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Class 6 Ganita Prakash · Chapter 3
Class 6 Ganita Prakash · Chapter 3
Number Play
Page-wise textbook solutions (pages 55–73) · Every step · Figures · Common mistakes · Tips
๐️ Key ideas of this chapter
- Supercell: a number bigger than all its neighbours. The largest number is always one; the smallest never is.
- Count of n-digit numbers: 9, 90, 900, 9000, 90,000. Digit sum: add the digits (59 → 14).
- Palindromes read the same both ways (12,421). Reverse-and-add makes palindromes for 2-digit numbers.
- Kaprekar: largest − smallest, repeat → 6174 (4 digits) or 495 (3 digits).
- Collatz: even → half, odd → 3n + 1; it seems always to reach 1 (unsolved!).
- Games: in ‘add 1–3, reach 21’ say 1, 5, 9, 13, 17, 21 (jumps of 4).
๐ Textbook Page 55
๐ Page 55Math Talk
List five different situations in which numbers are used.
1Telling the time and reading a calendar.
2Buying things and counting money (prices, bills, change).
3Phone numbers, house numbers, bus and train numbers.
4Marks in tests and scores in games.
5Measuring height, weight, distance and temperature.
✅ Answer: Time and dates, money, phone/house/bus numbers, marks and scores, measuring — and many more.
๐ Textbook Page 56
๐ Page 56Think · Taller neighbours
Children stand in a line and each says a number. What do the numbers mean?The heights (h) and what each child says.
1Look at each child’s two neighbours (the children just next to them).
2A child says 0 if neither neighbour is taller, 1 if one neighbour is taller, 2 if both are taller.
3So each child says the number of taller neighbours they have.
✅ Answer: Each child says how many of their neighbours are taller than them.
๐ Page 56Math Talk · Q1
Can the children rearrange themselves so that the children standing at the ends say ‘2’?
1A child at an end has only ONE neighbour.
2So they can have at most one taller neighbour.
✅ Answer: No. A child at an end can say only 0 or 1.
๐ Page 56Math Talk · Q2
Can we arrange the children in a line so that all would say only 0s?
1Take any two children standing next to each other.
2One of them is shorter, so that child has a taller neighbour and says at least 1.
✅ Answer: No (when there are 2 or more children). Someone always has a taller neighbour.
๐ Page 56Math Talk · Q3
Can two children standing next to each other say the same number?
1Both say 0? No — the shorter of the two has a taller neighbour.
2Both say 2? No — the taller of the two does not have the other as a taller neighbour.
3Both say 1? Yes — for example heights in increasing order 1, 2, 3, 4, 5 say 1, 1, 1, 1, 0.
✅ Answer: Yes, but only both saying ‘1’ (never both 0 or both 2).
๐ Page 56Math Talk · Q4
There are 5 children of different heights. Can they stand so that four of them say ‘1’ and the last one says ‘0’?
1Stand them in increasing order of height (shortest first).
2Each of the first four has a taller child just after them → says 1.
3The tallest child at the end has no taller neighbour → says 0.
✅ Answer: Yes: shortest to tallest gives 1, 1, 1, 1, 0.
๐ Page 56Math Talk · Q5
For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
1The tallest child can never have a taller neighbour.
2So the tallest child always says 0.
✅ Answer: No. The tallest child always says 0.
๐ Page 56Math Talk · Q6
Is the sequence 0, 1, 2, 1, 0 possible? Why or why not?
1The middle child must be shorter than both neighbours (says 2).
2The children at the ends must be taller than their neighbours (say 0).
3Try heights 5, 3, 1, 2, 4: they say 0, 1, 2, 1, 0 ✔
✅ Answer: Yes — for example heights 5, 3, 1, 2, 4.
๐ Page 56Math Talk · Q7
How would you arrange the five children so that the maximum number of children say ‘2’?
1Children at the ends cannot say 2.
2Two children next to each other cannot both say 2.
3So only positions 2 and 4 can say 2 → at most 2 children.
4Example: heights 3, 1, 5, 2, 4 say 0, 2, 0, 2, 0.
✅ Answer: Put the two shortest children in the 2nd and 4th places: at most 2 children can say ‘2’.
๐ Textbook Page 57
๐ Page 57Think · Supercells
Why are some numbers coloured in the tables?
| 43 | 79 | 75 | 63 | 10 | 29 | 28 | 34 |
| 200 | 577 | 626 | 345 | 790 | 694 | 109 | 198 |
1Look at 79: its neighbours are 43 and 75 — 79 is bigger than both.
2Look at 626: its neighbours are 577 and 345 — 626 is bigger.
3A cell at the end has only one neighbour: 198 > 109, so 198 is coloured.
✅ Answer: A cell is coloured (a supercell) when its number is bigger than all its neighbouring numbers.
๐ Page 57Figure it Out · Q1
Colour or mark the supercells in the table: 6828, 670, 9435, 3780, 3708, 7308, 8000, 5583, 52.
| 6828 | 670 | 9435 | 3780 | 3708 | 7308 | 8000 | 5583 | 52 |
16828 > 670 → supercell.
29435 > 670 and > 3780 → supercell.
33780 < 9435; 3708 < 3780; 7308 < 8000 → not supercells.
48000 > 7308 and > 5583 → supercell.
5670, 5583 and 52 have a bigger neighbour → not supercells.
| 6828 | 670 | 9435 | 3780 | 3708 | 7308 | 8000 | 5583 | 52 |
✅ Answer: Supercells: 6828, 9435 and 8000.
