Chapter 6: Number Play (Numbers)

Counting, for each child, the taller children standing before them (to their left):

0, 0, 1, 0, 3, 0, 3

Give the seven children heights 1 (shortest) to 7 (tallest). One arrangement for each (left to right):

Check (a): the child of height 4 (4th) has 7 and 5 in front taller → 2 ✓; the child of height 1 (5th) has all four in front taller → 4 ✓.

E.g. (a) 7, 3, 5, 4, 1, 6, 2 (b) 1, 2, 3, 4, 5, 6, 7 (c) 7, 6, 5, 4, 3, 2, 1 (d) 2, 1, 4, 3, 6, 5, 7 (e) 7, 1, 2, 3, 4, 5, 6 (f) 5, 6, 7, 1, 2, 3, 4 (1 = shortest).

(a) Only sometimes true. '0' means no one in front is taller; there may be taller people behind (e.g. the first person always says 0).

(b) Always true. No one at all is taller than the tallest person.

(c) Always true. There is no one in front of the first person.

(d) Only sometimes true. A person in the middle says 0 whenever they are taller than everyone in front (e.g. heights 1, 2, 3, …).

(e) Only sometimes true. For heights 3, 1, 2 the calls are 0, 1, 1: the shortest calls the largest number (shared). But for heights 1, 3, 2 the calls are 0, 0, 1: the shortest (first) says 0, while someone else calls the largest number.

(f) 7, said by the last person when all 7 people in front are taller (i.e. the shortest person stands last).

(a) Sometimes (b) Always (c) Always (d) Sometimes (e) Sometimes (f) 7

Odd numbers pair up two at a time (their leftover ones join into a pair):

(a) 4 odd numbers → even (b) 5 odd numbers → odd (c) 6 odd numbers → even

An even count of odd numbers gives an even sum; an odd count gives an odd sum.

Yes: an odd number is also one less than a collection of pairs, e.g. $7=8-1$ (four pairs with one missing).

(a) even (b) odd (c) even; yes, e.g. 7 = 8 − 1.

No. Their ages are consecutive numbers, so one is even and the other odd. Even + odd = odd, but 112 is even. (For example, 55 + 56 = 111 and 56 + 57 = 113.)

No: two consecutive numbers always have an odd sum.

Even numbers never leave a leftover; odd numbers leave one each, and two leftovers form a pair.

(a) Even (two leftovers make a pair) (b) Even (c) Even (d) Even (eight leftovers make four pairs)

All four sums are even.

• ₹1 coins: an odd number of ₹1 → odd amount.
• ₹5 coins: odd × 5 → odd amount.
• ₹10 coins: always even.

Total = odd + odd + even = even. But ₹205 is odd, so he made a mistake.

Yes: the total must be even, but 205 is odd.

(d) even (e) even (f) odd (g) odd

(Subtracting removes pairs, plus possibly one leftover; e.g. $9-5=4$, $8-3=5$.)

(d) even (e) even (f) odd (g) odd

Number of squares = rows × columns. A product is odd only when both numbers are odd; if either is even, the product is even (it is made of an even number of equal groups, or groups of even size).

(a) odd × odd → odd (b) even × even → even (c) odd × even → even

(a) odd (b) even (c) even

• Always even: $2n$, $100p$, $48w-2$, $6k+4$
• Always odd: $2n+1$, $2n-1$, $4m+3$, $10t+5$
• Either parity: $n+1$, $3n+4$, $5n$ (depends on whether $n$ is odd or even)

$2n$ ($n=1,2,3,\dots$) lists all even numbers, and $2n-1$ lists all odd numbers.

The 100th odd number is $2\times 100-1=$ 199; the nth odd number is $2n-1$.

Even: 2n; odd: 2n − 1 (or 2n + 1); either: 3n + 4. 2n gives all evens and 2n − 1 all odds; the 100th odd number is 199.

Grid 1: the first column needs 24 with a 9 at the top, so the other two cells are 8 and 7.

Grid 2: the bottom row needs 6, which can only be 1 + 2 + 3; with 3 at the right it is 1, 2, 3. The first column then gives $12-4-1=7$.

(Grid 2 also works with the top row 7, 9, 8 and the middle row 4, 5, 6.)

Grid 1: 9 1 3 / 8 2 4 / 7 6 5. Grid 2: 7 8 9 / 4 6 5 / 1 2 3.

The three row sums together add every number in the grid exactly once: $1+2+\cdots +9=45$. The column sums also add every number once: 45. So all six circled numbers add to $45+45=90$.

Each set of three sums adds all the numbers 1–9 once, i.e. 45; rows + columns = 90.

The magic sum is $45÷3=15$. If the centre is $c$, then each pair of numbers opposite each other through the centre adds to $15-c$.

• 9 needs a partner $15-c-9=6-c$, which must be at least 1, so $c\le 5$.
• 1 needs a partner $14-c$, which must be at most 9, so $c\ge 5$.

