Number Play
Chapter 3 — Full Solutions
Supercells, palindromes, the Kaprekar constant, the Collatz conjecture, and every Math Talk and Figure It Out question — answered step‑by‑step with clean worked numbers and original diagrams.
In‑Text Questions
Math Talk, Explore and Try This prompts asked right in the middle of the story — before you reach a formal “Figure It Out” box.
Five everyday situations where numbers are used:
- Time — reading a clock or setting an alarm.
- Calendar — tracking dates, days, and months.
- Counting objects / marks — counting items or recording test scores.
- Measurement — height, weight, and length.
- Money — prices, change, and savings.
(There could be many more — comparing your list with classmates is the real point of this activity.)
3.1 Numbers can Tell us Things
Page 55–56
Hint check: heights are exactly what decide the number! The rule (revealed on the next page) is:
- A child says 1 if exactly one neighbour is taller.
- A child says 2 if both neighbours are taller.
- A child says 0 if neither neighbour is taller.
In short — each child announces how many of their immediate neighbours are taller than them.
No. A child at either end of the line has only one neighbour, not two. Since “2” requires both neighbours to be taller, an end child can never say 2 — the maximum an end child can say is 1.
Yes — but only if every child in the line is exactly the same height. If no neighbour is ever taller than any other child, every single child will say ‘0’.
Yes. Look back at the very first picture on page 55 — you can find neighbouring children who both say ‘1’, or both say ‘0’. Saying the same number just means both children happen to have the same “taller‑neighbour count”, which can easily happen.
Yes — arrange the 5 children in strictly ascending order of height (shortest to tallest). Then every child except the tallest has exactly one taller neighbour (the one right after them), so they say ‘1’. The tallest child at the end has no taller neighbour, so they say ‘0’.
No. The tallest child in the whole group can never have a taller neighbour, so wherever the tallest child stands, they must say ‘0’ — never ‘1’. Since one position is forced to be 0, the sequence “1,1,1,1,1” is impossible.
Yes, it’s possible. Arrange heights so they rise to a peak in the middle and fall back down symmetrically (short, medium, tallest, medium, short). The two end children have no taller neighbour on their outside (0), the middle child has two taller neighbours (2), and the two children beside the middle each have exactly one taller neighbour (1).
At most 2 children can say ‘2’ at the same time. Arrange them so the heights alternate short–tall–short–tall–short (a “zig‑zag”): the two shorter children tucked between two taller neighbours each say ‘2’, while the taller children and the two ends say ‘0’.
Why not more? A child can only say ‘2’ if both neighbours are taller — but two “2”-sayers can never be next to each other (they’d need to be taller than each other), which limits the count to 2 out of 5.
Neighbours here mean the cell immediately left, right, above and below. One valid filled table:
| 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 in the table = 96,310 · Smallest even number = 10,396 · Smallest number greater than 50,000 = 60,193
3.3 Patterns of Numbers on the Number Line
Page 59
Following the same idea for every number (look at the thousands digit to find the right gap, then the hundreds digit to place it within that gap):
- Between 1000–2000: 1050, 1500
- Between 2000–3000: 2180, 2754
- Between 3000–4000: 3050, 3600
- Between 5000–6000: 5030, 5300
- Between 8000–9000: 8400
- Between 9000–10000: 9590, 9950
3.4 Playing with Digits
Page 60–61
| 1‑digit | 2‑digit | 3‑digit | 4‑digit | 5‑digit |
|---|---|---|---|---|
| 9 | 90 | 900 | 9,000 | 90,000 |
Pattern: each step is exactly $10\times$ the one before it!
$68 \to 6+8=14$, $176 \to 1+7+6=14$, $545 \to 5+4+5=14$. Even though the numbers themselves are completely different, their digit sum (the sum of all the digits) can be exactly the same — this is the idea explored further in the Figure It Out questions below.
1–100 → the digit 7 occurs 20 times.
1–1000 → $20 \times 10 = 300$ — the digit 7 occurs 300 times. (1000 itself contributes none.)
3.5 Pretty Palindromic Patterns
Page 61–62
A 3‑digit palindrome has the form (first digit)(middle digit)(first digit again) — the outer digits must match, and the middle digit can be anything.
9 palindromes in total: 3 choices for the outer digit × 3 choices for the middle digit.
