NCERT Solutions Ganita Prakash (Part 2) Chapter 6 – 1476.6 Decoding Divisibility Tricks — Figure it Out

Book page 145 Updated on2026-09-05

Q1.

In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient?

Answer

The quotient is the difference of the two digits.

Difference of the numbers = 9 × (larger digit – smaller digit)
Quotient = |a – b|

Number	Reversed	Difference	Quotient	Digits differ by
74	47	27	3	7 – 4 = 3
52	25	27	3	5 – 2 = 3
81	18	63	7	8 – 1 = 7
96	69	27	3	9 – 6 = 3

Why it happens: the difference of the two numbers is exactly 9(a – b) or 9(b – a), so dividing by 9 undoes the 9 and hands back the digit gap. There is a nice consequence: the quotient does not depend on the digits themselves, only on how far apart they are. That is why 74, 52 and 96 all give the same quotient 3 — and why the quotient can only ever be one of 1, 2, …, 8.

Try This: ask a friend for the quotient instead of the number. You will know the gap between their digits at once — a good start for guessing the number itself.

Q2.

In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? For example: We start with 31. After reversing we get 13. Adding 31 and 13, we get 44. We start with 28. After reversing we get 82. Adding 28 and 82, we get 110. We start with 12. After reversing we get 21. Adding 12 and 21, we get 33. Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra?

Answer

Yes, it is always true, and the algebra is one line.

(10a + b) + (10b + a)
= 10a + a + 10b + b
= 11a + 11b
= 11(a + b)

So the sum is always a multiple of 11, and the quotient is a + b, the sum of the digits.

Number	Reversed	Sum	Sum ÷ 11	a + b
31	13	44	4	3 + 1 = 4
28	82	110	10	2 + 8 = 10
12	21	33	3	1 + 2 = 3

Why it happens: in the two numbers together, each digit occupies the tens place once and the units place once. So each digit contributes 10 of itself plus 1 of itself, that is 11 of itself. Subtracting compared the digits and produced 9s; adding pools them and produces 11s. Both facts come from the same source — the place values 10 and 1 differ by 9 and add to 11.

Tip: here the two digits need not be different, and the digits may even be equal. 55 + 55 = 110 = 11 × 10 ✓.

Q3.

Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3? [Hint: Look at some multiples of 37.]

Answer

Write all three numbers by place value and add.

abc = 100a + 10b + c
bca = 100b + 10c + a
cab = 100c + 10a + b
─────────────────────
Sum = (100a + a + 10a) + (10b + 100b + b) + (c + 10c + 100c)
= 111a + 111b + 111c
= 111(a + b + c)

Now factorise 111. Looking at multiples of 37 — 37, 74, 111 — we see that

111 = 3 × 37
So Sum = 3 × 37 × (a + b + c)

The sum is therefore always divisible by 37, and always divisible by 3 as well — in fact it is divisible by 111.

Number	The three cycles	Sum	÷ 37	÷ 3
253	253 + 532 + 325	1110	30	370
147	147 + 471 + 714	1332	36	444
908	908 + 089 + 890	1887	51	629

Why it happens: cycling the digits sends each digit through the hundreds, the tens and the units place exactly once. So over the three numbers every digit is counted 100 + 10 + 1 = 111 times, whatever the digits are. The whole result is decided by that 111 — and since 111 = 3 × 37, both divisibility claims follow at once. Note the quotient on dividing by 111 is a + b + c, the digit sum.

Tip: in the last row 089 is not really a three-digit number, but the arithmetic is unaffected — 111(9 + 0 + 8) = 111 × 17 = 1887 ✓.

Q4.

Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the digits, that is abcabc. Divide this number by 7, then by 11, and finally by 13. What do you get? Try this with other numbers. Figure out why it works. [Hint: Multiply 7, 11 and 13.]

Answer

You get back the original three-digit number abc, every time — and none of the three divisions leaves a remainder.

abcabc = abc × 1000 + abc = abc × 1001
And 7 × 11 × 13 = 77 × 13 = 1001
So abcabc = abc × 7 × 11 × 13

Dividing by 7, then 11, then 13 peels off exactly those three factors and leaves abc.