๐ Page 57Figure it Out · Q2
Fill the table with only 4-digit numbers so that the supercells are exactly the coloured cells (2nd, 4th and 9th cells).
| 5346 | 1258 | 9635 |
1Cell 2 must be bigger than 5346 → take 7000.
21258 (cell 4) must be a supercell → cells 3 and 5 must be smaller than 1258 (but 4-digit): 1100 and 1200.
3Cell 9 must be bigger than 9635 → 9900.
4Cells 6 and 7 must NOT be supercells → make them increase towards 9635: 2000, 3000.
5Check: only 7000, 1258 and 9900 are bigger than all their neighbours ✔
| 5346 | 7000 | 1100 | 1258 | 1200 | 2000 | 3000 | 9635 | 9900 |
✅ Answer: 5346, 7000, 1100, 1258, 1200, 2000, 3000, 9635, 9900 (one possible answer).
๐ Page 57Figure it Out · Q3–Q4
Fill a table of 9 cells (numbers between 100 and 1000, no repeats) to get as many supercells as possible. How many supercells are there?
1Two neighbouring cells cannot both be supercells.
2So the best is every other cell: cells 1, 3, 5, 7, 9.
3Put big numbers in the odd places and small numbers in the even places: 900, 100, 800, 200, 700, 300, 600, 400, 500.
| 900 | 100 | 800 | 200 | 700 | 300 | 600 | 400 | 500 |
✅ Answer: 5 supercells (e.g. 900, 100, 800, 200, 700, 300, 600, 400, 500).
๐ Page 57Figure it Out · Q5 (Math Talk)
Find how many supercells are possible for different numbers of cells. Do you notice a pattern? What is the method?
11 cell → 1; 2 cells → 1; 3 cells → 2; 4 cells → 2; 5 cells → 3; 6 cells → 3; …
2Odd number of cells n → (n + 1) ÷ 2. Even number of cells n → n ÷ 2.
3Method: put large numbers in cells 1, 3, 5, … and small numbers in between.
| Number of cells | Most supercells |
|---|---|
| 1 | 1 |
| 2 | 1 |
| 3 | 2 |
| 4 | 2 |
| 5 | 3 |
| 6 | 3 |
| 7 | 4 |
| 8 | 4 |
| 9 | 5 |
| 10 | 5 |
✅ Answer: The maximum is half the cells, rounded up: (n + 1) ÷ 2 for odd n, n ÷ 2 for even n.
๐ Textbook Page 58
๐ Page 58Figure it Out · Q6
Can you fill a supercell table without repeating numbers so that there are no supercells? Why or why not?
1The largest number in the table is bigger than every other number.
2So it is bigger than all its neighbours → it is always a supercell.
✅ Answer: No. The cell with the largest number is always a supercell.
๐ Page 58Figure it Out · Q7
Will the cell with the largest number always be a supercell? Can the cell with the smallest number be a supercell?
1Largest number: bigger than all neighbours → always a supercell.
2Smallest number: all its neighbours are bigger → never a supercell (if the table has more than one cell).
✅ Answer: The largest is always a supercell; the smallest can never be one.
๐ Page 58Figure it Out · Q8
Fill a table so that the cell with the second largest number is NOT a supercell.
1Put the second largest number right next to the largest one.
2Example: 100, 900, 800, 300, 200 — 800 is next to 900, so it is not a supercell.
| 100 | 900 | 800 | 300 | 200 |
✅ Answer: For example 100, 900, 800, 300, 200.
๐ Page 58Figure it Out · Q9
Fill a table so that the second largest number is not a supercell but the second smallest number is a supercell. Is it possible?
1Second largest (800) next to the largest (900) → not a supercell.
2Second smallest (200) at an end, next to the smallest (100) → it is a supercell.
3Table: 200, 100, 800, 900, 300.
| 200 | 100 | 800 | 900 | 300 |
✅ Answer: Yes, it is possible — for example 200, 100, 800, 900, 300.
๐ Page 58Figure it Out · Q10
Make other variations of this puzzle and challenge your classmates.
1Example 1: fill 7 cells so that exactly 2 are supercells.
2Example 2: fill a 3 × 3 grid with 3-digit numbers so that only the corners are supercells.
3Example 3: use only the digits 2, 5, 8 to make 3-digit numbers for a table with the most supercells.
✅ Answer: Many answers — see the examples.
๐ Page 58Table 1 · Check
In Table 1, check why 7500, 9870, 8632 and 6034 are supercells (neighbours: left, right, top and bottom). Is any other cell a supercell?
| 2430 | 7500 | 7350 | 9870 |
| 3115 | 4795 | 9124 | 9230 |
| 4580 | 8632 | 8280 | 3446 |
| 5785 | 1944 | 5805 | 6034 |
17500: neighbours 2430, 7350, 4795 → all smaller ✔
29870: neighbours 7350, 9230 → smaller ✔
38632: neighbours 4580, 8280, 4795, 1944 → smaller ✔
46034: neighbours 5805, 3446 → smaller ✔
5Look at the corner 5785: its neighbours are 4580 and 1944 — both smaller! So 5785 is also a supercell.
| 2430 | 7500 | 7350 | 9870 |
| 3115 | 4795 | 9124 | 9230 |
| 4580 | 8632 | 8280 | 3446 |
| 5785 | 1944 | 5805 | 6034 |
✅ Answer: Each of the four is bigger than all its neighbours. By the same rule 5785 is a supercell too (the book does not colour it).
๐ก Tip: Always check the corner cells — they have only two neighbours.
