NCERT Solutions Ganita Prakash Chapter 3 Figure it Out — Games and Winning Strategies

Book page 72 & 73 Updated on2026-09-05

Q1.

There is only one supercell (number greater than all its neighbours) in this grid: 16,200 | 39,344 | 29,765 / 23,609 | 62,871 | 45,306 / 19,381 | 50,319 | 38,408. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap.

Answer

At present only 62,871 (the centre cell) is a supercell — it is bigger than 39,344, 23,609, 45,306 and 50,319. Because this one number towers over the whole grid, it blocks everybody else.

Swap the digits 6 and 1 of 62,871 — that is, the first digit and the last digit — to get 12,876.

16,200	39,344 ✓	29,765
23,609 ✓	12,876	45,306 ✓
19,381	50,319 ✓	38,408

Now check each of the four:

39,344 > 16,200, 29,765 and 12,876 ✓
23,609 > 16,200, 12,876 and 19,381 ✓
45,306 > 29,765, 12,876 and 38,408 ✓
50,319 > 12,876, 19,381 and 38,408 ✓

Answer: swap the 6 and the 1 in 62,871 to make it 12,876 — the grid then has 4 supercells.

Why it works: every one of the four “middle-edge” cells touches the centre. Once the centre number becomes small (12,876), each of them wins over all its own neighbours at the same time.

Q2.

How many rounds does your year of birth take to reach the Kaprekar constant?

Answer

Take the year 2014 as an example (use your own year in the same way).

Round	Largest	Smallest	Difference
1	4210	1024	3186
2	8631	1368	7263
3	7632	2367	5265
4	6552	2556	3996
5	9963	3699	6264
6	6642	2466	4176
7	7641	1467	6174

2014 takes 7 rounds. Two more examples:

2013: 3210 − 1023 = 2187; 8721 − 1278 = 7443; 7443 − 3447 = 3996; 9963 − 3699 = 6264; 6642 − 2466 = 4176; 7641 − 1467 = 6174 → 6 rounds
1980: 9810 − 1089 = 8721; 8721 − 1278 = 7443; 7443 − 3447 = 3996; 9963 − 3699 = 6264; 6642 − 2466 = 4176; 7641 − 1467 = 6174 → 6 rounds

Tip: when a year contains a 0 (as 2014 does), the smallest 4-digit number cannot begin with that 0 — for the digits 2, 0, 1, 4 the smallest number is 1024.

Q3.

We are the group of 5-digit numbers between 35,000 and 75,000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50,000?

Answer

Only the digits 1, 3, 5, 7, 9 may be used, and the number must lie between 35,000 and 75,000.

Largest → 73,999

The first digit can be at most 7 (9 is too big — the number must stay below 75,000).
With 7 in front, the next digit must be less than 5, so it is 3.
The rest can be the biggest odd digit, 9 → 73,999

Smallest → 35,111

The first digit must be at least 3.
With 3 in front, the number must be more than 35,000, so the next digit is 5.
The rest are the smallest odd digit, 1 → 35,111

Closest to 50,000 → 51,111

Below 50,000 the best we can do is 39,999 (4 is not an odd digit!) → 50,000 − 39,999 = 10,001 away
Above 50,000 the smallest is 51,111 → 51,111 − 50,000 = 1,111 away
1,111 < 10,001, so 51,111 is the closest

Why 4 is the villain: every number in the 40,000s starts with the digit 4, which is even. So the group has no member at all between 39,999 and 51,111 — a gap of more than eleven thousand.

Q4.

Estimate the number of holidays you get in a year including weekends, festivals and vacation. Then, try to get an exact number and see how close your estimate is.

Answer

Estimate first — count the big pieces:

Type of holiday	Rough count
Sundays	52
Second Saturdays	12
Festivals (Diwali, Holi, Eid, Christmas, Pongal …)	about 15
Summer vacation	about 40
Winter and other breaks	about 15
Estimated total	about 134 days

Now the exact count. A school year has about 220 working days, so

365 − 220 = 145 holidays

The estimate 134 is quite close to 145 — off by about 11 days, which is less than a tenth. That is good estimation!

Check it yourself: some festival holidays fall on a Sunday, so counting them twice is a common mistake. Cross them off your calendar to avoid it.

Q5.

Estimate the number of liters a mug, a bucket and an overhead tank can hold.

