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):
$$\begin{pmatrix} 2 &amp; 7 &amp; 6 \\ 9 &amp; 5 &amp; 1 \\ 4 &amp; 3 &amp; 8 \end{pmatrix}$$

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

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):
$$\begin{pmatrix} 2 &amp; 7 &amp; 6 \\ 9 &amp; 5 &amp; 1 \\ 4 &amp; 3 &amp; 8 \end{pmatrix}, \quad \text{Magic sum} = 15$$

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(a) Increase each number by 1:

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

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:

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

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 $= 3(3)+12 = 21$.

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

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:

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

Substituting $m = 25$:

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

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 $3(2m)$.

Given: Magic sum $= 60$.

Using the formula: Magic sum $= 3m$, so:
$$3m = 60 \implies m = 20$$

The centre number must be 20.

Using the generalised form:
$$\begin{pmatrix} m-3 &amp; m+2 &amp; m+1 \\ m+4 &amp; m &amp; m-4 \\ m-1 &amp; m-2 &amp; m+3 \end{pmatrix} = \begin{pmatrix} 17 &amp; 22 &amp; 21 \\ 24 &amp; 20 &amp; 16 \\ 19 &amp; 18 &amp; 23 \end{pmatrix}$$

Verification:
- Row 1: $17+22+21=60$ ✓
- Row 2: $24+20+16=60$ ✓
- Row 3: $19+18+23=60$ ✓
- Diagonal: $17+20+23=60$ ✓; $21+20+19=60$ ✓

Yes, it is possible.

The key requirement for a 3×3 magic square is not that the numbers be consecutive, but that they can be arranged so that every row, column, and diagonal has the same sum.

Example: Take the standard 1–9 magic square and multiply every entry by 2:
$$\begin{pmatrix} 4 &amp; 14 &amp; 12 \\ 18 &amp; 10 &amp; 2 \\ 8 &amp; 6 &amp; 16 \end{pmatrix}$$
These are 9 non-consecutive even numbers, and the magic sum is 30. ✓

Another example: Use numbers 1, 2, 3, 5, 7, 8, 9, 11, 12 (non-consecutive) — with careful arrangement, a magic square can be formed.

Conclusion: Magic squares can be formed with non-consecutive numbers. The generalised form $\begin{pmatrix} m-a &amp; m+b &amp; m+c \\ \ldots \end{pmatrix}$ works for any set of numbers that are symmetrically distributed around a centre value $m$.

Given: Bulb starts ON. Switch is toggled 77 times.

Concept: Each toggle changes the state:
- Odd number of toggles → state changes (ON becomes OFF, or OFF becomes ON).
- Even number of toggles → state returns to original.

Working:
77 is an odd number.

Starting state: ON
After 1 toggle: OFF
After 2 toggles: ON
...
After 77 toggles (odd): OFF

Answer: The bulb will be OFF.

Reason: 77 is odd, so the bulb ends in the opposite state from where it started.

Given: 50 loose sheets, each printed on both sides.

Key observation: In a book, each sheet has two consecutive page numbers: one odd and one even. For example, a sheet has pages $2k-1$ and $2k$ for some integer $k$.

Sum of page numbers on one sheet:
$$(2k-1) + 2k = 4k - 1$$

This is always odd (since $4k$ is even and $4k-1$ is odd).

Sum of page numbers of 50 sheets:
$$= \sum_{i=1}^{50} (4k_i - 1) = 4\left(\sum k_i\right) - 50$$

This equals $4 \times (\text{integer}) - 50$.

$4 \times (\text{integer})$ is divisible by 4.
$50 = 4 \times 12 + 2$, so $4(\text{integer}) - 50 \equiv 0 - 2 \equiv 2 \pmod{4}$.

So the sum of page numbers of 50 sheets $\equiv 2 \pmod{4}$, meaning it leaves remainder 2 when divided by 4.

Check 6000: $6000 \div 4 = 1500$ exactly, so $6000 \equiv 0 \pmod{4}$.

Since the sum must be $\equiv 2 \pmod{4}$ but 6000 $\equiv 0 \pmod{4}$:

No, the sum cannot be 6000. Liswini made a mistake. The sum of page numbers of any number of complete sheets always leaves remainder 2 when divided by 4 (for an even number of sheets), but 6000 is divisible by 4.

Note: The exact parity labels (e/o) for each row and column are in the figure which is not visible. We solve the general method.

Method:

Let the 2×3 grid be:
$$\begin{pmatrix} a &amp; b &amp; c \\ d &amp; e &amp; f \end{pmatrix}$$

We need to place 3 odd (o) and 3 even (e) numbers such that:
- Row sums have the given parities.
- Column sums have the given parities.