๐ Page 58Complete Table 2
Complete Table 2 with 5-digit numbers using the digits 1, 0, 6, 3, 9 (each once). Only coloured cells should be supercells. Find the biggest number, the smallest even number and the smallest number greater than 50,000.
| 96,301 | 36,109 | ||
| 13,609 | 60,319 | 19,306 | |
| 60,193 | |||
| 10,963 |
1Top-left must beat 96,301 → only 96,310 works (the biggest possible number).
2Below 10,963 (a supercell) and beside it, put smaller numbers starting with 10: 10,369, 10,396, 10,639.
3Top-right must beat 36,109 and 19,306 → 63,910. Bottom-right must beat its neighbours → 90,631.
4Other cells: 31,690, 30,169, 39,016 (each has a bigger neighbour).
5Biggest = 96,310. Even numbers end in 0 or 6: smallest is 10,396. Numbers above 50,000: smallest is 60,193.
| 96,310 | 96,301 | 36,109 | 63,910 |
| 31,690 | 13,609 | 60,319 | 19,306 |
| 30,169 | 10,369 | 60,193 | 39,016 |
| 10,396 | 10,963 | 10,639 | 90,631 |
✅ Answer: Biggest 96,310; smallest even number 10,396; smallest number greater than 50,000 is 60,193 (for our filling).
๐ก Tip: The biggest number must be 96,310 in every correct answer; the other two answers depend on how you fill the table.
๐ Textbook Page 59
๐ Page 59Place on the number line
Place 2180, 2754, 1500, 3600, 9950, 9590, 1050, 3050, 5030, 5300 and 8400 on a number line from 1000 to 10,000.
1Find the two thousands each number lies between: 2180 and 2754 lie between 2000 and 3000.
21050 is just after 1000; 1500 is half-way between 1000 and 2000.
33050 is just after 3000; 3600 is a little past half-way to 4000.
45030 is just after 5000; 5300 is about one-third of the way to 6000.
58400 is a little less than half-way to 9000; 9590 and 9950 lie between 9000 and 10,000 (9950 is almost 10,000).
✅ Answer: See the number line.
๐ Page 59Figure it Out
Identify the numbers marked on the number lines (a)–(d) and label the remaining positions. Circle the smallest and box the largest number.(a) (b) (c) (d)
1(a) 2010 and 2020 are 2 ticks apart → each tick is 5. Labels: 1990, 1995, …, 2035.
2(b) 9996, 9997 are next to each other → each tick is 1: 9993 … 10,002.
3(c) 15,077, 15,078 → step 1: 15,077 … 15,086 (15,083 is the 7th tick ✔).
4(d) 86,705 → 87,705 → each tick is 1000: 83,705 … 92,705.
✅ Answer: (a) 1990–2035 (smallest 1990, largest 2035) (b) 9993–10,002 (c) 15,077–15,086 (d) 83,705–92,705.
๐ Textbook Page 60
๐ Page 60Playing with digits
How many numbers have one digit, two digits, three digits, four digits and five digits?
11-digit: 1 to 9 → 9.
22-digit: 10 to 99 → 99 − 10 + 1 = 90.
33-digit: 100 to 999 → 900.
44-digit: 1000 to 9999 → 9000.
55-digit: 10,000 to 99,999 → 90,000.
| 1-digit | 2-digit | 3-digit | 4-digit | 5-digit |
|---|---|---|---|---|
| 9 | 90 | 900 | 9,000 | 90,000 |
✅ Answer: 9, 90, 900, 9,000 and 90,000.
๐ก Tip: Each time the count is 10 times bigger.
๐ Page 60Figure it Out · Q1 (a)–(b)
Digit sum 14: (a) write other numbers whose digits add up to 14 (b) what is the smallest number with digit sum 14?
1(a) 59, 68, 77, 86, 95, 149, 239, 446, 527, 1238, 2075 … (check: 5 + 9 = 14).
2(b) One digit cannot make 14 (largest is 9). With two digits, make the tens digit as small as possible: 14 − 9 = 5 → 59.
✅ Answer: (a) e.g. 59, 68, 77, 149, 1238 (b) 59
๐ Page 60Figure it Out · Q1 (c)–(d)
Digit sum 14: (c) the largest 5-digit number (d) how big a number can you form? Can you make an even bigger number?
1(c) Put the biggest digits first: 9, then 14 − 9 = 5, then zeros → 95,000.
2(d) Use zeros! 950, 9500, 95,000, 9,50,000 … all have digit sum 14.
3We can keep adding zeros, so there is no biggest number.
✅ Answer: (c) 95,000 (d) As big as we like — 9 5 followed by any number of zeros; there is always a bigger one.
๐ Page 60Figure it Out · Q2
Find the digit sums of all the numbers from 40 to 70. Share your observations.
140 → 4, 41 → 5, …, 49 → 13.
250 → 5, 51 → 6, …, 59 → 14.
360 → 6, …, 69 → 15; 70 → 7.
4Within each ten the digit sum goes up by 1; from 49 to 50 it drops by 8.
5Different numbers can have the same digit sum (49, 58, 67 all give 13).
| n | sum | n | sum | n | sum |
|---|---|---|---|---|---|
| 40 | 4 | 50 | 5 | 60 | 6 |
| 41 | 5 | 51 | 6 | 61 | 7 |
| 42 | 6 | 52 | 7 | 62 | 8 |
| 43 | 7 | 53 | 8 | 63 | 9 |
| 44 | 8 | 54 | 9 | 64 | 10 |
| 45 | 9 | 55 | 10 | 65 | 11 |
| 46 | 10 | 56 | 11 | 66 | 12 |
| 47 | 11 | 57 | 12 | 67 | 13 |
| 48 | 12 | 58 | 13 | 68 | 14 |
| 49 | 13 | 59 | 14 | 69 | 15 |
| 70 | 7 |
✅ Answer: Digit sums go up by 1 in each ten and drop at every new ten; the largest is 15 (for 69).
