At first glance there’s no obvious pattern — the numbers don’t match height order, position, or anything visually obvious. That’s the point of the activity: you’re meant to guess the rule from a few examples, the same way the second picture (a different arrangement, different numbers) gives you a second clue to test your guess against.

Comparing both pictures side by side is the key move: whatever rule explains the numbers must work for both arrangements at once. Checking each child against their neighbours (rather than the group as a whole) reveals it — each child’s number depends only on the taller children standing ahead of them.

1. Label the children 1st to 7th from left to right (the front of the line).
2. For each child, look only at the children before them (to their left) and count how many are taller.
3. The very first child always says 0 — there’s nobody ahead of them to compare with.
4. Go child by child: 1st has nobody ahead → 0. 2nd has one shorter child ahead → 0. 3rd has one taller child among the two ahead → 1. 4th is taller than everyone ahead → 0. 5th has three taller children ahead → 3. 6th is taller than everyone ahead → 0. 7th has three taller children ahead → 3.

Every single card shown is odd (13, 9, 1, 7, 11, 5, 3 are all odd). Picking any 5 of them and adding is the same as adding 5 odd numbers — and as the chapter goes on to prove, 5 odd numbers can never add up to an even number like 30. So it is not possible, no matter which 5 cards you choose — you don’t even need to try combinations to know this.

The sum is always even, no matter how many even numbers you add. Every even number can be arranged fully into pairs with nothing left over, so combining several even numbers just combines complete pairs — the total can still be arranged into pairs with no leftovers, which is exactly what makes a number even.

Here it does matter how many you add. Each odd number is “one pair short of complete” (one dot left over). Add two odd numbers and the two leftover dots pair up with each other — giving a fully-paired (even) total. This is why the sum of two odd numbers is always even.

No. Two of the three odd numbers pair up their leftover dots (as before, giving an even amount), but the third odd number still has its own leftover dot with no partner. So the sum of 3 odd numbers is always odd.

1. (a) 4 odd numbers: the 4 leftover dots pair up into 2 pairs, nothing left over → even.
2. (b) 5 odd numbers: 4 of the leftover dots pair up, 1 stays unpaired → odd.
3. (c) 6 odd numbers: all 6 leftover dots pair up into 3 pairs → even.

1. Being born exactly one year apart means their ages today are two consecutive whole numbers (like 51 and 52).
2. Counting numbers alternate even, odd, even, odd, … so any two consecutive numbers are one even and one odd.
3. An even number plus an odd number can never be arranged fully into pairs — it’s always odd.
4. $112$ is even, but the sum of two consecutive ages must be odd.

1. The number of small squares in an $m \times n$ grid is the product $m \times n$.
2. If either $m$ or $n$ is even, the product is even (an even factor makes the whole product even).
3. If both $m$ and $n$ are odd, the product is odd (odd $\times$ odd $=$ odd).

1. Always even: any expression built entirely around a factor of 2, e.g. $100p$, $48w-2$, $6k+4$ — every term is a multiple of 2 (or a sum of multiples of 2), so the value is always even, whatever whole number $p$, $w$, or $k$ is.
2. Always odd: take an always-even expression and add 1, e.g. $2m+1$, $8a+3$, $6n+5$ — an even number plus 1 (or plus any odd constant) is always odd.
3. Either odd or even (like $3n+4$): any expression with an odd coefficient on the variable, e.g. $5n+2$ or $7n-1$ — when $n$ is even the odd-coefficient term is even (parity depends on $n$), so the overall parity flips depending on whether $n$ is odd or even.

1. Every even number has a factor of 2, so it can be written as $2 \times (\text{some whole number})$. Letting $n=1,2,3,\ldots$, the expression $\boxed{2n}$ lists every even number: $2,4,6,8,\ldots$
2. Every odd number is one less than an even number, so $\boxed{2n-1}$ lists every odd number: $1,3,5,7,\ldots$ (unlike $6k+2$, nothing is skipped).

