: Use the Lehmer code / factorial number system interpretation.

For sequence 0,1,1,2,4,1,5: One valid arrangement (heights as ranks 0–6 from shortest to tallest):

Front→Back: rank 5, rank 3, rank 4, rank 2, rank 0, rank 6, rank 1
(i.e., 2nd tallest, 4th tallest, 3rd tallest, 5th tallest, shortest, tallest, 6th tallest)

- Pos 1 (rank 5): 0 in front → 0 ✓
- Pos 2 (rank 3): rank 5 > rank 3 → 1 ✓
- Pos 3 (rank 4): rank 5 > rank 4; rank 3 < rank 4 → 1 ✓
- Pos 4 (rank 2): rank 5>2, rank 3>2, rank 4>2 → 3 (need 2) ✗

Try front→back: rank 4, rank 2, rank 3, rank 0, rank 1, rank 5, rank 6 — child 5 (rank 1): ranks 4,2,3,0 in front; 4>1, 2>1, 3>1, 0<1 → 3 taller (need 4). ✗

Correct approach using Lehmer code directly:
For sequence $c_1,c_2,\ldots,c_n$, place the child at position $i$ such that exactly $c_i$ of the children already placed (in front) are taller.

For (a) 0,1,1,2,4,1,5 with 7 children (heights 1–7):
- Pos 1: $c=0$ → place the child who will have 0 taller in front. Start with available heights {1,2,3,4,5,6,7}. $c_1=0$: pick any height for pos 1 (no one in front). Say we pick height 6.
- Pos 2: $c=1$ → 1 of the children in front (just pos 1, height 6) is taller. So pos 2's height < 6. Pick height 4 (6>4, count=1). ✓
- Pos 3: $c=1$ → 1 of {6,4} is taller. So pos 3's height is between 4 and 6: pick height 5 (6>5, 4<5, count=1). ✓
- Pos 4: $c=2$ → 2 of {6,4,5} are taller. Heights taller than pos 4: need exactly 2. So pos 4 < 5 and pos 4 < 6 but pos 4 > 4: not possible since 4<pos4<5 has no integer. Try pos 4 < 4: then all 3 are taller (count=3≠2). Try pos 4 between 4 and 5: no integer. Hmm.

Let me restart with a cleaner method. Assign heights 1–7.

For (a) 0,1,1,2,4,1,5:
Build from back to front using the sequence in reverse: 5,1,4,2,1,1,0.
Available slots (sorted heights): 1,2,3,4,5,6,7.
- Last child (pos 7), $c_7=5$: 5 children in front are taller → this child is the 6th tallest among those in front. Since pos 7 is last, "in front" means positions 1–6. Among 7 children, 5 in front are taller means this child has rank 2 from bottom among all 7 → height = 2.
- Pos 6, $c_6=1$: 1 child in front (positions 1–5) is taller → among remaining heights {1,3,4,5,6,7}, this child has 1 taller in front of them among positions 1–5. This child's height should be the 2nd from top of remaining = height 6 (so only 1 person, height 7, will be taller). Assign height 6.
- Pos 5, $c_5=4$: 4 children in front (pos 1–4) are taller → among remaining {1,3,4,5,7}, this child is 5th from top = height 1.
- Pos 4, $c_4=2$: 2 of pos 1–3 are taller → among remaining {3,4,5,7}, this child is 3rd from top = height 4.
- Pos 3, $c_3=1$: 1 of pos 1–2 is taller → among remaining {3,5,7}, this child is 2nd from top = height 5.
- Pos 2, $c_2=1$: 1 of pos 1 is taller → among remaining {3,7}, this child is 2nd from top = height 3 (so height 7 at pos 1 is taller). Assign height 3.
- Pos 1, $c_1=0$: remaining height = 7. ✓ (0 in front)

Final arrangement for (a): Front→Back heights: 7, 3, 5, 4, 1, 6, 2

Verification:
- Pos 1 (7): 0 taller in front → 0 ✓
- Pos 2 (3): 7>3 → 1 ✓
- Pos 3 (5): 7>5, 3<5 → 1 ✓
- Pos 4 (4): 7>4, 3<4, 5>4 → 2 ✓
- Pos 5 (1): 7>1, 3>1, 5>1, 4>1 → 4 ✓
- Pos 6 (6): 7>6, 3<6, 5<6, 4<6, 1<6 → 1 ✓
- Pos 7 (2): 7>2, 3>2, 5>2, 4>2, 1<2, 6>2 → 5 ✓

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(b) 0, 0, 0, 0, 0, 0, 0

Every child says 0 → no child in front of any child is taller → each successive child is taller than all before them.