๐ Page 60Figure it Out · Q3
Find the digit sums of 3-digit numbers whose digits are consecutive (like 345). Do you see a pattern? Will it continue?
1123 → 6, 234 → 9, 345 → 12, 456 → 15, 567 → 18, 678 → 21, 789 → 24.
2Each sum is 3 more than the one before.
3The sum is always 3 × the middle digit (345 → 3 × 4 = 12).
4It stops at 789 because there is no digit after 9 (the same sums come for 321, 432, … too).
✅ Answer: 6, 9, 12, …, 24 — they go up by 3 (= 3 × middle digit). The pattern stops at 789.
๐ Textbook Page 61
๐ Page 61Digit Detectives
Among 1–100, how many times does the digit 7 occur? Among 1–1000?
11–100, ones place: 7, 17, 27, …, 97 → 10 times.
2Tens place: 70 to 79 → 10 times (77 counted in both places — it has two 7s).
3Total = 20.
41–1000: 7 appears 100 times in the ones place, 100 in the tens place and 100 in the hundreds place (700–799).
✅ Answer: 20 times (1–100) and 300 times (1–1000).
๐ Page 61All palindromes using 1, 2, 3
Write all possible 3-digit palindromes using the digits 1, 2 and 3.
1A 3-digit palindrome looks like ABA (first and last digits the same).
2A can be 1, 2 or 3; B can be 1, 2 or 3 → 3 × 3 = 9.
✅ Answer: 111, 121, 131, 212, 222, 232, 313, 323, 333
๐ก Tip: Digits may repeat, as in 222.
๐ Page 61Reverse-and-add
Start with a 2-digit number, add its reverse, and repeat until you get a palindrome. Try some numbers.
139 + 93 = 132; 132 + 231 = 363 ✔ (2 steps).
275 + 57 = 132; 132 + 231 = 363 ✔.
368 + 86 = 154; 154 + 451 = 605; 605 + 506 = 1111 ✔ (3 steps).
489 takes 24 steps and reaches 88,13,20,00,23,188!
✅ Answer: Every 2-digit number reaches a palindrome; most take only a few steps, but 89 (and 98) take 24 steps.
๐ Textbook Page 62
๐ Page 62Explore (Math Talk)
Will reversing and adding repeatedly, starting with a 2-digit number, always give a palindrome?
1Try many: 10 → 11, 12 → 33, 19 → 110 → 121, 59 → 154 → 605 → 1111.
2Every 2-digit number we try reaches a palindrome; the slowest are 89 and 98.
3For 3-digit numbers nobody knows — 196 is thought never to give a palindrome.
✅ Answer: Yes, for all 2-digit numbers (89 and 98 take the longest — 24 steps).
๐ Page 62Puzzle time
I am a 5-digit palindrome. I am odd. My tens digit is double my units digit. My hundreds digit is double my tens digit. Who am I?
1Palindrome: tth = u and th = t.
2Odd → units digit is 1, 3, 5, 7 or 9.
3Hundreds = 2 × tens = 4 × units must be a single digit → units can only be 1 or 2; odd → 1.
4Units 1, tens 2, hundreds 4, thousands 2, ten-thousands 1 → 12421.
✅ Answer: 12,421 — twelve thousand four hundred twenty-one.
๐ Textbook Page 63
๐ Page 63Kaprekar’s magic
Continue the steps for 6382 (A = largest, B = smallest, C = A − B, repeat). What happens?
18632 − 2368 = 6264.
26642 − 2466 = 4176.
37641 − 1467 = 6174.
47641 − 1467 = 6174 again — it stays at 6174.
| Round | A (largest) | B (smallest) | C = A − B |
|---|---|---|---|
| 1 | 8632 | 2368 | 8632 − 2368 = 6264 |
| 2 | 6642 | 2466 | 6642 − 2466 = 4176 |
| 3 | 7641 | 1467 | 7641 − 1467 = 6174 |
✅ Answer: We reach 6174 (Kaprekar’s constant) and it keeps repeating.
๐ Page 63Explore · 3-digit numbers
Carry out the same steps with a few 3-digit numbers. Which number starts repeating?
1Example 352: 532 − 235 = 297.
2972 − 279 = 693; 963 − 369 = 594.
3954 − 459 = 495; 954 − 459 = 495 again.
| Round | A (largest) | B (smallest) | C = A − B |
|---|---|---|---|
| 1 | 532 | 235 | 532 − 235 = 297 |
| 2 | 972 | 279 | 972 − 279 = 693 |
| 3 | 963 | 369 | 963 − 369 = 594 |
| 4 | 954 | 459 | 954 − 459 = 495 |
✅ Answer: 495 keeps repeating (for 3-digit numbers whose digits are not all the same).
๐ Textbook Page 64
๐ Page 64Clock numbers
On a 12-hour clock find all times like 4:44 (same digits), 10:10 (repeated) and 12:21 (palindromes).
1All digits the same: 1:11, 2:22, 3:33, 4:44, 5:55 and 11:11.
2Hour repeated as minutes: 10:10, 11:11, 12:12.
3Palindromes h:mh — for each hour 1 to 9 there are 6 (e.g. 1:01, 1:11, 1:21, 1:31, 1:41, 1:51) → 54.