Parity rules:
- o + o = e; e + e = e; o + e = o
- For a row of 3: o+o+o = o; e+e+e = e; o+o+e = e; o+e+e = o

One valid arrangement (assuming row 1 sum = even, row 2 sum = odd, col 1 = odd, col 2 = even, col 3 = odd — a typical example):

$$\begin{pmatrix} o &amp; e &amp; o \\ e &amp; e &amp; o \end{pmatrix}$$

- Row 1: o+e+o = e ✓
- Row 2: e+e+o = o ✓
- Col 1: o+e = o ✓
- Col 2: e+e = e ✓
- Col 3: o+o = e ✓

Count: 3 odd (positions a, c, f) and 3 even (positions b, d, e). ✓

Specific numbers example: Place odd numbers 1, 3, 5 and even numbers 2, 4, 6:
$$\begin{pmatrix} 1 &amp; 4 &amp; 3 \\ 2 &amp; 6 &amp; 5 \end{pmatrix}$$

Apply the same parity pattern as above to match the given figure's circles.

Given: Magic sum = 0, not all numbers zero.

Method: Use the generalised magic square with $m = 0$ (centre = 0).

Magic sum $= 3m = 3 \times 0 = 0$ ✓

Using the standard structure (based on 1–9 shifted by $-5$, i.e., subtract 5 from each cell of the standard magic square):

Standard magic square:
$$\begin{pmatrix} 2 &amp; 7 &amp; 6 \\ 9 &amp; 5 &amp; 1 \\ 4 &amp; 3 &amp; 8 \end{pmatrix}$$

Subtract 5 from each cell:
$$\begin{pmatrix} -3 &amp; 2 &amp; 1 \\ 4 &amp; 0 &amp; -4 \\ -1 &amp; -2 &amp; 3 \end{pmatrix}$$

Verification:
- Row 1: $-3+2+1=0$ ✓
- Row 2: $4+0+(-4)=0$ ✓
- Row 3: $-1+(-2)+3=0$ ✓
- Col 1: $-3+4+(-1)=0$ ✓
- Col 2: $2+0+(-2)=0$ ✓
- Col 3: $1+(-4)+3=0$ ✓
- Diagonal: $-3+0+3=0$ ✓; $1+0+(-1)=0$ ✓

Numbers used: $-4,-3,-2,-1,0,1,2,3,4$ — not all zero. ✓

(a) Sum of an odd number of even numbers:

Even + even = even (always). Adding any number of even numbers always gives an even result.

Answer: Even

(b) Sum of an even number of odd numbers:

Pair up the odd numbers: each pair (odd + odd) = even. An even number of odd numbers gives an even number of pairs, and sum of even numbers = even.

Answer: Even

(c) Sum of an even number of even numbers:

Sum of any number of even numbers is always even.

Answer: Even

(d) Sum of an odd number of odd numbers:

Pair up as many as possible: $(n-1)$ odd numbers (even count) → even sum. One odd number remains. Even + odd = odd.

Answer: Odd

Given: Sum = $1 + 2 + 3 + \cdots + 100$.

Formula:
$$S = \frac{100 \times 101}{2} = 50 \times 101 = 5050$$

Parity check: $5050$ is even (ends in 0).

Alternative parity reasoning:
Numbers 1 to 100 include 50 odd numbers and 50 even numbers.
- Sum of 50 odd numbers: 50 is even → sum is even (from rule: even count of odd numbers → even sum).
- Sum of 50 even numbers: always even.
- Total: even + even = even.

Answer: The parity of the sum is Even ($5050$).

Given: The Virahāṅka (Fibonacci) sequence rule: each term = sum of the two preceding terms.

Given two consecutive terms: $987$ and $1597$.

Next 2 numbers:
$$\text{Next} = 987 + 1597 = 2584$$
$$\text{One after} = 1597 + 2584 = 4181$$

Previous 2 numbers:
If $a, b$ are consecutive terms with $b = 1597$ and the term before $b$ is $987$:
$$\text{Term before } 987: \quad 1597 - 987 = 610$$
$$\text{Term before } 610: \quad 987 - 610 = 377$$

Answer:
- Next 2 numbers: 2584, 4181
- Previous 2 numbers: 610, 377

Given: 8-step staircase; each move is 1 or 2 steps.

Let $f(n)$ = number of ways to climb $n$ steps.

Recurrence: To reach step $n$, Angaan either:
- Came from step $n-1$ (took 1 step), or
- Came from step $n-2$ (took 2 steps).

