Chapter 6: Algebra Play (Algebra)

The answer is half of what is added: $\frac{2x+k}{2}-x=\frac{k}{2}$.

• Answer 3: add 6 instead of 4.
• Answer 5: add 10.

A longer trick:

1. Think of a number: $x$
2. Multiply by 3: $3x$
3. Add 15: $3x+15$
4. Divide by 3: $x+5$
5. Add 7: $x+12$
6. Subtract the number: $12$

The answer is always 12.

Add 6 to get 3, add 10 to get 5 (the answer is always half the number added).

Subtract 165. The last two digits are then the day and the digits before them are the month:

1. $1269-165=1104$, which gives 4 November
2. $394-165=229$, which gives 29 February (a leap-year date)
3. $296-165=131$, which gives 31 January

(i) 4 November (ii) 29 February (iii) 31 January

Yes. Any steps that end in $100M+(\text{a fixed number})+D$ will work. For example:

1. Multiply the month by 5: $5M$
2. Add 7: $5M+7$
3. Multiply by 4: $20M+28$
4. Add 3: $20M+31$
5. Multiply by 5: $100M+155$
6. Add the day: $100M+155+D$

Now subtract 155 to read off the date.

The month must end up multiplied by 100, so that the day (at most 31) fits in the last two digits.

Yes, e.g. ×5, +7, ×4, +3, ×5, + day gives 100M + 155 + D; subtract 155.

Tops: 8, 14 and 32.

Use letters for the empty bottom boxes and write each box in terms of them.

(a) Bottom $4,x,6,y$:

• Third row: $10+2x$ and $12+x+y$.
• $12+x+y=22$, so $x+y=10$.
• Top: $(10+2x)+22=50$, so $x=9$ and $y=1$.

(b) Bottom $5,x,7,y$:

• Second row: $7+y=9$, so $y=2$.
• Third row: $12+2x=40$, so $x=14$.
• The top is then $40+30=70$.

(c) Bottom $3,5,x,y$:

• Second row: $x+y=7$.
• Top: $(13+x)+(12+x)=35$, so $x=5$ and $y=2$.

(a) bottom 4, 9, 6, 1 (b) bottom 5, 14, 7, 2 with top 70 (c) bottom 3, 5, 5, 2.

• 2 rows: $a+b$
• 3 rows: $a+2b+c$
• 4 rows: $a+3b+3c+d$

Each bottom number is counted as many times as there are paths from it up to the top. These counts (1 1; 1 2 1; 1 3 3 1; …) are the rows of Pascal's triangle, which the Indian mathematician Piṅgala knew as the Meru-prastāra.

The top is the bottom row weighted by 1, 2, 1 (3 rows) or 1, 3, 3, 1 (4 rows), the rows of the Meru-prastāra (Pascal's triangle).

Use $a+2b+c$:

• $4+26+8=$ 38
• $7+22+3=$ 32
• $10+28+25=$ 63

38, 32, 63

With bottom $a,b,c,d$:

• Second row: $a+b,b+c,c+d$
• Third row: $a+2b+c,b+2c+d$
• Top: $a+3b+3c+d$

a + 3b + 3c + d

Use $a+3b+3c+d$:

• $8+57+63+13=$ 141
• $7+54+57+6=$ 124
• $9+21+15+11=$ 56

141, 124, 56

Bottom 1, 2, 3 → second row 3, 5 → top 8.

Yes, they are all Virahāṅka-Fibonacci numbers. Two neighbours in the sequence add up to the next term, so each row is just the sequence moved on.

The numbers are 1, 2, 3, 3, 5, 8; the top is 8, and all of them are Virahāṅka-Fibonacci numbers.

Number the sequence as $V_1=1,V_2=2,V_3=3,V_4=5,\dots$ Then $V_k+V_{k+1}=V_{k+2}$, so each row is the row below shifted two places along the sequence.

(i) The rows are:

• 1, 2, 3, 5
• 3, 5, 8
• 8, 13
• top 21 $=V_7$

(ii) Every box is again a Virahāṅka-Fibonacci number. The top is $V_{57}$, the 57th number of the sequence, which is 591,286,729,879.

Every entry is a Virahāṅka-Fibonacci number. (i) Top = 21 (ii) top = the 57th number, 591,286,729,879.

All the numbers in the pyramid are Virahāṅka-Fibonacci numbers. The $r$th row from the bottom holds $V_{2r-1},V_{2r},\dots$

So the top is $V_{2n-1}$, the $(2n-1)$th number of the sequence. For example, $n=3$ gives $V_5=8$ and $n=4$ gives $V_7=21$.

All entries are Virahāṅka-Fibonacci numbers, and the top is the (2n − 1)th one.

3 × 3 square: if the top-left number is $a$, the nine numbers add up to

$$9a+(0+1+2+7+8+9+14+15+16)=9a+72=9(a+8)$$

The centre is $a+8$, so the sum is 9 times the centre. If a friend says 117, the centre is 13, and you can name all nine numbers.

Plus shape (a centre $c$ with the numbers above, below, left and right of it): the sum is $(c-7)+(c+7)+(c-1)+(c+1)+c=5c$. Divide by 5 to find the centre.

Vertical strip of 3 (on the 10-wide grid in the book): $a+(a+10)+(a+20)=3a+30$.