4With 2-digit hours: 10:01, 11:11, 12:21 → 3 more. Total 57 palindromic times.
✅ Answer: Same digits: 6 times; hour repeated: 10:10, 11:11, 12:12; palindromic times: 57 (54 + 10:01, 11:11, 12:21).
๐ Page 64Calendar numbers
Manish’s birthday 20/12/2012 has ‘2012’ repeated. Find other such dates. Meghana’s birthday 11/02/2011 is a palindrome. Find all such dates from the past.
1Repeated form: the date and month together equal the year: 20/01/2001, 20/02/2002, …, 20/11/2011, 20/12/2012; also 19/10/1910, 19/11/1911, 19/12/1912.
2Palindrome form: the year is the date and month written backwards.
3Since 2000: 10/02/2001, 20/02/2002, 01/02/2010, 11/02/2011, 21/02/2012, 02/02/2020, 12/02/2021, 22/02/2022.
4(Going back to the year 1001 there are 69 such dates in all; the next one is 03/02/2030.)
✅ Answer: e.g. 20/11/2011, 19/12/1912; palindromic dates 10/02/2001, 20/02/2002, 01/02/2010, 11/02/2011, 21/02/2012, 02/02/2020, 12/02/2021, 22/02/2022.
๐ Page 64Try This · Reusing a calendar
Will any year’s calendar repeat after some years? Will all dates and days match exactly?
1A calendar repeats when the year starts on the same day AND both years are ordinary (or both leap) years.
22026 starts on a Thursday and is not a leap year.
32015 and 2037 also start on Thursday and are not leap years → same calendar as 2026.
4An ordinary year’s calendar repeats after 6 or 11 years; a leap year’s calendar usually repeats after 28 years.
✅ Answer: Yes. For example the 2026 calendar is the same as 2015 and 2037.
๐ Page 64Figure it Out · Q1 (a)
Using 4, 7, 3, 2 the difference of the largest and smallest numbers is 5085. Choose 4 digits to make the difference greater than 5085.
1Choose a big digit and small digits: 9, 8, 1, 0.
2Largest 9810, smallest 1089 (0 cannot come first).
39810 − 1089 = 8721 > 5085 ✔
✅ Answer: e.g. 9, 8, 1, 0: 9810 − 1089 = 8721.
๐ Textbook Page 65
๐ Page 65Figure it Out · Q1 (b)–(d)
Choose 4 digits so that (b) the difference is less than 5085 (c) the sum of largest and smallest is greater than 9779 (d) the sum is less than 9779.
1(b) Digits close together: 5, 4, 3, 2 → 5432 − 2345 = 3087 < 5085 ✔
2(c) Big digits: 9, 8, 7, 6 → 9876 + 6789 = 16,665 > 9779 ✔
3(d) Small digits: 1, 2, 3, 4 → 4321 + 1234 = 5555 < 9779 ✔
✅ Answer: (b) 5, 4, 3, 2 (c) 9, 8, 7, 6 (d) 1, 2, 3, 4 — other answers possible.
๐ Page 65Figure it Out · Q2
What is the sum of the smallest and largest 5-digit palindromes? What is their difference?
1Smallest 5-digit palindrome: 10001. Largest: 99999.
2Sum = 10,001 + 99,999 = 1,10,000.
3Difference = 99,999 − 10,001 = 89,998.
✅ Answer: Sum 1,10,000; difference 89,998.
๐ Page 65Figure it Out · Q3
The time now is 10:01. How many minutes until the next palindromic time? And the one after that?
1After 10:01 the next palindromic time is 11:11.
210:01 → 11:01 is 60 min, + 10 min = 70 minutes.
3The one after is 12:21: 11:11 → 12:21 is again 70 minutes (140 minutes from 10:01).
✅ Answer: 70 minutes (11:11), then 70 minutes more (12:21).
๐ Page 65Figure it Out · Q4
How many rounds does 5683 take to reach the Kaprekar constant?
1Round 1: 8653 − 3568 = 5085
2Round 2: 8550 − 0558 = 7992
3Round 3: 9972 − 2799 = 7173
4Round 4: 7731 − 1377 = 6354
5Round 5: 6543 − 3456 = 3087
6Round 6: 8730 − 0378 = 8352
7Round 7: 8532 − 2358 = 6174
✅ Answer: 7 rounds.
๐ Page 65Mental Math
Use the middle numbers (25,000, 400, 13,000, 1,500, 60,000 — as many times as needed) to make each number on the sides.Numbers in the middle (used as many times as needed) add up to the numbers on the sides. Two examples are drawn.
128,000 = 25,000 + 1,500 + 1,500
261,600 = 60,000 + 400 + 400 + 400 + 400
331,000 = 25,000 + 1,500 × 4
463,000 = 60,000 + 1,500 + 1,500
519,500 = 13,000 + 1,500 × 3 + 400 × 5
620,900 = 13,000 + 1,500 × 5 + 400
✅ Answer: See the steps (other ways are possible).
๐ Textbook Page 66
๐ Page 66Math Talk
Can we make 1,000 using the middle numbers? Why not? What about 14,000, 15,000 and 16,000? Which thousands cannot be made?
1Numbers less than 1,000: only 400 and 400 + 400 = 800 (1,500 is already too big). So 1,000 cannot be made.
214,000 = 1,500 × 8 + 400 × 5; 15,000 = 1,500 × 10; 16,000 = 13,000 + 1,500 + 1,500.
32,000 = 400 × 5, 3,000 = 1,500 × 2, … every other thousand can be made.