Start	÷ 7	÷ 11	÷ 13
253253	36179	3289	253
481481	68783	6253	481
907907	129701	11791	907

Why it happens: writing abc twice shifts the first copy three places to the left, which multiplies it by 1000, and then adds one more copy — so the six-digit number is abc × (1000 + 1) = abc × 1001. Nothing about the actual digits matters. The surprise is really a fact about 1001: it happens to be the product of the three consecutive primes 7, 11 and 13. The order of the divisions does not matter either, since multiplication can be done in any order.

Try This: repeat a two-digit number instead — abab = ab × 101, and 101 is prime, so there is no similar chain of three divisions. Repeating a four-digit block gives × 10001 = 73 × 137.

Q5.

There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person has some flowers. He dips them all in the first pond and then places some flowers in shrine 1. Next, he dips the remaining flowers in the second pond and places some flowers in shrine 2. Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine?

Answer

He started with 7 flowers and placed 8 flowers in each shrine.

Let x be the number of flowers he started with and k the number placed at each shrine.

After pond 1: 2x → leaves k → carries 2x – k
After pond 2: 2(2x – k) = 4x – 2k → leaves k → carries 4x – 3k
After pond 3: 2(4x – 3k) = 8x – 6k → leaves all of it at shrine 3

So 8x – 6k = k
8x = 7k (adding 6k to both sides)

x and k are whole numbers, so 7 must divide 8x, and therefore 7 divides x. The smallest possibility is x = 7, which gives k = 8.

Stage	Flowers
Start	7
Dip in pond 1	14
Leave 8 at shrine 1	6
Dip in pond 2	12
Leave 8 at shrine 2	4
Dip in pond 3	8
Leave all 8 at shrine 3	0

Why it happens: the condition 8x = 7k says the answer is not a single pair of numbers but a whole family — 7 and 8, 14 and 16, 21 and 24, and so on, all in the ratio 7 : 8. This is what happens when one equation carries two unknowns: it fixes the ratio, not the sizes. Requiring whole flowers is what pins down the smallest, and most natural, answer.

Tip: working backwards is quicker if you only want the smallest case. He ends with 0 after leaving k at shrine 3, so before that dip he had k/2; before the second offering, k/2 + k = 3k/2; before the second dip, 3k/4; and so on. For all these to be whole numbers k must be a multiple of 8.

Q6.

A farm has some horses and hens. The total number of heads of these animals is 55 and the total number of legs is 150. How many horses and how many hens are on the farm? Can you solve this without letter-numbers? [Hint: If all the 55 animals were hens, then how many legs would there be? Using the difference between this number and 150, can you find the number of horses?]

Answer

20 horses and 35 hens.

With letter-numbers. Let h be the number of horses and n the number of hens. Each animal has one head; a horse has 4 legs and a hen has 2.

h + n = 55 … (heads)
4h + 2n = 150 … (legs)

Halve the second equation: 2h + n = 75 (dividing both sides by 2)
Subtract the first from this: (2h + n) – (h + n) = 75 – 55
h = 20
n = 55 – 20 = 35

Without letter-numbers (the hint’s method):

If all 55 animals were hens, legs = 55 × 2 = 110
But there are 150 legs — that is 150 – 110 = 40 legs too few counted
Turning one hen into a horse adds 4 – 2 = 2 legs
Number of horses = 40 ÷ 2 = 20, so hens = 55 – 20 = 35

Why it happens: the two methods are the same argument in different clothing. “Assume all hens” is exactly the step of halving the leg equation and subtracting the head equation: 2h + n – (h + n) = h. Each horse counted as a hen loses 2 legs, so the shortfall of 40 legs, shared 2 at a time, counts the horses.

Check it yourself: 20 + 35 = 55 heads ✓ and 20 × 4 + 35 × 2 = 80 + 70 = 150 legs ✓.

Q7.

A mother is 5 times her daughter’s age. In 6 years’ time, the mother will be 3 times her daughter’s age. How old is the daughter now?

Answer

The daughter is 6 years old now, and the mother is 30.