Arrangement: Heights in strictly increasing order from front to back.
Example: Front→Back: 1, 2, 3, 4, 5, 6, 7 (shortest at front, tallest at back).

Verification: Each child has 0 taller children in front (all in front are shorter). ✓

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(c) 0, 1, 2, 3, 4, 5, 6

Each child says a number equal to their position minus 1 → every child in front of them is taller.

Arrangement: Heights in strictly decreasing order from front to back.
Example: Front→Back: 7, 6, 5, 4, 3, 2, 1 (tallest at front, shortest at back).

Verification:
- Pos 1 (7): 0 taller → 0 ✓
- Pos 2 (6): 7>6 → 1 ✓
- Pos 3 (5): 7>5, 6>5 → 2 ✓
- ... and so on. ✓

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(d) 0, 1, 0, 1, 0, 1, 0

Alternating 0 and 1. Children at odd positions (1,3,5,7) say 0; children at even positions (2,4,6) say 1.

Using the same back-to-front method with heights 1–7:
- Pos 7, $c=0$: 0 taller in front among pos 1–6 → this child is tallest among all 7 → height 7.
- Pos 6, $c=1$: 1 taller in front → among remaining {1,2,3,4,5,6}, 2nd from top → height 5.
- Pos 5, $c=0$: 0 taller in front → tallest of remaining {1,2,3,4,6} → height 6.
- Pos 4, $c=1$: 1 taller in front → among remaining {1,2,3,4}, 2nd from top → height 3.
- Pos 3, $c=0$: tallest of remaining {1,2,4} → height 4.
- Pos 2, $c=1$: 1 taller in front → among remaining {1,2}, 2nd from top → height 1.
- Pos 1, $c=0$: remaining → height 2.

Arrangement: Front→Back: 2, 1, 4, 3, 6, 5, 7

Verification:
- Pos 1 (2): 0 → 0 ✓
- Pos 2 (1): 2>1 → 1 ✓
- Pos 3 (4): 2<4, 1<4 → 0 ✓
- Pos 4 (3): 2<3, 1<3, 4>3 → 1 ✓
- Pos 5 (6): 2<6,1<6,4<6,3<6 → 0 ✓
- Pos 6 (5): 2<5,1<5,4<5,3<5,6>5 → 1 ✓
- Pos 7 (7): all 6 in front < 7 → 0 ✓

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(e) 0, 1, 1, 1, 1, 1, 1

First child says 0; all others say 1 → exactly 1 child in front of each of them is taller.

Using back-to-front:
- Pos 7, $c=1$: 1 taller in front → 2nd from top of all 7 → height 6.
- Pos 6, $c=1$: 1 taller in front among pos 1–5 → among remaining {1,2,3,4,5,7}, 2nd from top → height 5.
- Pos 5, $c=1$: among remaining {1,2,3,4,7}, 2nd from top → height 4.
- Pos 4, $c=1$: among remaining {1,2,3,7}, 2nd from top → height 3.
- Pos 3, $c=1$: among remaining {1,2,7}, 2nd from top → height 2.
- Pos 2, $c=1$: among remaining {1,7}, 2nd from top → height 1.
- Pos 1, $c=0$: remaining → height 7.

Arrangement: Front→Back: 7, 1, 2, 3, 4, 5, 6

Verification:
- Pos 1 (7): 0 → 0 ✓
- Pos 2 (1): 7>1 → 1 ✓
- Pos 3 (2): 7>2, 1<2 → 1 ✓
- Pos 4 (3): 7>3, 1<3, 2<3 → 1 ✓
- Pos 5 (4): 7>4, 1<4, 2<4, 3<4 → 1 ✓
- Pos 6 (5): 7>5, rest <5 → 1 ✓
- Pos 7 (6): 7>6, rest <6 → 1 ✓

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(f) 0, 0, 0, 3, 3, 3, 3

First three say 0; last four say 3.