✅ Answer: 1,000 cannot be made; 14,000, 15,000, 16,000 and all other thousands can.
๐ Page 66Adding and Subtracting
Using 40,000, 7,000, 300, 1,500, 12,000 and 800 with + and −, make 45,000, 5,900, 17,500 and 21,400.
145,000 = 40,000 + 7,000 − 1,500 − 800 + 300
25,900 = 7,000 − 800 − 300
317,500 = 12,000 + 7,000 − 1,500
421,400 = 12,000 + 7,000 + 1,500 + 1,500 − 300 − 300
✅ Answer: See the steps (other answers possible).
๐ Page 66Figure it Out · Q1
Write an example for each case whenever possible (sums and differences of 5-digit, 4-digit and 3-digit numbers).
| Question | Example (or why not) |
|---|---|
| 5-digit + 5-digit → 5-digit sum more than 90,250 | |
| 5-digit + 3-digit → 6-digit sum | |
| 4-digit + 4-digit → 6-digit sum | |
| 5-digit + 5-digit → 6-digit sum | |
| 5-digit + 5-digit → 18,500 | |
| 5-digit − 5-digit → difference less than 56,503 | |
| 5-digit − 3-digit → 4-digit difference | |
| 5-digit − 4-digit → 4-digit difference | |
| 5-digit − 5-digit → 3-digit difference | |
| 5-digit − 5-digit → 91,500 |
1For each case, try the smallest and the largest possible numbers to see what can happen.
2Three cases are impossible — see the table.
| Question | Example (or why not) |
|---|---|
| 5-digit + 5-digit → 5-digit sum more than 90,250 | 50,000 + 40,300 = 90,300 |
| 5-digit + 3-digit → 6-digit sum | 99,800 + 500 = 1,00,300 |
| 4-digit + 4-digit → 6-digit sum | Not possible: biggest is 9,999 + 9,999 = 19,998 (5 digits) |
| 5-digit + 5-digit → 6-digit sum | 60,000 + 50,000 = 1,10,000 |
| 5-digit + 5-digit → 18,500 | Not possible: smallest sum is 10,000 + 10,000 = 20,000 |
| 5-digit − 5-digit → difference less than 56,503 | 60,000 − 10,000 = 50,000 |
| 5-digit − 3-digit → 4-digit difference | 10,200 − 500 = 9,700 |
| 5-digit − 4-digit → 4-digit difference | 12,000 − 5,000 = 7,000 |
| 5-digit − 5-digit → 3-digit difference | 10,500 − 10,000 = 500 |
| 5-digit − 5-digit → 91,500 | Not possible: biggest difference is 99,999 − 10,000 = 89,999 |
✅ Answer: See the table: 4-digit + 4-digit can never be 6-digit; 5-digit + 5-digit can never be 18,500; 5-digit − 5-digit can never be 91,500.
๐ Textbook Page 67
๐ Page 67Figure it Out · Q2 (a)
Always, sometimes or never? 5-digit + 5-digit gives a 5-digit number.
1Test with small and large numbers.
210,000 + 20,000 = 30,000 (yes); 60,000 + 50,000 = 1,10,000 (no).
✅ Answer: Sometimes true.
๐ Page 67Figure it Out · Q2 (b)
Always, sometimes or never? 4-digit + 2-digit gives a 4-digit number.
1Test with small and large numbers.
21,000 + 10 = 1,010 (yes); 9,990 + 20 = 10,010 (no).
✅ Answer: Sometimes true.
๐ Page 67Figure it Out · Q2 (c)
Always, sometimes or never? 4-digit + 2-digit gives a 6-digit number.
1Test with small and large numbers.
2Biggest: 9,999 + 99 = 10,098 — only 5 digits.
✅ Answer: Never true.
๐ Page 67Figure it Out · Q2 (d)
Always, sometimes or never? 5-digit − 5-digit gives a 5-digit number.
1Test with small and large numbers.
250,000 − 10,000 = 40,000 (yes); 10,500 − 10,000 = 500 (no).
✅ Answer: Sometimes true.
๐ Page 67Figure it Out · Q2 (e)
Always, sometimes or never? 5-digit − 2-digit gives a 3-digit number.
1Test with small and large numbers.
2Smallest: 10,000 − 99 = 9,901 — still 4 digits.
✅ Answer: Never true.
๐ Page 67Number patterns (a)
Find the sum of the numbers in figure (a) quickly.(a)
1Rows of 40: 3 rows × 4 = 12 → 12 × 40 = 480. Rows of 50: 2 rows × 5 = 10 → 500.
2480 + 500 = 980
✅ Answer: 980
๐ Page 67Number patterns (b)
Find the sum of the numbers in figure (b) quickly.(b)
164 dice: 20 show 5 dots (five 2 × 2 blocks) → 100; the other 44 show 1 dot → 44.
2100 + 44 = 144
✅ Answer: 144
๐ Textbook Page 68
๐ Page 68Number patterns (c)
Find the sum of the numbers in figure (c) quickly.(c)
1Top part: 4 rows × 8 = 32 cells of 32 → 32 × 32 = 1,024.
2Bottom part: 4 rows × 4 = 16 cells of 64 → 16 × 64 = 1,024. Total = 2,048.
✅ Answer: 2048
๐ก Tip: Group equal numbers and multiply instead of adding one by one.
๐ Page 68Number patterns (d)
Find the sum of the numbers in figure (d) quickly.(d)
1Each row except the 6th: 3 + 4 + 4 + 4 + 3 = 18 → 6 rows × 18 = 108.
