Chapter 3 — Number Play
Ganita Prakash · Grade 6 Mathematics
Numbers help us organise everyday life. We already use them to count and to do addition, subtraction, multiplication and division. In this chapter we keep going — looking at numbers around us, noticing patterns, and using numbers in fresh new ways.
3.1 Numbers Can Tell Us Things
Some children stand in a line. Each child calls out a number. What could these numbers mean? Here is the rule they follow:
- A child says '1' if only one taller child stands next to them.
- A child says '2' if both neighbours are taller.
- A child says '0' if no neighbour is taller.
In short: each child says how many taller neighbours they have.
Questions & Answers
Q1. Can the children stand so that the two children at the ends both say '2'?
No. A child at an end has only one neighbour, so it can never have two taller neighbours.
Q2. Can we arrange them so everyone says '0'?
Yes. If every child is the same height, no one has a taller neighbour, so all say 0.
Q3. Can two children standing next to each other say the same number?
Yes, this is possible (as seen in the pictures).
Q4. Five children of different heights — can four say '1' and one say '0'?
Yes. Line them up in increasing height. The four shorter ones each have one taller neighbour (say 1), and the tallest at the end has no taller neighbour (says 0). Sequence: 1, 1, 1, 1, 0.
Q5. For 5 children, is the sequence 1, 1, 1, 1, 1 possible?
No. The tallest child cannot have a taller neighbour, so the tallest can never say 1.
Q6. Is the sequence 0, 1, 2, 1, 0 possible?
Yes. Place two tall children at the ends (say 0), a short child in the middle (says 2 because both sides are taller), and medium-height children in between saying 1.
Q7. How can we arrange five children so that the maximum number say '2'?
At most 2 children can say 2. Arrangement like 0, 2, 0, 2, 0 works — two short children each sit between two taller ones.
3.2 Supercells
A cell is a supercell if the number inside it is larger than every number touching it (its neighbours).
Example: in a row, 626 is a supercell because it is bigger than 577 and 345 on either side. A cell at the end has only one neighbour, so it needs to beat just that one.
Q1. Mark the supercells in the row.
| 6828 | 670 | 9435 | 3780 | 3708 | 7308 | 8000 | 5583 | 52 |
Supercells (shaded): 6828, 9435, 8000.
Q2. Fill with 4-digit numbers so supercells are exactly the coloured cells.
| 5346 | 5347 | 1000 | 1258 | 1100 | 1200 | 1300 | 9635 | 9636 |
One possible answer (many exist). Each shaded cell is larger than its neighbours.
Q3. Fill 9 cells (numbers 100–1000, no repeats) to get as many supercells as possible.
| 110 | 100 | 150 | 130 | 280 | 200 | 230 | 210 | 270 |
Making the 1st, 3rd, 5th, 7th and 9th cells the big ones gives 5 supercells.
Q4. How many supercells are in the table above?
5 supercells.
Q5. How many supercells are possible for different numbers of cells? Any pattern?
- For an even number of cells n: maximum supercells = n ÷ 2 (2→1, 4→2, 6→3 …).
- For an odd number of cells n: maximum = (n+1) ÷ 2 (1→1, 3→2, 5→3, 7→4 …).
Q6. Can we fill a table (no repeats) with no supercells at all?
No. Whatever the arrangement, the largest number chosen will always beat its neighbours, so it must become a supercell.
Q7. Is the largest number always a supercell? Can the smallest ever be one?
Largest: Yes, always a supercell — no neighbour can beat it.
Smallest: Never a supercell — every neighbour is bigger than it.
Smallest: Never a supercell — every neighbour is bigger than it.
Q8. Fill a table so the second-largest number is NOT a supercell.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 9 | 8 |
Here 8 (second largest) sits next to 9, so it is not a supercell.
Q9. Fill a table so the second-largest is NOT a supercell but the second-smallest IS. Possible?
| 2 | 1 | 3 | 4 | 5 | 6 | 7 | 9 | 8 |
Yes. The second-smallest (2) sits at the start next to 1, so it is a supercell. The second-largest (8) sits next to 9, so it is not.
Q10. Make your own variations.
For example: "Fill 9 cells so there are more than 5 supercells" or "so there are exactly 4 supercells". Challenge a friend!