Answer

Container	Estimate	How to check
Mug	about 1 litre	a 1-litre milk packet just fills it
Bucket	about 15 – 20 litres	about 15 mugs fill one bucket
Overhead tank	about 500 – 1000 litres	about 40 – 60 buckets fill it

1 bucket ≈ 15 mugs ≈ 15 litres
1 tank ≈ 50 buckets ≈ 50 × 15 = 750 litres

The method: start from something you know for sure — a 1-litre milk packet. Then measure the bucket in mugs, and the tank in buckets. Each step multiplies, so you can estimate very big amounts from a very small one.

Q6.

Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.

Answer

Choose the two 3-digit numbers first, then work out the 5-digit number by subtraction.

Let the 3-digit numbers be 670 and 500.
670 + 500 = 1,170
18,670 − 1,170 = 17,500
Check: 17,500 + 670 + 500 = 18,670 ✓

Answer: 17,500 + 670 + 500 = 18,670. Two more correct answers:

18,000 + 370 + 300 = 18,670 ✓
16,870 + 900 + 900 = 18,670 ✓

Tip: the two 3-digit numbers can add up to anything from 200 to 1998, so the 5-digit part can be anything from 16,672 to 18,470. There are thousands of correct answers.

Q7.

Choose a number between 210 and 390. Create a number pattern similar to those shown in Section 3.9 that will sum up to this number.

Answer

Choose 250. Build it from four 25s and three 50s:

4 × 25 = 100
3 × 50 = 150
100 + 150 = 250 ✓

A pattern like those in Section 3.9 — two rows of two 25s with a row of three 50s in between. The numbers add up to 250.

Another pattern for the same total: a 5 × 5 square of 10s.

25 tiles × 10 = 250 ✓

Try This: make one for 360 — six rows of six 10s (36 × 10 = 360), or three 100s and three 20s. Then ask a friend to find the sum quickly.

Q8.

Recall the sequence of Powers of 2 from Chapter 1, Table 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence?

Answer

The powers of 2 are 1, 2, 4, 8, 16, 32, 64, 128, … — and every one of them is even.

The Collatz rule says: if the number is even, halve it. Half of a power of 2 is the power of 2 just before it, which is again even. So the sequence only halves, step after step, and never uses the “3 × it + 1” part at all.

64 → 32 → 16 → 8 → 4 → 2 → 1
128 → 64 → 32 → 16 → 8 → 4 → 2 → 1

Why it must reach 1: 2 × 2 × 2 × 2 × 2 × 2 (that is 64) loses one 2 at every halving. After exactly six halvings no 2 is left and only 1 remains. In general 2 taken n times reaches 1 in exactly n steps — so the Collatz conjecture is certainly true for every power of 2.

Did you notice? Every Collatz sequence ends in 16, 8, 4, 2, 1 — that is, as soon as any sequence lands on a power of 2, the rest of the journey is pure halving.

Q9.

Check if the Collatz Conjecture holds for the starting number 100.

Answer

Apply the rule step by step — even numbers get halved, odd numbers become 3 × it + 1.

100 → 50 → 25 → 76 (25 is odd: 25 × 3 + 1 = 76) → 38 → 19 → 58 → 29 → 88 → 44 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

The full sequence is

100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

Yes, the Collatz conjecture holds for 100. It reaches 1 in 25 steps.

Check it yourself: from 22 onwards this is exactly sequence (d) printed on page 68 of the book — 22, 11, 34, 17, 52, … 1.

Q10.

Starting with 0, players alternate adding numbers between 1 and 3. The first person to reach 22 wins. What is the winning strategy now?

Answer

The largest addition allowed is 3, so the magic step is 3 + 1 = 4. Working backwards from 22 in steps of 4:

22, 18, 14, 10, 6, 2

The first player wins. The strategy is:

Start by saying 2.
After that, whatever your friend adds (1, 2 or 3), you add the number that makes the round total 4 — if they add 1 you add 3, if they add 2 you add 2, if they add 3 you add 1.
The totals you announce will be 2, 6, 10, 14, 18 and finally 22 — you win.

You say	2	6	10	14	18	22 🏆
Friend can only reach	3, 4 or 5	7, 8 or 9	11, 12 or 13	15, 16 or 17	19, 20 or 21	—

Why 2 is the right start: 22 ÷ 4 = 5 with remainder 2. That remainder is exactly the number the first player must say on the very first move. If the remainder had been 0 (as with the target 20), the second player would win instead.

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