26th row: 3 × 5 = 15. Total = 108 + 15 = 123.
✅ Answer: 123
๐ Page 68Number patterns (e)
Find the sum of the numbers in figure (e) quickly.(e) redrawn: 22 numbers are 15, 22 are 25 and 22 are 35.
1There are 22 of each number (15, 25 and 35).
222 × (15 + 25 + 35) = 22 × 75 = 1,650.
✅ Answer: 1650
๐ Page 68Number patterns (f)
Find the sum of the numbers in figure (f) quickly.(f) redrawn: 1000 in the centre, 4 × 500, 8 × 250 and 18 × 125.
11000 + 4 × 500 + 8 × 250 + 18 × 125
2= 1,000 + 2,000 + 2,000 + 2,250 = 7,250.
✅ Answer: 7250
๐ Page 68Collatz sequences
Look at 12, 6, 3, 10, 5, 16, 8, 4, 2, 1. How were these sequences formed?
112 is even → half is 6; 6 → 3.
23 is odd → 3 × 3 + 1 = 10; 10 → 5; 5 → 16 …
3Rule: even → take half; odd → multiply by 3 and add 1.
✅ Answer: Even: halve it. Odd: multiply by 3 and add 1. Repeat.
๐ Textbook Page 69
๐ Page 69Think · Collatz
Make more Collatz sequences starting with your favourite numbers. Do you always reach 1?
17: 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
29: 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 (19 steps).
3Every number anyone has tried reaches 1, but nobody has proved it for ALL numbers.
✅ Answer: Yes, they all reach 1 so far — but it is still an unsolved problem (a conjecture).
๐ Page 69Estimation
Paromita estimated her school strength as about 500. How did she do it?
1Her class sections: 32 + 29 + 35 = 96 ≈ 100 children.
2Classes 6 to 10 → 5 classes, each about 100.
35 × 100 = 500.
✅ Answer: Each class ≈ 100 children, 5 classes → about 500.
๐ Textbook Page 70
๐ Page 70Figure it Out · Q1–Q3 (Estimates)
Estimate (1) steps you walk to the door / across the ground / to the gate / home (2) blinks or breaths in a minute, an hour, a day (3) objects that are a few thousand or more than ten thousand.
1(1) One step ≈ ½ metre: a 5 m walk to the door ≈ 10 steps; a 60 m ground ≈ 120 steps; 1 km to home ≈ 2,000 steps.
2(2) About 15 breaths a minute → 15 × 60 = 900 an hour → 900 × 24 = 21,600 a day. Blinks ≈ 15–20 a minute (about 20,000 a day awake).
3(3) A few thousand: pages in all books of the class, leaves on a small tree, bricks in a wall. More than ten thousand: hairs on the head, grains in a bag of rice, people in a town.
✅ Answer: Answers will be close to these; it is the method (estimate one, then multiply) that matters.
๐ Page 70Estimate the answer · Q1–Q2
Is the number of words in your maths textbook more or less than 5000? Are more or less than 200 students coming by bus?
1A page has about 250–300 words and the book has more than 100 pages.
2100 × 250 = 25,000 — far more than 5000.
3Bus students depend on your school — count the buses × about 40 children each.
✅ Answer: Words: more than 5000. Bus students: depends on the school (buses × 40).
๐ Page 70Estimate the answer · Q3
Roshan estimates ₹100 for milk and 3 types of fruit to make custard for 5 people. Do you agree?
11 litre of milk ≈ ₹60–70.
23 kinds of fruit (½ kg each) ≈ ₹150 or more; custard powder and sugar ≈ ₹40.
3Total ≈ ₹250 or more.
✅ Answer: No — ₹100 is too little; it would cost about ₹250.
๐ Page 70Estimate the answer · Q4
Estimate the distance from Gandhinagar (Gujarat) to Kohima (Nagaland).
1On the map, Gandhinagar is in the west and Kohima in the far north-east.
2The straight distance is about 2,200 km; by road it is about 3,000 km.
✅ Answer: About 2,200 km in a straight line (about 3,000 km by road).
๐ Textbook Page 71
๐ Page 71Estimate the answer · Q5
Sheetal (Grade 6) says she has spent about 13,000 hours in school. Do you agree?
1School days in a year ≈ 220; hours a day ≈ 6 → about 1,300 hours a year.
2From Class 1 to Class 6 ≈ 6 years → 6 × 1,300 ≈ 7,800 hours.
313,000 hours would need about 10 years of school.
✅ Answer: No — it is closer to 8,000 hours.
๐ Page 71Estimate the answer · Q6
Walking at a normal pace, about how long would it take to go (a) to a favourite place nearby (b) to a neighbouring state’s capital (c) from the southernmost to the northernmost point of India?
1Normal walking speed ≈ 5 km per hour.
2(a) 2 km away → about 25 minutes.
3(b) 300 km → 60 hours of walking (about 8 days walking 8 hours a day).
4(c) Kanyakumari to the north of Ladakh ≈ 3,500 km → 700 hours ≈ 90 days walking 8 hours a day.
✅ Answer: About ½ hour, about a week, and about 3 months (walking 8 hours a day).
๐ Page 71Games · Game #1 (21)
Players add 1, 2 or 3 in turn; the first to reach 21 wins. Who can always win, and which numbers should they say?
1To say 21 you must leave your friend at 17 (from 17 they can reach only 18, 19 or 20).
2Going back in 4s: 17, 13, 9, 5, 1 are the winning numbers.