Supercells with More Rows (Table 2)
Now neighbours are the cells directly left, right, top and bottom. Fill Table 2 with 5-digit numbers using digits 1, 0, 6, 3, 9 in some order, so only coloured cells beat all their neighbours.
| 96,310 | 96,301 | 36,109 | 39,160 |
| 96,103 | 13,609 | 60,319 | 19,306 |
| 13,906 | 10,396 | 60,193 | 60,931 |
| 10,369 | 10,963 | 10,936 | 69,031 |
Biggest number: 96,310 ·
Smallest even number: 10,396 ·
Smallest number greater than 50,000: 60,193.
3.3 Patterns of Numbers on the Number Line
We can place numbers at their correct spots on a number line. Numbers like 1050, 1500, 2180, 2754, 3050, 3600, 5030, 5300, 8400, 9590 and 9950 each sit between the right thousand marks.
Figure it Out — Identify & label the marked positions
| Line | Numbers (smallest circled ⭘ · largest boxed ▢) | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| a | ⭘1990 | 1995 | 2000 | 2005 | 2010 | 2015 | 2020 | 2025 | 2030 | ▢2035 |
| b | ⭘9993 | 9994 | 9995 | 9996 | 9997 | 9998 | 9999 | 10000 | 10001 | ▢10002 |
| c | ⭘15077 | 15078 | 15079 | 15080 | 15081 | 15082 | 15083 | 15084 | 15085 | ▢15086 |
| d | ⭘83705 | 84705 | 85705 | 86705 | 87705 | 88705 | 89705 | 90705 | 91705 | ▢92705 |
In (a) the steps go up by 5, in (b) and (c) by 1, and in (d) by 1000.
3.4 Playing with Digits
How many numbers have 1, 2, 3, 4 and 5 digits?
| 1-digit | 2-digit | 3-digit | 4-digit | 5-digit |
|---|---|---|---|---|
| 9 | 90 | 900 | 9,000 | 90,000 |
Digit Sums of Numbers
The digit sum means adding all the digits of a number. For example, 68 → 6+8 = 14, and so do 176 → 1+7+6 = 14 and 545 → 5+4+5 = 14.
Q1. Digit sum 14
a. Other numbers: 248, 653, 356, 815, 833, 12335, 23351.
b. Smallest number with digit sum 14 = 59.
c. Largest 5-digit number with digit sum 14 = 95000.
d. You can keep making bigger numbers: 95, 9005, 900005, 90000005 … There is no largest one — you can always add more zeros in the middle.
b. Smallest number with digit sum 14 = 59.
c. Largest 5-digit number with digit sum 14 = 95000.
d. You can keep making bigger numbers: 95, 9005, 900005, 90000005 … There is no largest one — you can always add more zeros in the middle.
Q3. Digit sums of 3-digit numbers with consecutive digits (like 345). Any pattern?
123→6, 234→9, 345→12, 456→15, 567→18, 678→21, 789→24.
Pattern: every sum is a multiple of 3, going up by 3 each time. But it cannot continue forever — after 789 there are no more 3-digit numbers with consecutive digits.
Pattern: every sum is a multiple of 3, going up by 3 each time. But it cannot continue forever — after 789 there are no more 3-digit numbers with consecutive digits.
Digit Detectives — Counting the digit '7'
How many times does '7' appear from 1–100? And from 1–1000?
From 1 to 100: 20 times. From 1 to 1000: 300 times.
3.5 Pretty Palindromic Patterns
A palindrome reads the same forwards and backwards, like 66, 848, 575, 1111.
Write all 3-digit palindromes using digits 1, 2, 3.
111, 121, 131, 212, 222, 232, 313, 323, 333.
Reverse-and-Add Palindromes
Steps: Take a 2-digit number, add it to its reverse. If you get a palindrome, stop; if not, reverse and add again.
| Start | Working | Result |
|---|---|---|
| 12 | 12 + 21 | 33 ✓ |
| 47 | 47 + 74 = 121 ✓ | 121 ✓ |
| 76 | 76+67=143 → 143+341 | 484 ✓ |
Explore: Will a 2-digit number always reach a palindrome by reversing and adding?
Yes, for every 2-digit number this method always ends in a palindrome.
ImpFor 3-digit numbers this is still unknown. It is suspected that starting with 196 may never give a palindrome — an unsolved puzzle!
Puzzle Time
I am a 5-digit palindrome and an odd number. My tens digit is double my units digit. My hundreds digit is double my tens digit. Who am I?
Let units = 1 (odd). Tens = 2×1 = 2. Hundreds = 2×2 = 4. As a palindrome the digits mirror: 1 2 4 2 1.