1. The 100th even number is $2\times100=200$.
2. Comparing term-by-term: Even numbers $2,4,6,8,\ldots$  vs  Odd numbers $1,3,5,7,\ldots$ — at every position, the odd number is exactly 1 less than the even number.
3. So the 100th odd number $= 200-1=199$.
4. In general: the $n$th even number is $2n$, so the $n$th odd number is $2n-1$.

The yellow circled numbers are the sums of the corresponding rows and columns: row 1 ($4+7+5=16$), row 2 ($6+1+2=9$), row 3 ($3+9+8=20$); column 1 ($4+6+3=13$), column 2 ($7+1+9=17$), column 3 ($5+2+8=15$).

Both completed grids are shown above (bold cells are the filled-in answers). Working from the given entries and target sums pins down each remaining cell uniquely — other arrangements of the same numbers within a row can also work as long as the row/column totals match, so try building your own version too!

1. The smallest possible sum of any 3 distinct numbers from 1–9 is $1+2+3=6$.
2. The largest possible sum of any 3 distinct numbers from 1–9 is $9+8+7=24$.
3. So every row/column sum in a valid grid must be between 6 and 24.

Every one of the numbers $1$ to $9$ appears in the grid exactly once, and every number belongs to exactly one row. So adding up all three row-sums counts each of the 9 numbers exactly once — the same as adding $1+2+3+\cdots+9$. That total is $1+2+3+4+5+6+7+8+9=45$. The same reasoning applies to the three column sums.

In a magic square, all 3 row sums are equal, and together they total 45. So each row sum $=45\div3=\bf{15}$. Observation 1: the magic sum of a 1–9 magic square must always be 15 — it can’t be any other number.

1. The centre cell belongs to 4 different lines: its row, its column, and both diagonals. Every one of those 4 lines must sum to 15.
2. If the centre number is $x$, each of those 4 lines needs a pair of the remaining 8 numbers that adds up to $15-x$ — and since all 8 remaining numbers are used exactly once across the 4 lines, we need 4 separate, non-overlapping pairs that all add up to $15-x$.
3. Try $x=9$: pairs from $\{1,\ldots,8\}$ summing to $15-9=6$ are only $(1,5)$ and $(2,4)$ — just 2 pairs, not the 4 we need. 9 fails.
4. Try $x=1$: pairs summing to $15-1=14$ are only $(6,8)$ and $(5,9)$ — again just 2. 1 fails.
5. Checking every other value the same way (2, 3, 4, 6, 7, 8) also comes up short of 4 valid pairs.
6. Try $x=5$: pairs from $\{1,2,3,4,6,7,8,9\}$ summing to $15-5=10$ are $(1,9),(2,8),(3,7),(4,6)$ — exactly 4 pairs, using all 8 numbers with nothing left over!

1. A corner cell also lies on 3 lines (its row, its column, and one diagonal), so it needs 3 different pairs (from the other 8 numbers) that each sum with it to 15.
2. For 1: the only triples containing 1 that sum to 15 are $\{1,5,9\}$ and $\{1,6,8\}$ — just 2 valid triples, not the 3 a corner needs.
3. For 9: the only triples containing 9 that sum to 15 are $\{9,1,5\}$ and $\{9,2,4\}$ — again just 2.

Since $1+5+9=15$ and 5 is fixed at the centre, the line “$1,5,9$” must be one complete row, column, or diagonal — which means 1 and 9 sit directly opposite each other through the centre. So the pair goes either in the top-middle & bottom-middle cells, or in the left-middle & right-middle cells (in either order).

Every cell can be written as $m$ plus or minus a fixed offset (shown above) — this pattern of offsets ($\pm1,\pm2,\pm3,\pm4$ around the centre) is exactly the same for any set of 9 consecutive numbers, which is why sliding the whole square up or down (changing $m$) always keeps it magic. This mirrors the $2\times2$ calendar-grid trick from the Algebraic Expressions chapter, where every date could be written relative to one reference date.

1. Chau̩tīs means 34 in Hindi — and every row, column, and diagonal of this square adds up to exactly 34, so the name literally describes its magic sum.
2. Beyond the obvious rows/columns/diagonals, other groups of four also sum to 34: the four corners $7+14+9+4=34$; the central $2\times2$ block $13+8+3+10=34$; and each set of four “corner cells of any $2\times2$ sub-square” within the grid.