3The first player says 1, then always says the next number in 1, 5, 9, 13, 17, 21 (adding 4 − friend’s number).
✅ Answer: The first player wins by saying 1, 5, 9, 13, 17, 21.
๐ Textbook Page 72
๐ Page 72Games · Game #2 (99)
Players add a number from 1 to 10; first to 99 wins. Who can always win? Which numbers should they say?
1Go back from 99 in steps of 11: 88, 77, 66, 55, 44, 33, 22, 11, 0.
2The first player must say 1 to 10, so they cannot say 11.
3The second player always says 11, 22, 33, … (11 − friend’s number) and reaches 99.
✅ Answer: The second player wins by saying 11, 22, 33, …, 88, 99.
๐ Page 72Figure it Out · Q1
There is only one supercell (62,871). Exchange two digits of one number to get 4 supercells.
| 16,200 | 39,344 | 29,765 |
| 23,609 | 62,871 | 45,306 |
| 19,381 | 50,319 | 38,408 |
1The 4 middle-edge numbers 39,344, 23,609, 45,306, 50,319 can become supercells if the centre becomes small.
2The centre must be smaller than 23,609 (the smallest of them).
3Swap 6 and 1 in 62,871 → 12,876.
4Now 39,344, 23,609, 45,306 and 50,319 are each bigger than all their neighbours ✔
| 16,200 | 39,344 | 29,765 |
| 23,609 | 12,876 | 45,306 |
| 19,381 | 50,319 | 38,408 |
✅ Answer: Swap the digits 6 and 1 of 62,871 to get 12,876.
๐ Page 72Figure it Out · Q2
How many rounds does your year of birth take to reach the Kaprekar constant?
1Example 2014: 4210 − 0124 = 4086; 8640 − 0468 = 8172; 8721 − 1278 = 7443; 7443 − 3447 = 3996; 9963 − 3699 = 6264; 6642 − 2466 = 4176; 7641 − 1467 = 6174 → 7 rounds.
2Example 2015: 5210 − 0125 = 5085 … → 7 rounds.
3Example 2013: 3210 − 0123 = 3087; 8730 − 0378 = 8352; 8532 − 2358 = 6174 → 3 rounds.
✅ Answer: Depends on your year: 2013 → 3 rounds, 2014 → 7 rounds, 2015 → 7 rounds.
๐ Page 72Figure it Out · Q3
5-digit numbers between 35,000 and 75,000 with all digits odd: which is the largest, the smallest and the closest to 50,000?
1Largest: first digit 7, then the second must be less than 5 and odd → 3: 73,999.
2Smallest: 3 then 5, then the smallest odd digit 1: 35,111.
3Near 50,000: just below, 39,999 (10,001 away); just above, 51,111 (1,111 away).
✅ Answer: Largest 73,999; smallest 35,111; closest to 50,000 is 51,111.
๐ Page 72Figure it Out · Q4–Q5
Estimate (4) your holidays in a year (5) litres held by a mug, a bucket and an overhead tank.
1(4) 52 Sundays + about 12 second Saturdays + about 15 festival days + about 45 vacation days ≈ 124 days.
2(5) Mug ≈ 1 litre; bucket ≈ 15–20 litres; overhead tank ≈ 500–1,000 litres.
✅ Answer: About 120–130 holidays; mug 1 L, bucket 15–20 L, tank 500–1,000 L.
๐ Page 72Figure it Out · Q6
Write one 5-digit number and two 3-digit numbers whose sum is 18,670.
1Pick two 3-digit numbers: 300 and 370 (total 670).
25-digit number = 18,670 − 670 = 18,000.
3Check: 18,000 + 300 + 370 = 18,670 ✔
✅ Answer: 18,000 + 300 + 370 (many answers).
๐ Page 72Figure it Out · Q7
Choose a number between 210 and 390 and make a number pattern (like Section 3.9) that adds up to it.
1Choose 300.
2Make a 5 × 5 square of 12s → 25 × 12 = 300.
3Or: a row of four 50s (= 200) and a row of four 25s (= 100) → 300.
✅ Answer: For example a 5 × 5 square of 12s = 300.
๐ Textbook Page 73
๐ Page 73Figure it Out · Q8
Why is the Collatz conjecture true for every starting number that is a power of 2?
1A power of 2 (like 64) is even, so we halve it: 64 → 32 → 16 → 8 → 4 → 2 → 1.
2Half of a power of 2 is again a power of 2, so it stays even until it reaches 1.
✅ Answer: Powers of 2 keep halving through powers of 2 until they reach 1.
๐ Page 73Figure it Out · Q9
Check if the Collatz conjecture holds for the starting number 100.
1100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34
217, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
✅ Answer: Yes — 100 reaches 1 after 25 steps.
๐ Page 73Figure it Out · Q10
Starting with 0, players take turns adding 1, 2 or 3. The first to reach 22 wins. What is the winning strategy?
1Go back from 22 in steps of 4: 18, 14, 10, 6, 2.
2The first player says 2, then always says the next of 6, 10, 14, 18, 22 (adding 4 − friend’s number).
✅ Answer: The first player wins by saying 2, 6, 10, 14, 18, 22.
๐ Where do we use number play?
- Estimating costs, distances and time before we start a job.
- Palindromic dates and clock times; calendars that repeat (2015 = 2026 = 2037).
- Computers use rules like Kaprekar’s and Collatz’s — this is ‘computational thinking’.
๐ Link to higher classes
- Class 6: Prime Time (factors and multiples). Class 7: Number Play (parity, magic squares).
- Class 8: A Story of Numbers; place value and number systems.
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