Answer: 12421 — "Twelve thousand four hundred twenty-one".
Answer: 12421 — "Twelve thousand four hundred twenty-one".
3.6 The Magic Number of Kaprekar
D. R. Kaprekar was a maths teacher from Devlali, Maharashtra. In 1949 he found a lovely pattern with 4-digit numbers.
Take a 4-digit number (at least two different digits)
↓
Make the largest number from its digits → call it A
↓
Make the smallest number from its digits → call it B
↓
Subtract: C = A − B, then repeat with C's digits
Example starting with 6382:
| Round | A (largest) | B (smallest) | C = A − B |
|---|---|---|---|
| 1 | 8632 | 2368 | 6264 |
| 2 | 6642 | 2466 | 4176 |
| 3 | 7641 | 1467 | 6174 |
ImpYou always reach the magic number 6174, called the Kaprekar constant. For 3-digit numbers, the repeating number is 495.
Carry out the steps with a 3-digit number. What repeats?
Start with 321: 321→198→792→693→594→495→495. The number 495 keeps repeating.
3.7 Clock and Calendar Numbers
Find clock times with fun patterns (like 4:44, 10:10, 12:21).
Repeated hour: 11:11, 12:12, 10:10, 09:09 …
All-same: 2:22, 3:33, 4:44, 5:55 …
Mirror times: 12:21, 10:01, 05:50 … (think of more!)
All-same: 2:22, 3:33, 4:44, 5:55 …
Mirror times: 12:21, 10:01, 05:50 … (think of more!)
Dates where digits repeat like 20/12/2012 (2-0-1-2).
Other examples: 20/04/2004, 20/06/2006 … try more.
Palindrome dates like 11/02/2011 (reads same both ways).
Examples: 01/02/2001, 02/02/2002 … think of more.
Will a year's calendar repeat exactly?
Yes. A calendar usually repeats after 6 years if one leap year falls in between, or after 5 years if two leap years fall in between.
Figure it Out
Q1. Using 4 digits, get sums/differences bigger or smaller than the examples (diff 5085, sum 9779).
a. difference > 5085: 7431 − 1347 = 6084
b. difference < 5085: 7433 − 3347 = 4086
c. sum > 9779: 7433 + 3347 = 10780
d. sum < 9779: 7431 + 1347 = 8778
b. difference < 5085: 7433 − 3347 = 4086
c. sum > 9779: 7433 + 3347 = 10780
d. sum < 9779: 7431 + 1347 = 8778
Q2. Sum and difference of the smallest and largest 5-digit palindrome.
Smallest = 10001, Largest = 99999.
Sum = 110000, Difference = 89998.
Sum = 110000, Difference = 89998.
Q3. Now it is 10:01. When is the next palindromic time, and the one after?
Next: 11:11 — that is 70 minutes (1 hr 10 min) later.
After that: 12:21 — that is 140 minutes (2 hr 20 min) from 10:01.
After that: 12:21 — that is 140 minutes (2 hr 20 min) from 10:01.
Q4. How many rounds does 5683 take to reach the Kaprekar constant?
5683→5085→3492→7083→5652→3996→6264→4176→6174.
It takes 8 rounds.
It takes 8 rounds.
3.8 Mental Math
Middle numbers can be added (used as many times as needed) to make the side numbers. Two worked examples:
- 38,800 = 25,000 + (400 × 2) + 13,000
- 3,400 = 1,500 + 1,500 + 400
Can we make 1,000? What about 14,000, 15,000, 16,000?
1,000 — No. The only number below 1,000 is 400, and 1,000 is not a multiple of 400.
14,000 = (1,500 × 8) + (400 × 5) = 12,000 + 2,000
15,000 = 13,000 + (400 × 5) = 13,000 + 2,000
16,000 = (1,500 × 8) + (400 × 10) = 12,000 + 4,000
Only 1,000 cannot be made.
14,000 = (1,500 × 8) + (400 × 5) = 12,000 + 2,000
15,000 = 13,000 + (400 × 5) = 13,000 + 2,000
16,000 = (1,500 × 8) + (400 × 10) = 12,000 + 4,000
Only 1,000 cannot be made.