E.g. any 3 × 3 square sums to 9 × its centre; a plus shape sums to 5 × its centre.

Grid 1: $2■+●=27$ and $2●+■=21$.

• Adding the two: $3(■+●)=48$, so $■+●=16$.
• Then $■=27-16=11$ and $●=5$.
• The last row is $●+■+●=5+11+5=21$.

Grid 2: $●+2◆=18$ and $◆+2●=15$.

• Adding the two: $3(●+◆)=33$, so $●+◆=11$.
• Then $◆=18-11=7$ and $●=4$.
• The third row is $◆+●+●=15$.
• The column totals are 18, 15, 15, and the corner is $18+15+15=48$.

Grid 1: ■ = 11, ● = 5, last row 21. Grid 2: ● = 4, ◆ = 7; third row 15; column totals 18, 15, 15; corner 48.

Use the largest digit as the multiplier, and the other two in decreasing order: $31\times 7=217$.

Check against the others: $71\times 3=213$, $73\times 1=73$, $13\times 7=91$.

31 × 7 = 217

$53\times 9=477$. The nearest rival is $93\times 5=465$.

53 × 9 = 477

The difference is $(10a+b)-(10b+a)=9a-9b=9(a-b)$, which is again divisible by 9.

The difference is 9(a − b), still a multiple of 9.

The difference is $9\times (\text{larger digit}-\text{smaller digit})$, so the quotient is the difference of the two digits. For example, $74-47=27$ and $27÷9=3=7-4$.

The quotient is the difference between the two digits.

Yes. $(10a+b)+(10b+a)=11a+11b=11(a+b)$.

The quotient is the sum of the digits. For example, $28+82=110=11\times 10$.

Yes: the sum is 11(a + b).

$$(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c=111(a+b+c)$$

Since $111=3\times 37$, the sum is divisible by 37 and also by 3.

For example, $123+231+312=666=111\times 6$.

The sum is 111(a + b + c) = 3 × 37 × (a + b + c): divisible by both 37 and 3.

You get abc back. For example, $583583÷7=83369$, then $÷11=7579$, then $÷13=583$.

The reason: $abcabc=abc\times 1000+abc=abc\times 1001$, and $1001=7\times 11\times 13$.

You get the original 3-digit number, because abcabc = abc × 1001 = abc × 7 × 11 × 13.

Start with $x$ flowers and place $k$ at each shrine:

• After pond 1 and shrine 1: $2x-k$
• After pond 2 and shrine 2: $4x-3k$
• After pond 3: $8x-6k$, all placed at shrine 3, so $8x-6k=k$

This gives $8x=7k$. The smallest answer is $x=7$ and $k=8$.

Check: 7 → 14 → leave 8 → 6 → 12 → leave 8 → 4 → 8 → leave 8 ✓

He started with 7 flowers and placed 8 at each shrine (or any multiple: 14 and 16, …).

Without letters: if all 55 animals were hens, there would be 110 legs. The extra 40 legs come from horses, which have 2 more legs each. So there are $40÷2=20$ horses and 35 hens.

Check: $20\times 4+35\times 2=80+70=150$ ✓

20 horses and 35 hens.

$5d+6=3(d+6)$, so $2d=12$ and $d=6$. (The mother is 30.)

6 years old (mother 30).

Let Gauri have $g$ cows. Then $2g-3=g+3$, so $g=6$.

Gauri 6, Naina 12.

(i) Costs $=5000+100\times 10=6000$. For a profit of 2000, sales must be 8000, so the price is $8000÷100=$ ₹80.

(ii) Each dosa earns $50-10=$ ₹40 above its cost. We need $40n-5000=2000$, so $n=175$.

(i) ₹80 per dosa (ii) 175 dosas.

The values are $\frac{1}{3}$, $\frac{4}{12}=\frac{1}{3}$, $\frac{9}{27}=\frac{1}{3}$, and so on: always $\frac{1}{3}$.

The top is the sum of the first $n$ odd numbers, which is $n^2$. The bottom is the sum of the next $n$ odd numbers, which is $(2n{)}^2-n^2=3n^2$. So the fraction is $\frac{n^2}{3n^2}=\frac{1}{3}$.

Every fraction equals 1/3, because it is n² / (4n² − n²).

(i) Start with $x$:

• After round 1: $2x-8$
• After round 2: $4x-24$
• After round 3: $8x-56=0$, so $x=7$

Check: 7 → 14 − 8 = 6 → 12 − 8 = 4 → 8 − 8 = 0 ✓

(ii) A round doubles $x$ and takes $c$, so Karim gains only if $2x-c>x$, i.e. $c<x$. The cost must be less than the number of coins he starts with. Then he gains every round, since his coins keep growing. For example, with 7 coins a cost of 6 or less works (7 → 8 → 10 → 14 at cost 6).

(iii) After three rounds he has $8x-7c$. To leave him with nothing, the genie needs $8x-7c=0$, so $c=\frac{8x}{7}$. That is 8 coins for $x=7$, and only works when $8x$ is a multiple of 7. In general, after $n$ rounds the cost must be $\frac{2^{n}x}{2^n-1}$.

(i) 7 coins (ii) only if the cost is less than the coins he starts with (iii) cost = 8x/7 for three rounds (8 coins when he has 7).