1. The Lo Shu Square (China, 2000+ years old) is the earliest magic square on record, legendarily carried on a turtle’s back.
2. A 3×3 magic square was found carved on a pillar in a temple in Palani, Tamil Nadu, dating back to the 8th century CE.
3. The Navagraha Yantra uses nine different 3×3 magic squares (one per graha/celestial body), each with its own magic sum.
4. The Kubera Yantra shown uses the numbers $20$–$28$: magic sum $=27+20+25=72$ (check any row/column/diagonal — they all agree).

Besides $2+2+2+2$, $1+1+1+1+1+1+1+1$, $1+2+2+1+2$, and $2+2+1+1+2$, many more combinations work — e.g. $2+1+2+1+2$, $1+1+2+2+2$, $2+2+2+1+1$, $1+2+1+2+2$, and so on. (The full, systematic count is worked out below: there are exactly 34 ways in total.)

1. Every rhythm for 5 either starts with a 1 (leaving 4 more to fill, in any of the 5 ways already found for $n=4$) or starts with a 2 (leaving 3 more to fill, in any of the 3 ways found for $n=3$).
2. So the count for $n=5$ is simply $5+3=8$ — no need to list every possibility from scratch.
3. In general: (ways for $n$) $=$ (ways for $n-1$) $+$ (ways for $n-2$).

By the same reasoning, (ways for 6) $=$ (ways for 5) $+$ (ways for 4) $=8+5=\bf{13}$. Yes — writing out all 6-beat rhythms (starting each with ‘1+’ from the 5-beat list, then ‘2+’ from the 4-beat list) confirms exactly 13 rhythms.

Continuing the pattern $1,2,3,5,8,13,21,\underline{34}$ — the 8th term of the Virahāṅka sequence is $\bf{34}$. So there are 34 rhythms of 8 beats (matching Kishor’s card puzzle from Section 6.2, where all the number cards were exactly these Virahāṅka numbers!).

1. Each term is the sum of the two before it: $34+55=89$ (given), $55+89=144$, $89+144=233$, $144+233=377$.
2. So the next 3 numbers are $144,\ 233,\ 377$.
3. Listing parities so far: $1(O),2(E),3(O),5(O),8(E),13(O),21(O),34(E),55(O),89(O),144(E),\ldots$ — grouping in 3s from the start ($1,2,3$), ($5,8,13$), ($21,34,55$)… each group reads odd, even, odd.
4. 377 is the 13th term; the 14th term (two after it) falls at the “even” spot of its group of three, so it must be even — and indeed $233+377=610$, which is even.

The parities repeat in fixed groups of 3: odd, even, odd | odd, even, odd | … This happens because odd+odd=even and odd+even=odd, so once two odds are added to make an even, that even plus the next odd is forced back to odd — locking in the same 3-term cycle forever.

The three daisies have 13, 21, and 34 petals — all Virahāṅka–Fibonacci numbers! This is a genuine, widely-observed pattern in nature: petal counts, seed spirals, and leaf arrangements very often land on numbers from this exact sequence.

1. $T+T+T=3T$ must end in the digit $T$ itself, so $3T$ and $T$ have the same units digit — meaning $3T-T=2T$ is a multiple of 10.
2. $2T$ is a multiple of 10 only when $T=0$ or $T=5$ (since $T$ is a single digit).
3. $T=0$ gives the trivial (not really 2-digit) case $0+0+0=0$, so we take $T=5$: check $3\times5=15$ — units digit is 5, matching $T$! And $T=2$ gives $3\times2=6$ (a 1-digit number, not 2-digit $U T$); $T=3$ gives $9$ (also 1-digit) — neither works.

1. Doubling a 2-digit number K2 gives at most $2\times99=198$, so the hundreds digit $H$ can only be $\bf{1}$ — it cannot be 2 or 3, since doubling a 2-digit number never reaches 200.
2. $2\times(10K+2)=100(1)+11M \Rightarrow 20K+4=100+11M \Rightarrow 20K-96=11M$.
3. Testing $K=5,6,7,8,9$: only $K=7$ gives a whole-number $M$: $20(7)-96=44=11\times4$, so $M=4$.
4. Check: $K2=72$, and $72+72=144$ — that’s $HMM=144$ with $H=1, M=4$. ✓

Each is solved the same way: write the addition in place-value form (e.g. $YY=11Y$), set it equal to the expanded target (e.g. $ZOO=100Z+11O$), then use the fact that every letter is a single digit (0–9) to narrow down the possibilities until only one combination fits.