Adding and Subtracting
Using the boxes (40,000 · 7,000 · 300 · 1,500 · 12,000 · 800) with both + and −:
| Target | One way to make it |
|---|---|
| 39,800 | 40,000 − 800 + 300 + 300 |
| 45,000 | 40,000 + 7,000 − 800 − 1,500 + 300 |
| 5,900 | 7,000 − 1,500 + 300 + 300 − 200 (adjust with boxes) |
| 17,500 | 12,000 + 7,000 − 1,500 |
| 21,400 | 12,000 + 7,000 + 1,500 + 800 + 300 − 200 |
Several answers are possible — the point is to mix addition and subtraction cleverly.
Digits and Operations — Figure it Out
Q1. Give one example for each case (or explain why impossible).
- 5-digit + 5-digit > 90,250 → 45,000 + 45,400 = 90,400 ✓
- 5-digit + 3-digit → 6-digit → 99,999 + 999 = 100,998 ✓
- 4-digit + 4-digit → 6-digit → Not possible (9999 + 9999 = 19,998, only 5 digits).
- 5-digit + 5-digit → 6-digit → 60,000 + 40,000 = 100,000 ✓
- 5-digit + 5-digit = 18,500 → Not possible (smallest sum 10,000+10,000 = 20,000).
- 5-digit − 5-digit < 56,503 → 80,000 − 50,000 = 30,000 ✓
- 5-digit − 3-digit → 4-digit → 10,000 − 999 = 9,001 ✓
- 5-digit − 4-digit → 4-digit → 12,000 − 2,500 = 9,500 ✓
- 5-digit − 5-digit → 3-digit → 50,999 − 50,000 = 999 ✓
- 5-digit − 5-digit = 91,500 → Not possible (biggest difference 99,999 − 10,000 = 89,999).
Q2. Always, Sometimes or Never true?
a. 5-digit + 5-digit gives 5-digit → Sometimes (e.g. 20,000 + 80,000 = 100,000, which is 6 digits).
b. 4-digit + 2-digit gives 4-digit → Sometimes (9,999 + 99 = 10,098 is 5 digits).
c. 4-digit + 2-digit gives 6-digit → Never (biggest possible is 9,999 + 99 = 10,098, only 5 digits).
d. 5-digit − 5-digit gives 5-digit → Sometimes (12,000 − 10,000 = 2,000 is 4 digits).
e. 5-digit − 2-digit gives 3-digit → Never (10,000 − 99 = 9,901, still 4 digits).
b. 4-digit + 2-digit gives 4-digit → Sometimes (9,999 + 99 = 10,098 is 5 digits).
c. 4-digit + 2-digit gives 6-digit → Never (biggest possible is 9,999 + 99 = 10,098, only 5 digits).
d. 5-digit − 5-digit gives 5-digit → Sometimes (12,000 − 10,000 = 2,000 is 4 digits).
e. 5-digit − 2-digit gives 3-digit → Never (10,000 − 99 = 9,901, still 4 digits).
3.9 Playing with Number Patterns
When numbers are arranged in neat patterns, we can find their total by counting how many of each number instead of adding one by one.
| Figure | Quick method | Total |
|---|---|---|
| (a) 40s and 50s grid | Count each value × its count, then add | Add all values together |
| (c) 32s (top) & 64s | (number of 32s × 32) + (number of 64s × 64) | Multiply then add |
The smart way: group equal numbers and multiply, which is faster than adding each box.
3.10 An Unsolved Mystery — the Collatz Conjecture
Take any whole number and apply this rule:
- If the number is even, take half of it.
- If the number is odd, multiply by 3 and add 1.
- Repeat.
Examples that all end at 1:
- 12 → 6 → 3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
- 21 → 64 → 32 → 16 → 8 → 4 → 2 → 1
ImpIn 1937 Lothar Collatz guessed that every starting number eventually reaches 1. No one has proved or disproved it yet — it is one of the most famous unsolved problems in mathematics.
Make Collatz sequences from your own numbers. Do you always reach 1?
28 → 14 → 7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
19 → 58 → 29 → 88 → 44 → 22 → 11 → … → 1
Yes, we always reach 1. Even numbers keep halving; odd numbers turn even after 3×+1, then halve again — and the smallest even number, 2, halves to 1.
19 → 58 → 29 → 88 → 44 → 22 → 11 → … → 1
Yes, we always reach 1. Even numbers keep halving; odd numbers turn even after 3×+1, then halve again — and the smallest even number, 2, halves to 1.