Yes! For every 2‑digit starting number, repeatedly reversing and adding always eventually produces a palindrome.
Every 2‑digit number reaches a palindrome — though a few (like 89) take many rounds! (Footnote: for 3‑digit starting numbers this is still an open question — it’s suspected that 196 never reaches a palindrome.)
The digit places are labelled tth (ten‑thousands), th (thousands), h (hundreds), t (tens), u (units).
The number is 1 2 4 2 1 → Twelve thousand, four hundred and twenty‑one.
3.6 The Magic Number of Kaprekar
Page 62–63
D.R. Kaprekar was a school mathematics teacher who loved exploring number patterns. In 1949 he discovered that repeating a simple “largest minus smallest” process on almost any 4‑digit number always leads to the same number: 6174, now called the Kaprekar constant.
Continuing the process keeps giving 6174 again and again — this is the Kaprekar constant. Every 4‑digit number with at least two different digits eventually reaches it.
Try starting with 321:
For 3‑digit numbers, the process always settles on 495 — the “Kaprekar constant” for 3 digits.
3.7 Clock and Calendar Numbers
Page 64
All-same-digit times (like 4:44): 1:11, 2:22, 3:33, 4:44, 5:55.
Repeated-pair times (like 10:10): 10:10, 11:11, 12:12, and also 01:01 through 09:09.
Palindromic times (like 12:21): 10:01, 11:11, 12:21, 05:50, 04:40, 03:30, 02:20, 01:10, and more — any time where reading the 4 digits backward gives the same time.
Dates where the day and month digits repeat as the first two digits of the year:
These are palindromic dates in DD/MM/YYYY form:
(Try writing the date as 8 digits and check it reads the same reversed — e.g. 11022011 reversed is 11022011.)
Yes — calendars do repeat!
- If exactly one leap year falls within the gap, the calendar repeats after 6 years.
- If two leap years fall within the gap, it repeats sooner, after 5 years.
This happens because a normal year shifts the weekday of a fixed date by 1 day, and a leap year shifts it by 2 days — the calendar realigns once these shifts add up to a multiple of 7.
3.8 Mental Math
Page 65–66
Each number on the sides is built by adding up a combination of the numbers in the middle column — and the same middle number can be reused as many times as needed. For example: $38{,}800 = 25{,}000 + 400\times2 + 13{,}000$ and $3{,}400 = 1{,}500+1{,}500+400$. It rewards spotting round‑number combinations instead of adding digit‑by‑digit.
1,000 is impossible. The smallest middle number is 400, and $1000$ is not a multiple of $400$ — and every other middle number (1500, 13000, 25000, 60000) is already bigger than 1000, so no combination of additions can land exactly on 1000.
Only the thousand 1,000 itself is impossible to build from this set.
Given example: $39{,}800 = 40{,}000-800+300+300$. Following the same style:
(Many other combinations work too — these are just one valid path each.)
3.9 Playing with Number Patterns
Page 67–68
The quicker way: group the repeated numbers first, then multiply.
Total = $480+500 = \mathbf{980}$ — far quicker than adding 22 separate numbers one by one!
This 8×8 grid uses a “dot count” code instead of printed digits: a plain dot = value 1, a cluster of six dots = value 6. Using the same grouping trick:
Total = $44+120=\mathbf{164}$.
Total = $1{,}024+1{,}024 = \mathbf{2{,}048}$.
Like figure (b), this grid encodes each cell’s value as a die‑face dot count. The method stays exactly the same even though this one isn’t perfectly rectangular in colour: sort cells into groups that share the same value, count each group, multiply, then add the group totals. That one habit — “group identical values before adding” — is the real takeaway of this whole section, and it’s what makes patterns (e) and (f) below manageable too.
The hexagon has clean 6‑fold rotational symmetry — all six big wedges are identical copies of each other, just rotated. So instead of reading all the numbers across the whole hexagon:
This symmetry shortcut — “solve one slice, then multiply by the number of slices” — turns a 40‑plus-number addition into one small sum and one multiplication.
This is the most elegant one — look at each ring separately, from the centre outward:
Every ring after the centre adds exactly 2,000 — because each ring has twice as many cells as the last, but holds half the value! Total = $1{,}000+2{,}000+2{,}000+2{,}000 = \mathbf{7{,}000}$.