So the centre must be 5; 1, 2, 3, 4, 6, 7, 8, 9 cannot be at the centre.

A corner number lies in three lines (row, column, diagonal). With 1 there are only two ways to make 15: $1+5+9$ and $1+6+8$. With 9 only $9+5+1$ and $9+4+2$. So 1 and 9 cannot be at corners; they go in the middle of opposite sides, with 5 between them.

Completing (1 at the bottom middle, 9 at the top middle): the top row is 9 with 4 and 2; the bottom row is 1 with 8 and 6:

Only 5 can be at the centre; 1 and 9 go in middle edge positions; e.g. 4 9 2 / 3 5 7 / 8 1 6.

8. All of them come from one square by rotating it (4 positions) and reflecting it (×2). So there is essentially only one magic square, in 8 orientations.

8 (one square with its rotations and reflections).

Add 1 to every number of a 1–9 magic square. Each line now gains 3, so the magic sum is $15+3=18$, and the centre is 6.

Add 1 to each number of a 1–9 square: 9 2 7 / 4 6 8 / 5 10 3 (magic sum 18).

Both are still magic squares.

(a) Every line has 3 numbers, so each sum increases by 3: magic sum $15\to 18$.

(b) Every line sum doubles: magic sum $15\to 30$.

Yes in both; (a) the magic sum increases by 3 (b) it doubles.

• Adding or subtracting the same number to every entry (magic sum changes by 3 times that number).
• Multiplying or dividing every entry by the same non-zero number.
• Rotating the square or reflecting it (in a middle line or a diagonal).
• Replacing each entry $x$ by $10-x$ (for 1–9), which gives another 1–9 magic square.

(Squaring each entry does not keep it magic.)

Adding/subtracting a constant, multiplying/dividing by a non-zero constant, rotations, reflections, and x → 10 − x.

Take the 1–9 square and add (first number − 1) to every entry. For 3–11 add 2; for 9–17 add 8:

The middle number of the set goes at the centre, and the magic sum is 3 times it (here $3\times 13=39$).

Add the same number to each entry of a 1–9 square; the middle number goes at the centre and the magic sum is 3 × centre.

From 8 1 6 / 3 5 7 / 4 9 2 (centre 5):

Observations: opposite numbers through the centre are $m+k$ and $m-k$, so each line adds to $3m$.

m + 3, m − 4, m + 1 / m − 2, m, m + 2 / m − 1, m + 4, m − 3

(Magic sum 75.)

28 21 26 / 23 25 27 / 24 29 22

E.g. $(m+3)+(m-4)+(m+1)=3m$. Every line gives $3m$.

3m

(a)

(the magic sum becomes $3m+3$)

(b)

(the magic sum becomes $6m$)

(a) each entry +1, sum 3m + 3 (b) each entry doubled, sum 6m

$3m=60$ gives $m=20$:

Centre 20: 23 16 21 / 18 20 22 / 19 24 17

Yes. Multiply each entry of the 1–9 square by 3 (numbers 3, 6, …, 27, which are not consecutive):

(Magic sum 45.)

Yes, e.g. 24 3 18 / 9 15 21 / 12 27 6 (magic sum 45).

"Chautīs" means 34, and every row, column and diagonal adds up to 34.

Other groups of four that make 34:

• Each 2 × 2 corner block: $7+12+2+13=34$, $1+14+8+11=34$, $16+3+9+6=34$, $10+5+15+4=34$.
• The central 2 × 2 block: $13+8+3+10=34$.
• The four corners: $7+14+9+4=34$.
• Broken diagonals, e.g. $12+8+5+9=34$ and $1+13+16+4=34$.
• Any 2 × 2 block anywhere, e.g. $12+1+13+8=34$.

Every row, column and diagonal sums to 34 (chautīs); so do all 2 × 2 blocks, the four corners and the broken diagonals.

$27+20+25=72$; the centre is 24 and $3\times 24=72$. Magic sum 72.

72

5 beats (8 ways): 1+1+1+1+1, 1+1+1+2, 1+1+2+1, 1+2+1+1, 2+1+1+1, 1+2+2, 2+1+2, 2+2+1

6 beats (13 ways): put "1+" before each 5-beat rhythm (8 ways) and "2+" before each 4-beat rhythm (5 ways):

1+1+1+1+1+1, 1+1+1+1+2, 1+1+1+2+1, 1+1+2+1+1, 1+2+1+1+1, 1+1+2+2, 1+2+1+2, 1+2+2+1, 2+1+1+1+1, 2+1+1+2, 2+1+2+1, 2+2+1+1, 2+2+2

8 beats: the sequence 1, 2, 3, 5, 8, 13, 21, 34, so 34 rhythms. (Other 8-beat rhythms: short-long-long-long-short = 1+2+2+2+1, long-short-short-long-long = 2+1+1+2+2, …)

8 ways for 5, 13 for 6, and 34 rhythms of 8 beats.