3.11 Simple Estimation
Sometimes we don't need an exact count — a good estimate is enough. Example: Paromita's 3 sections have 32, 29 and 35 children (about 100). With Classes 6–10, each having 3 sections, she estimated about 500 students in the school.
Q3. Objects that are (a) a few thousand, (b) more than ten thousand in number.
A few thousand: vehicle registration numbers, 4-digit PINs.
More than ten thousand: monthly salaries, mobile numbers.
More than ten thousand: monthly salaries, mobile numbers.
Q3 (Estimate). Roshan buys milk + 3 fruits for fruit custard for 5 people, estimating ₹100. Fair?
Yes, with small quantities and one of each fruit it is possible. But not if the fruits are costly or the servings are large.
Q4 (Estimate). Distance from Gandhinagar (Gujarat) to Kohima (Nagaland).
Roughly 2,500 kilometres.
Q5. Sheetal (Grade 6) says she has spent about 13,000 hours in school. Agree?
No. With about 6 school hours a day and 200 school days a year, 13,000 ÷ (6 × 200) ≈ 10.8 years. But from Nursery/KG up to Grade 6 she has been in school only about 8 years, so 13,000 hours is too high.
Q7. Make your own estimation questions.
Examples: "How many students are in your school?" or "How many hours does a person sleep over a lifetime, on average?"
3.12 Games and Winning Strategies
Game #1 — Reach 21
First player says 1, 2 or 3. Players take turns adding 1, 2 or 3. Whoever reaches 21 first wins.
ImpWinning idea: aim to say the numbers 1, 5, 9, 13, 17, 21 (each is a multiple of 4, plus 1). If you can say these, you win. The first player can always win by starting with 1.
Game #2 — Reach 99
Players add any number from 1 to 10 each turn, trying to reach 99 first.
Winning targets are the numbers 0, 11, 22, 33, 44, 55, 66, 77, 88, 99 (multiples of 11). Reach these and control the game.
Figure it Out
Q1. Only one supercell in this grid. Swap two digits of one number to get 4 supercells.
| 16,200 | 39,344 | 29,765 |
| 23,609 | 12,876 | 45,306 |
| 19,381 | 50,319 | 38,408 |
Swap the digits 1 and 6 in 62,871 to make it 12,876. Now the corners 39,344, 29,765, 50,319 and 45,306 all become supercells — giving 4 supercells.
Q2. How many rounds does your birth year take to reach the Kaprekar constant?
Example — birth year 1980:
9810 − 1089 = 8721 → 8721 − 1278 = 7443 → 7443 − 3447 = 3996 → 6264 → 4176 → 6174.
That is 6 rounds. (Try it for your own year of birth.)
9810 − 1089 = 8721 → 8721 − 1278 = 7443 → 7443 − 3447 = 3996 → 6264 → 4176 → 6174.
That is 6 rounds. (Try it for your own year of birth.)
Q3. Group of 5-digit numbers between 35,000 and 75,000 with all odd digits. Largest? Smallest? Closest to 50,000?
| Digits may repeat | Digits all different | |
|---|---|---|
| Largest | 73,999 | 73,951 |
| Smallest | 35,111 | 35,179 |
| Closest to 50,000 | 51,111 | 51,379 |
Q6. One 5-digit and two 3-digit numbers with sum 18,670.
18,000 + 300 + 370 = 18,670. (Many other combinations work too.)
Q7. Pick a number between 210 and 390 and build a pattern that sums to it.
Chosen number: 250.
Pattern A: two 25s + three 50s + two 25s = (4×25) + (3×50) = 100 + 150 = 250. ✓
Pattern B: a 5×5 grid of 10s = 25 × 10 = 250. ✓
Pattern A: two 25s + three 50s + two 25s = (4×25) + (3×50) = 100 + 150 = 250. ✓
Pattern B: a 5×5 grid of 10s = 25 × 10 = 250. ✓
Q8. Why is the Collatz conjecture true for all powers of 2?
A power of 2 (like 128 = 2×2×2×2×2×2×2) is even, so it keeps halving each step, staying a power of 2. It goes 128 → 64 → 32 → … → 2 → 1, always reaching 1.
Q9. Does the Collatz Conjecture hold for 100?
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, it reaches 1.
Q10. Start at 0, add 1–3 each turn, first to reach 22 wins. Winning strategy?
Aim to say the multiples of 4 leading to 22 — that is 2, 6, 10, 14, 18, 22. The first player can win by starting with 2 and always keeping to these targets.