Using back-to-front:
- Pos 7, $c=3$: 3 taller in front → 4th from top → height 4.
- Pos 6, $c=3$: among remaining {1,2,3,5,6,7}, 4th from top → height 3.
- Pos 5, $c=3$: among remaining {1,2,5,6,7}, 4th from top → height 2.
- Pos 4, $c=3$: among remaining {1,5,6,7}, 4th from top → height 1.
- Pos 3, $c=0$: tallest of remaining {5,6,7} → height 7.
- Pos 2, $c=0$: tallest of remaining {5,6} → height 6.
- Pos 1, $c=0$: remaining → height 5.

Arrangement: Front→Back: 5, 6, 7, 1, 2, 3, 4

Verification:
- Pos 1 (5): 0 → 0 ✓
- Pos 2 (6): 5<6 → 0 ✓
- Pos 3 (7): 5<7, 6<7 → 0 ✓
- Pos 4 (1): 5>1, 6>1, 7>1 → 3 ✓
- Pos 5 (2): 5>2, 6>2, 7>2, 1<2 → 3 ✓
- Pos 6 (3): 5>3, 6>3, 7>3, 1<3, 2<3 → 3 ✓
- Pos 7 (4): 5>4, 6>4, 7>4, 1<4, 2<4, 3<4 → 3 ✓

(a) If a person says '0', then they are the tallest in the group.

Only Sometimes True.

Reasoning: A person says '0' if no one in front of them is taller. This means they are taller than everyone in front of them, but someone behind them could be taller. For example, in arrangement (b) [0,0,0,0,0,0,0], every person says 0 but only the last person is the tallest overall. So saying '0' does not necessarily mean being the tallest in the whole group.

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(b) If a person is the tallest, then their number is '0'.

Always True.

Reasoning: The tallest person is taller than everyone else. Therefore, no one in front of them can be taller than them. So the tallest person always says 0, regardless of their position in the line.

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(c) The first person's number is '0'.

Always True.

Reasoning: The first person (at the front of the line) has no one standing in front of them. So the count of taller people in front is always 0. The first person always says 0.

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(d) If a person is not first or last in line, then they cannot say '0'.

Only Sometimes True (i.e., it is NOT always true — they CAN say '0').

Reasoning: A middle person says '0' if they are taller than everyone in front of them. This is possible — for example, in arrangement (b) [0,0,0,0,0,0,0], every person (including middle ones) says 0. So middle persons CAN say 0. The statement is Never True as a restriction — middle persons can indeed say 0.

*(More precisely: the statement "cannot say 0" is Never True — middle persons can always potentially say 0.)*

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(e) The person who calls out the largest number is the shortest.

Only Sometimes True.

Reasoning: The person who calls the largest number has the most people taller than them in front. This often means they are quite short, but it is not guaranteed they are the absolute shortest. For example, if the shortest person is at the front, they say 0, and someone slightly taller but with many tall people in front could say the largest number. So it is not always the shortest person.

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(f) What is the largest number possible in a group of 8 people?

The largest number possible is 7.

Reasoning: In a group of 8 people, the last person (position 8) has 7 people in front of them. If all 7 people in front are taller, the last person says 7. This is the maximum possible since there are only 7 people who can stand in front of the 8th person.

$$\text{Maximum number} = 8 - 1 = 7$$

Key rules:
- even + even = even
- odd + odd = even
- even + odd = odd

(a) Sum of 2 even numbers and 2 odd numbers

$$\text{even} + \text{even} = \text{even}$$
$$\text{odd} + \text{odd} = \text{even}$$
$$\text{even} + \text{even} = \text{even}$$

Parity: Even

(b) Sum of 2 odd numbers and 3 even numbers

$$\text{odd} + \text{odd} = \text{even}$$
$$\text{even} + \text{even} + \text{even} = \text{even}$$
$$\text{even} + \text{even} = \text{even}$$

Parity: Even

(c) Sum of 5 even numbers

Sum of any number of even numbers is always even (even + even = even, repeated).