$55+89=$ 144, $89+144=$ 233, $144+233=$ 377. The next number is $233+377=610$, which is even.

Parities: odd, even, odd, odd, even, odd, odd, even, … After the first number, the pattern odd, odd, even repeats: every third number (2, 8, 34, 144, 610, …) is even, since odd + odd = even, then even + odd = odd, and odd + even = odd.

144, 233, 377; the next (610) is even; every 3rd term (2nd, 5th, 8th, …) is even.

13, 21 and 34 petals: all Virahāṅka numbers.

13, 21, 34

(i) $3T$ ends in T, so $2T$ ends in 0: $T=5$ (T = 0 gives 0). $5+5+5=15$: T = 5, U = 1.

(ii) $2+2=4$, so M = 4. Tens: $2K=14$, so K = 7, H = 1: $72+72=144$.

(iii) The sum is at most $99+9=108$, so Z = 1 and ZOO $=100$: O = 0, YY $=99$, Y = 9: $99+1=100$.

(iv) A 3-digit sum of two 2-digit numbers starts with 1: E = 1. Units: $5+D$ ends in 5, so D = 0. Tens: $B+3=10$, so B = 7: $75+30=105$.

(v) P = 1 (3-digit sum). Units: $1+1=2$, so R = 2. Tens: $2K=12$, so K = 6: $61+61=122$.

(vi) The sum is at most $91+9=100$, so 1FF $=100$: F = 0, and $10C+1+C=100$, so $11C=99$, C = 9: $91+9=100$.

(i) 5 + 5 + 5 = 15 (ii) 72 + 72 = 144 (iii) 99 + 1 = 100 (iv) 75 + 30 = 105 (v) 61 + 61 = 122 (vi) 91 + 9 = 100

Every 2 toggles bring it back to ON. 77 is odd ($38$ pairs + 1), so after 77 toggles the bulb is OFF.

Off, because 77 is odd.

Each sheet carries two consecutive page numbers, an odd one and the next even one: $(1,2)$, $(3,4)$, … So each sheet adds an odd amount, and 50 odd amounts give an even total. Parity alone does not rule out 6000.

But look closer: a sheet with pages $2k-1$ and $2k$ adds $4k-1$, which is 1 less than a multiple of 4. The total of 50 sheets is (a multiple of 4) $-50$, which leaves remainder 2 when divided by 4 (like 6, 10, 14, …). Since $6000$ is a multiple of 4, the sum cannot be 6000.

The sum is even (50 odd sheet-totals), so parity allows it; but each sheet adds 4k − 1, making the total 2 more than a multiple of 4, so 6000 is impossible.

For example: 1, 2, 4 in the top row and 3, 6, 5 in the bottom row (sums 7 and 14; columns 4, 8, 9).

Top row o, e, e; bottom row o, e, o (e.g. 1 2 4 / 3 6 5).

Use the generalised form with $m=0$:

3 −4 1 / −2 0 2 / −1 4 −3

(a) even (b) even (c) even (d) odd

(a) even (b) even (c) even (d) odd

From 1 to 100 there are 50 odd numbers; an even count of odd numbers gives an even sum, and the evens add an even amount. So the sum is even (indeed $1+2+\cdots +100=5050$).

Even (5050).

Next: $987+1597=$ 2584, $1597+2584=$ 4181.

Previous: $1597-987=$ 610, $987-610=$ 377.

Next 2584, 4181; previous 610, 377.

This is the number of ways of writing 8 as a sum of 1s and 2s, the 8th Virahāṅka number: 34 ways. (By the number of 2-steps: no 2s: 1 way; one 2: 7 ways; two 2s: 15; three 2s: 10; four 2s: 1; total 34.)

34 ways

The even terms are the 2nd, 5th, 8th, 11th, 14th, 17th, 20th, … (every third term from the 2nd). So the 20th term is even (it is 10946).

Even

(a) True: $4m$ is even, and even − 1 is odd.

(b) False: $6j-4$ gives 2, 8, 14, …; it misses 4, 6, 10, ….

(c) False for counting numbers $p,q=1,2,3,\dots$: $2q-1$ gives all odd numbers 1, 3, 5, …, but $2p+1$ starts at 3 and misses 1. (It becomes true only if $p$ is allowed to be 0.)

(d) False: $2f$ is even, so $2f+3$ is always odd.

Only (a) is true. [(c) is true only if p may be 0.]

The sum of two 2-digit numbers is less than 200, so T = 1. Units: $T+A=1+A$ must end in T = 1, so A = 0. Tens: $U+T=U+1$ must give "TA" = 10, so U = 9.

$91+10=101$ ✓

U = 9, T = 1, A = 0 (91 + 10 = 101)

Yes. Each marriage joins two people, so the number of people who got married (to each other, within the count) must be even. 14,70,369 is odd, so the figure must be wrong or rounded. (It could only be odd if, for example, some spouses were not counted because they married someone outside the region.)

Yes: marriages come in pairs, so the count of people married should be even; 14,70,369 is odd.