Work left to right. Each new child’s height only has to satisfy one condition: exactly the target number of taller children must already be standing before them. Concretely — (b) is simply increasing height order (nobody new is ever shorter than someone earlier, so the count of taller-before-them is always 0); (c) is the exact opposite, strictly decreasing height order; (d) alternates a short dip and a new tallest-so-far; (e) opens with the single tallest child, then continues in increasing order for everyone else; (f) starts with 3 increasing tall children, then continues with 4 increasing short children.

1. (a) A ‘0’ just means “nobody taller stood before me” — a short child at the very front also says 0 (nobody to compare to yet), so saying 0 doesn’t force you to be the tallest overall.
2. (b) The tallest person in the whole line is automatically taller than everyone ahead of them, so among those ahead, exactly 0 can be taller — this one is a genuine certainty.
3. (c) The first person has nobody ahead of them at all, so their count is always 0, no matter their height.
4. (d) A middle child says 0 exactly when they happen to be taller than everyone before them (see part (b) reasoning) — possible but not guaranteed, so only sometimes.
5. (e) A large number just means many taller people happened to stand ahead of you — it says nothing definite about your own height compared to the rest of the line.
6. (f) In a group of 8, the largest possible number is 7 — the last person could have all 7 others taller than them (if the line is arranged tallest-to-shortest).

1. An odd count of ₹1 coins contributes an odd amount (odd × 1 = odd).
2. An odd count of ₹5 coins contributes an odd amount too (odd × 5 = odd, since odd×odd=odd).
3. An even count of ₹10 coins always contributes an even amount (any count × 10 is even).
4. Total parity $=$ odd $+$ odd $+$ even $=$ even $+$ even $=$ even.

Subtraction follows the same pairing logic as addition: removing a whole number of pairs from an even number leaves it even (d); two odd numbers each have one leftover dot, and removing one from the other cancels both leftovers, leaving pairs only (e); mixing an even and an odd always leaves exactly one unpaired dot, so the result is odd either way (f, g).

There is exactly one essentially different magic square using 1–9 — every other valid arrangement is just this one rotated or reflected (giving 8 look-alikes in total, but only 1 truly distinct pattern).

Strategy: add 1 to every number of the 1–9 magic square. Since every row/column/diagonal already summed equally, adding the same constant (1) to every one of the 3 numbers in each line increases every line’s sum by exactly $3\times1=3$, keeping all lines equal — so it’s still magic, now with sum $15+3=18$.

1. (a) Adding 1 to every cell: still a magic square (as above); the magic sum increases by $3\times1=3$ (since each line has 3 cells).
2. (b) Doubling every cell: still a magic square — doubling every one of the 3 numbers in a line doubles that line’s sum too, so all lines stay equal; the magic sum is simply doubled (from 15 to 30).

Any operation applied identically to every cell keeps it magic, as long as it preserves the equal-sum property line by line — for example: subtracting a constant from every cell, multiplying every cell by any fixed number (including negative numbers), or combining both (e.g. “double every number, then add 1”). Rotating or reflecting the whole grid also always produces another magic square.

Start from the 1–9 magic square and add the same constant to every cell: to get 2–10, add 1; to get 3–11, add 2; to get 9–17, add 8 — in general, to get the set starting at $s$, add $(s-1)$ to every cell of the 1–9 square. The layout (which number goes where) never has to change, only the constant added.

Substitute $m=25$ into each offset cell ($m+3,\,m-4,\,m+1,\,\ldots$) to get the grid above. Magic sum $=3m=75$.

Every line’s three offsets always cancel out (e.g. $(m{+}3)+(m{-}4)+(m{+}1)=3m$; $(m{-}2)+m+(m{+}2)=3m$). So every row, column, and diagonal sums to $\boxed{3m}$ — confirming the magic sum is always 3 times the centre number.