Let the daughter’s present age be d. Then the mother’s present age is 5d.
In 6 years: daughter = d + 6, mother = 5d + 6.
The condition says: 5d + 6 = 3(d + 6)
5d + 6 = 3d + 18
2d + 6 = 18 (subtracting 3d from both sides)
2d = 12 (subtracting 6 from both sides)
d = 6 (dividing both sides by 2)
Mother = 5 × 6 = 30

Why it happens: the modelling step is the one to watch. “In 6 years” adds 6 to both ages, so the mother’s future age is 5d + 6 — not 5(d + 6). And the multiple changes: 5 times now, only 3 times later, because the same 6 years is a much bigger share of the daughter’s life than of the mother’s. Writing the condition as one equation and then removing 3d from both sides is what turns a puzzle about two people into a single statement about d.

Check it yourself: now 30 and 6, and 30 = 5 × 6 ✓. In 6 years, 36 and 12, and 36 = 3 × 12 ✓.

Q8.

Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. Gauri says to Naina, “You have twice as many cows as I do”. Naina says, “That’s true, but if I gave you three of my cows, we would each have the same number of cows”. How many cows do Gauri and Naina have?

Answer

Gauri has 6 cows and Naina has 12.

Let Gauri have g cows. Naina’s remark that she has twice as many gives Naina = 2g.
If Naina gives 3 cows to Gauri:
Naina then has 2g – 3, Gauri then has g + 3, and these are equal:
2g – 3 = g + 3
g – 3 = 3 (subtracting g from both sides)
g = 6 (adding 3 to both sides)
Naina = 2 × 6 = 12

Why it happens: giving away 3 cows is a double move — Naina goes down by 3 and Gauri goes up by 3, so the gap between them closes by 6, not 3. Since the gap has to close completely, the original gap must have been 6. And the original gap is 2g – g = g, which is why g = 6 falls out so quickly.

Check it yourself: 12 = 2 × 6 ✓. After the gift, Naina has 9 and Gauri has 9 ✓.

Q9.

I run a small dosa cart and my expenses are as follows: Rent for the dosa cart is ₹5000 per day. The cost of making one dosa (including all the ingredients and fuel) is ₹10. (i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹2000? (ii) If my customers are willing to pay only ₹50 for a dosa, how many dosas should I aim to sell in a day to make a profit of ₹2000?

Answer

(i) ₹80 per dosa. (ii) 175 dosas a day.

(i) Let the selling price be ₹p per dosa.

Money taken in = 100p
Money spent = rent + making cost = 5000 + 100 × 10 = ₹6000
Profit = 100p – 6000 = 2000
100p = 8000 (adding 6000 to both sides)
p = ₹80 (dividing both sides by 100)

(ii) Let the number of dosas sold be n, each at ₹50.

Money taken in = 50n Money spent = 5000 + 10n
Profit = 50n – (5000 + 10n) = 2000
40n – 5000 = 2000
40n = 7000 (adding 5000 to both sides)
n = 175 dosas (dividing both sides by 40)

Why it happens: the two costs behave differently, and the algebra makes that visible. The ₹5000 rent is fixed — it does not change with n — while the ₹10 per dosa grows with n. So each dosa sold at ₹50 contributes 50 – 10 = ₹40 towards the rent and the profit. First ₹5000 of that goes on rent; the ₹2000 profit needs 2000 ÷ 40 = 50 more dosas after the 125 that cover the rent. 125 + 50 = 175.

Check it yourself: (i) 100 × 80 = ₹8000 taken in; 5000 + 1000 = ₹6000 spent; profit ₹2000 ✓. (ii) 175 × 50 = ₹8750 taken in; 5000 + 1750 = ₹6750 spent; profit ₹2000 ✓.

Q10.

Evaluate the following sequence of fractions: 1/3, (1 + 3)/(5 + 7), (1 + 3 + 5)/(7 + 9 + 11). What do you observe? Can you explain why this happens? [Hint: Recall what you know about the sum of the first n odd numbers.]

Answer

Every one of them equals 1/3.

1/3 = 1/3
(1 + 3)/(5 + 7) = 4/12 = 1/3
(1 + 3 + 5)/(7 + 9 + 11) = 9/27 = 1/3

The pattern continues: the next fraction is (1 + 3 + 5 + 7)/(9 + 11 + 13 + 15) = 16/48 = 1/3.