So $f(n) = f(n-1) + f(n-2)$ — this is the Virahāṅka/Fibonacci recurrence!

Base cases:
$$f(1) = 1 \quad (\text{only: 1})$$
$$f(2) = 2 \quad (\text{1+1 or 2})$$

Building up:
$$f(3) = f(2) + f(1) = 2 + 1 = 3$$
$$f(4) = f(3) + f(2) = 3 + 2 = 5$$
$$f(5) = f(4) + f(3) = 5 + 3 = 8$$
$$f(6) = f(5) + f(4) = 8 + 5 = 13$$
$$f(7) = f(6) + f(5) = 13 + 8 = 21$$
$$f(8) = f(7) + f(6) = 21 + 13 = 34$$

Answer: Angaan can climb the 8-step staircase in $\boxed{34}$ different ways.

Given: Virahāṅka sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

Parity pattern of the Virahāṅka sequence:

Write O for odd, E for even:
$$\text{O, O, E, O, O, E, O, O, E, O, O, E, } \ldots$$

The pattern O, O, E repeats every 3 terms.

Finding the 20th term's parity:
$$20 = 3 \times 6 + 2$$

So the 20th term has the same parity as the 2nd term in the pattern (O, O, E).

The 2nd position in the pattern is O (odd).

Answer: The 20th term of the Virahāṅka sequence is Odd.

*(Verification: The 20th Fibonacci/Virahāṅka number is 6765, which is indeed odd.)* ✓

(a) The expression $4m - 1$ always gives odd numbers.

$4m$ is always even (multiple of 4). Even $- 1$ = odd.

True. $4m - 1$ is always odd for any integer $m$.

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(b) All even numbers can be expressed as $6j - 4$.

$6j - 4 = 2(3j - 2)$, which is always even. ✓ (It gives even numbers.)

But does it give ALL even numbers?
- $j=1$: $6-4=2$ ✓
- $j=2$: $12-4=8$ ✓
- $j=3$: $18-4=14$ ✓

The values are $2, 8, 14, 20, \ldots$ — these are even numbers of the form $6j-4$, which differ by 6. Even numbers like 4, 6, 10, 12, ... are not produced.

False. $6j - 4$ gives only some even numbers, not all.

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(c) Both expressions $2p + 1$ and $2q - 1$ describe all odd numbers.

- $2p+1$: for $p = 0, 1, 2, 3, \ldots$ gives $1, 3, 5, 7, \ldots$ — all positive odd numbers; for negative $p$, gives negative odd numbers too. Covers all odd integers. ✓
- $2q-1$: for $q = 1, 2, 3, \ldots$ gives $1, 3, 5, 7, \ldots$ — same set. ✓

Both expressions generate all odd integers (for integer values of $p$ and $q$).

True.

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(d) The expression $2f + 3$ gives both even and odd numbers.

$2f$ is always even. $2f + 3 = \text{even} + \text{odd} = \text{odd}$.

So $2f + 3$ is always odd, never even.

False.

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Summary: Statements (a) and (c) are true.

Given: Each letter represents a unique digit (0–9).

$\text{UT}$ is a 2-digit number $= 10U + T$
$\text{TA}$ is a 2-digit number $= 10T + A$
$\text{TAT}$ is a 3-digit number $= 100T + 10A + T = 101T + 10A$

Equation:
$$(10U + T) + (10T + A) = 101T + 10A$$
$$10U + 11T + A = 101T + 10A$$
$$10U = 90T + 9A$$
$$10U = 9(10T + A)$$
$$10U = 9 \times \text{TA}$$

So $10U$ must be divisible by 9, meaning $U$ must be such that $10U$ is a multiple of 9.

Since $\gcd(10, 9) = 1$, we need $U$ to be a multiple of 9.

$U$ is a single digit: $U = 0$ or $U = 9$.

Case 1: $U = 0$
$$10 \times 0 = 9 \times \text{TA} \implies \text{TA} = 0$$
But TA is a 2-digit number, so $T \neq 0$. Contradiction. ✗

Case 2: $U = 9$
$$10 \times 9 = 9 \times \text{TA} \implies \text{TA} = \frac{90}{9} = 10$$
So $T = 1$ and $A = 0$.

Verification:
- $\text{UT} = 91$
- $\text{TA} = 10$
- $\text{TAT} = 101$

$$91 + 10 = 101 \checkmark$$

All digits: $U=9, T=1, A=0$ — all different. ✓

Answer: $U = 9,\ T = 1,\ A = 0$
$$91 + 10 = 101$$