1. Magic sum $=3m$, so $3m=60 \Rightarrow m=20$.

Yes. Take the original 1–9 magic square and multiply every cell by 3 — the result (shown above, magic sum $45$) uses the non-consecutive multiples-of-3 set $\{3,6,9,\ldots,27\}$ and is still magic, since multiplying every cell by the same constant always preserves equal line-sums.

1. Every toggle flips the state: ON→OFF or OFF→ON.
2. After an even number of toggles, the bulb returns to its starting state (ON); after an odd number, it ends in the opposite state.
3. $77$ is odd.

1. Each physical sheet carries 2 consecutive page numbers, e.g. $(n, n{+}1)$ — one even, one odd — summing to $2n+1$, which is odd.
2. 50 sheets means adding 50 such odd numbers together.
3. An even count (50) of odd numbers always sums to an even total.
4. $6000$ is even — consistent with what 50 sheets can produce.

One valid filling is shown above: row 1 $=$ o,e,e (sum parity odd ✓); row 2 $=$ o,e,o (sum parity even ✓); columns $=$ (o,o)→even ✓, (e,e)→even ✓, (e,o)→odd ✓. Any filling that keeps the same odd/even pattern in each cell works equally well — only the parities matter, not the exact numbers.

This is exactly the generalised offset grid with centre $m=0$: every cell is just its offset ($\pm1,\pm2,\pm3,\pm4$). Every row, column, and diagonal sums to $3m=0$. ✓

$1+2+\cdots+100=\dfrac{100\times101}{2}=5050$, which is Even. (Quicker route: numbers 1–100 contain exactly 50 odd numbers; an even count of odds sums to even, and adding the 50 even numbers — always even — keeps the total even.)

1. Going forward: $987+1597=2584$, then $1597+2584=4181$.
2. Going backward, use $\text{previous}=\text{next}-\text{current}$: $1597-987=610$, then $987-610=377$.

1. This is exactly the “sum of 1’s and 2’s” counting problem from Section 6.4 — each way to climb 8 steps corresponds to one way of writing 8 as an ordered sum of 1’s and 2’s.
2. That count is the 8th Virahāṅka number.

1. Recall the parity cycle repeats every 3 terms: odd, even, odd (positions $1,2,3,4,5,6,\ldots \to O,E,O,O,E,O,\ldots$, i.e. every position that is $2 \pmod 3$ is even).
2. $20 = 18+2$, so $20 \equiv 2 \pmod 3$.

1. (a) $4m$ is always even, so $4m-1$ is always odd, for any whole number $m$. True.
2. (b) $6j-4$ only ever produces $2,8,14,20,\ldots$ — it skips many even numbers like $4, 6, 10$. False as a claim about all even numbers.
3. (c) If $p$ and $q$ are both taken to range over $1,2,3,\ldots$ (natural numbers), then $2p+1$ gives $3,5,7,9,\ldots$ — it misses the odd number 1. Only $2q-1$ (giving $1,3,5,7,\ldots$) captures every odd number starting from 1. Since $2p+1$ alone doesn’t cover all odd numbers over that domain, the joint claim is False.
4. (d) $2f$ is always even, so $2f+3$ (even $+$ odd) is always odd — it never produces an even number. False.

1. Write in place value: $UT=10U+T$, $TA=10T+A$, and $TAT=100T+10A+T=101T+10A$.
2. $(10U+T)+(10T+A)=101T+10A \Rightarrow 10U+11T+A=101T+10A \Rightarrow 10U=90T+9A \Rightarrow 10U=9(10T+A)$.
3. Since the result TAT is a 3-digit number starting with T, and UT+TA is at most $99+99=198$, we need $T=1$.
4. With $T=1$: $10U=9(10+A) \Rightarrow 10U=90+9A$. Testing $A=0$: $10U=90 \Rightarrow U=9$. (Other values of $A$ from 1–9 don’t give a whole-number, single-digit $U$.)
5. Check: $UT=91$, $TA=10$, sum $=101=TAT$ with $T=1,A=0$. ✓