Why, in general. Recall that the sum of the first n odd numbers is n².

Numerator = 1 + 3 + … + (2n – 1) = n²
Denominator = the next n odd numbers
= (sum of the first 2n odd numbers) – (sum of the first n odd numbers)
= (2n)² – n²
= 4n² – n² = 3n²

Fraction = n² ÷ 3n² = 1/3 for every n

Why it happens: the clever step is not adding the odd numbers one at a time but recognising the second group as “the first 2n odd numbers, minus the first n”. That turns an awkward sum into a subtraction of two squares. Because both the numerator and the denominator come out as multiples of n², the n² cancels — which is exactly why the answer does not depend on how many terms you take.

Check it yourself: for n = 4, numerator = 4² = 16 and denominator = 8² – 4² = 64 – 16 = 48, and 16/48 = 1/3 ✓.

Q11.

Karim and the Genie. Karim was taking a nap under a tree. He had a dream about a magical lamp and a genie… The genie said, “Do you see the banyan tree over there? All you have to do is go around it once. The money in your pocket will double”, and then, “Since I am bringing you great riches, you should share some of your gains with me. You must give me 8 coins each time you go around the tree.” He went around the tree once and the number of coins doubled; he gave 8 coins to the genie. He made another round; again the number doubled and he gave 8 more coins. He went around the tree for the third time. The number of coins doubled again, but to his horror, he was left with only 8 coins, exactly the number of coins he owed the genie! (i) How many coins did Karim initially have? (ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has? (iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim’s coins?

Answer

(i) Karim started with 7 coins. Let x be the number of coins he began with.

After round 1: 2x, then pay 8 → 2x – 8
After round 2: 2(2x – 8) = 4x – 16, then pay 8 → 4x – 24
After round 3 (the doubling): 2(4x – 24) = 8x – 48

He was left with exactly 8 coins: 8x – 48 = 8
8x = 56 (adding 48 to both sides)
x = 7 (dividing both sides by 8)

Stage	Coins
Start	7
Round 1 doubles	14
Pay the genie 8	6
Round 2 doubles	12
Pay the genie 8	4
Round 3 doubles	8
Pay the genie 8	0

(ii) The cost per round must be less than the number of coins he has. If he holds x coins and the cost is c, then after a round he holds 2x – c, and

2x – c > x ⟺ c < x (subtracting x from both sides)

In the story Karim had 7 coins, so any cost of 6 coins or less would have made him richer. With c = 7 he would stay at 7 forever, and with c = 8 he lost everything.

Why it happens: once c < x holds at the start, it keeps holding — after the round he has 2x – c, which is more than x, so it is still more than c. His pile then grows every single round, without limit. The number c is a tipping point: below it Karim wins forever, above it he is ruined, and exactly at it nothing ever changes.

(iii) The genie must set the cost at c = 2ⁿ x ÷ (2ⁿ – 1), where x is what Karim starts with and n is the number of rounds the genie wants it to take.

Coins after n rounds = 2ⁿx – c(2^n–1 + … + 2 + 1) = 2ⁿx – c(2ⁿ – 1)
For this to be 0: c(2ⁿ – 1) = 2ⁿx
c = 2ⁿx ÷ (2ⁿ – 1)

Rounds n	Cost c	With x = 7
1	2x	14
2	4x/3	28/3 — not a whole number
3	8x/7	8 ✓ — the genie’s choice

Why it happens: the fastest way is simply c = 2x — Karim doubles to 2x and hands over all of it in one round. To drag it out over n rounds the genie must charge a little less than double, namely 2x × 2ⁿ/(2ⁿ – 1), and that fraction shrinks towards 2x as n grows. In the story x = 7 and n = 3 give c = 8 × 7 ÷ 7 = 8, exactly the price the genie named — it knew Karim’s 7 coins and chose the cost that emptied his pocket in precisely three rounds.

Tip: the deal is worth taking only when the cost is below your holding, so the genie’s trap is that 8 was just one coin above Karim’s 7. Had he carried 9 coins, the same deal would have made him richer round after round.

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