Parity: Even

(d) Sum of 8 odd numbers

Pair the 8 odd numbers: $4$ pairs, each pair (odd + odd) = even.
Sum of 4 even numbers = even.

Parity: Even

*General rule:* Sum of an even number of odd numbers is always even.

Given:
- Number of ₹1 coins = odd → contribution = odd × 1 = odd
- Number of ₹5 coins = odd → contribution = odd × 5 = odd (since odd × odd = odd)
- Number of ₹10 coins = even → contribution = even × 10 = even

Total = odd + odd + even

$$\text{odd} + \text{odd} = \text{even}$$
$$\text{even} + \text{even} = \text{even}$$

So the total must be even.

But ₹205 is odd.

Conclusion: Yes, Lakpa made a mistake. The total of an odd number of ₹1 coins, an odd number of ₹5 coins, and an even number of ₹10 coins must always be even. Since 205 is odd, it is impossible. He made an error in his calculation.

We use the fact that subtraction has the same parity rules as addition (since $a - b = a + (-b)$ and $-b$ has the same parity as $b$).

(d) even − even

Example: $6 - 4 = 2$ (even); $8 - 2 = 6$ (even).

$$\text{even} - \text{even} = \boxed{\text{even}}$$

(e) odd − odd

Example: $7 - 3 = 4$ (even); $9 - 5 = 4$ (even).

$$\text{odd} - \text{odd} = \boxed{\text{even}}$$

(f) even − odd

Example: $6 - 3 = 3$ (odd); $8 - 5 = 3$ (odd).

$$\text{even} - \text{odd} = \boxed{\text{odd}}$$

(g) odd − even

Example: $7 - 4 = 3$ (odd); $9 - 2 = 7$ (odd).

$$\text{odd} - \text{even} = \boxed{\text{odd}}$$

Given: We use numbers 1 to 9, each exactly once, in a 3×3 grid.

Key facts:
- The magic sum for numbers 1–9 is $\dfrac{9 \times 10}{2 \div 3} = \dfrac{45}{3} = 15$.
- The centre must be 5 (the middle value).
- There is essentially only 1 fundamental magic square using 1–9 (up to rotation and reflection).

However, counting rotations (4) and reflections (2 per rotation), there are:
$$4 \times 2 = 8 \text{ different magic squares}$$

Answer: There are 8 different magic squares using numbers 1–9 (all are rotations and reflections of one basic arrangement).

Strategy: Add 1 to each number in the standard 1–9 magic square.

Standard 1–9 magic square (one version):
$$\begin{pmatrix} 2 & 7 & 6 \\ 9 & 5 & 1 \\ 4 & 3 & 8 \end{pmatrix}$$

Add 1 to each entry:
$$\begin{pmatrix} 3 & 8 & 7 \\ 10 & 6 & 2 \\ 5 & 4 & 9 \end{pmatrix}$$

Verification (magic sum):
- Row 1: $3+8+7=18$ ✓
- Row 2: $10+6+2=18$ ✓
- Row 3: $5+4+9=18$ ✓
- Col 1: $3+10+5=18$ ✓
- Col 2: $8+6+4=18$ ✓
- Col 3: $7+2+9=18$ ✓
- Diagonal: $3+6+9=18$ ✓; $7+6+5=18$ ✓

Magic sum = 18 (which is $15 + 3 = 15 + 3 \times 1$, since we added 1 to each of 3 numbers per row).

Comparison: The structure (pattern of differences between numbers) is identical to the 1–9 magic square. Only the values are shifted by 1. The magic sum increases by 3 (one for each cell in a row/column/diagonal).

Starting magic square (1–9):
$$\begin{pmatrix} 2 & 7 & 6 \\ 9 & 5 & 1 \\ 4 & 3 & 8 \end{pmatrix}, \quad \text{Magic sum} = 15$$

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(a) Increase each number by 1:

$$\begin{pmatrix} 3 & 8 & 7 \\ 10 & 6 & 2 \\ 5 & 4 & 9 \end{pmatrix}$$

Each row/column/diagonal sum increases by $1+1+1 = 3$.

**New magic sum = $15 + 3 = 18$.

Yes, it is still a magic square. Adding a constant to every cell preserves the magic property because every line sum increases by the same amount.

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(b) Double each number:**

$$\begin{pmatrix} 4 & 14 & 12 \\ 18 & 10 & 2 \\ 8 & 6 & 16 \end{pmatrix}$$

Each row/column/diagonal sum is doubled.

**New magic sum = $15 \times 2 = 30$.

Yes, it is still a magic square.** Multiplying every cell by a constant preserves the magic property because every line sum is multiplied by the same factor.

The following operations on a magic square always yield another magic square:

1. Adding a constant to every cell: The magic sum increases by $3 \times$ (constant).

2. Multiplying every cell by a constant (non-zero): The magic sum is multiplied by that constant.

3. Subtracting a constant from every cell: The magic sum decreases by $3 \times$ (constant).

4. Rotating the square by 90°, 180°, or 270°: Produces a different valid magic square.

5. Reflecting the square (horizontally, vertically, or along a diagonal): Produces another valid magic square.

6. Combining the above: e.g., multiply by $k$ then add $c$ — still a magic square.

Key principle: Any operation that treats all cells uniformly (same arithmetic operation applied to every cell) preserves the magic property.

Method: Start with the standard 1–9 magic square and shift all values.

If the desired set of 9 consecutive numbers starts at $n$ (i.e., $n, n+1, \ldots, n+8$), then:

Step 1: Take the standard 1–9 magic square.

Step 2: Add $(n-1)$ to every cell.

This gives a magic square using numbers $n$ to $n+8$.

Magic sum for numbers $n$ to $n+8$:
$$\text{Sum of all 9 numbers} = 9n + (0+1+2+\cdots+8) = 9n + 36$$
$$\text{Magic sum} = \frac{9n+36}{3} = 3n + 12$$

Example for 3–11 ($n=3$):
- Add 2 to each cell of the standard magic square.
- Magic sum $= 3(3)+12 = 21$.

$$\begin{pmatrix} 4 & 9 & 8 \\ 11 & 7 & 3 \\ 6 & 5 & 10 \end{pmatrix}$$

Verification: $4+9+8=21$, $11+7+3=21$, $6+5+10=21$ ✓

Given: Centre number $m = 25$.

The generalised 3×3 magic square (based on the standard 1–9 structure where centre = 5) has the form:

$$\begin{pmatrix} m-3 & m+2 & m+1 \\ m+4 & m & m-4 \\ m-1 & m-2 & m+3 \end{pmatrix}$$

Substituting $m = 25$:

$$\begin{pmatrix} 22 & 27 & 26 \\ 29 & 25 & 21 \\ 24 & 23 & 28 \end{pmatrix}$$

Magic sum $= 3m = 3 \times 25 = 75$.

Verification:
- Row 1: $22+27+26=75$ ✓
- Row 2: $29+25+21=75$ ✓
- Row 3: $24+23+28=75$ ✓
- Diagonal: $22+25+28=75$ ✓; $26+25+24=75$ ✓

Given: In the generalised magic square, the centre is $m$ and the other cells are expressed as $m \pm d$ for various values of $d$.

For the standard generalised form, each row, column, and diagonal contains three terms that sum to:

$$\text{Magic sum} = m + (m-d) + (m+d) = 3m$$

(The $+d$ and $-d$ cancel out, leaving $3m$.)

**The magic sum = $3m$**, where $m$ is the centre number.

This is because each line passes through the centre or is balanced symmetrically around it.

**Generalised magic square with centre $m$ and magic sum $3m$.

(a) Adding 1 to every term:**

New centre = $m + 1$.
Every cell becomes $(\text{original cell}) + 1$.
New magic sum $= 3m + 3 = 3(m+1)$.

The result is still a magic square with centre $(m+1)$ and magic sum $3(m+1)$.

(b) Doubling every term:

New centre = $2m$.
Every cell becomes $2 \times (\text{original cell})$.
New magic sum $= 2 \times 3m = 6m = 3(2m)$.

The result is still a magic square with centre $2m$ and magic sum $3(2m)$.