3.10 An Unsolved Mystery — The Collatz Conjecture
Page 68–69
Yes — check sequence (a): start at 12 (even) → $12\div2=6$ (even) → $6\div2=3$ (odd) → $3\times3+1=10$ (even) → $10\div2=5$ (odd) → $5\times3+1=16$ → $16\to8\to4\to2\to1$. Every sequence given follows this same even‑halve / odd‑triple‑plus‑one rule, and all four examples eventually reach 1.
Yes, both reach 1 — and every whole number ever tested reaches 1 eventually. Since even numbers keep shrinking by half and odd numbers are forced back into an even number ($3n+1$ is always even when $n$ is odd), the sequence keeps getting nudged downward, though nobody has ever proven it must always reach 1 — which is exactly why it’s still an unsolved conjecture!
3.11 Simple Estimation
Page 69–71
This is the heart of estimation: round each messy number to a friendly nearby number, then combine — you don’t need the exact headcount to get a useful answer.
3.12 Games and Winning Strategies
Page 71–72
The first player can always force a win by making sure they are the one to say 17, 13, 9, 5, and 1 — numbers 4 apart, counting down from 21.
Winning numbers to claim: 1, 5, 9, 13, 17, 21.
Same idea, bigger step. Since each turn can add 1 through 10, a full “round” between both players can always be forced to add up to 11 (opponent adds $x$, you reply with $11-x$).
The first player should claim 1, 12, 23, 34, 45, 56, 67, 78, 89, then finally 99 — each 11 apart — to guarantee the win.
Exercise Questions (Figure It Out)
Every formally‑headed “Figure It Out” box from the chapter, section by section.
3.2 Supercells — Figure It Out
Page 57–58
A cell is a supercell if it’s bigger than every number directly next to it. Checking each cell against its neighbour(s):
| 6828 | 670 | 9435 | 3780 | 3708 | 7308 | 8000 | 5583 | 52 |
Supercells: 6828, 9435, 8000 — each is bigger than the number(s) immediately beside it.
One valid way — make each coloured cell bigger than its neighbours by exactly 1 more, and keep the “filler” cells low and flat:
| 5346 | 5347 | 1000 | 1258 | 1100 | 1200 | 1300 | 9635 | 9636 |
Wait — check the pattern carefully: the coloured cells in the question are the 2nd, 4th, and 8th cells. Placing 5347 (just above 5346), 1258 as a peak between two 1000‑ish neighbours, and 9636 as the final peak keeps exactly those three positions as supercells.
Alternate high, low, high, low, … — every “high” cell then automatically beats its low neighbours on both sides:
| 110 | 100 | 150 | 130 | 280 | 200 | 230 | 210 | 270 |
This gives 5 supercells out of 9 cells — the maximum possible for 9 cells in a row.
| Number of cells (n) | Maximum supercells |
|---|---|
| Even ($n$) | $n \div 2$ |
| Odd ($n$) | $(n+1)\div 2$ |
Strategy: start by making the first cell a supercell, then alternate high–low–high–low all the way along the row. This “every other cell” pattern always achieves the maximum possible count.
No, it’s impossible. Whichever cell ends up holding the single greatest number in the whole table will automatically be bigger than every one of its neighbours — no matter where you place it, that cell is forced to become a supercell.
Largest number: Yes, always — it is automatically bigger than every neighbour, since nothing in the table can exceed it.
Smallest number: Never — every one of its neighbours will be bigger than it (or equal, if repeats were allowed), so it can never satisfy the “bigger than all neighbours” rule.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 9 | 8 |
The second‑largest number here is 8, but it sits right next to 9 (the largest) — so 8 is not a supercell, since it has a bigger neighbour.
Yes, it’s possible!
| 2 | 1 | 3 | 4 | 5 | 6 | 7 | 9 | 8 |
Here, 2 (the second‑smallest number) sits between 1 and 3 — both smaller than it — so 2 is a supercell. Meanwhile 8 (the second‑largest) sits next to 9, so it is not.
A few ideas to try:
- Can you fill a 9‑cell table so there are more than 5 supercells?
- Can you fill a 9‑cell table with exactly 4 supercells?
- What’s the minimum number of supercells a filled table can have?
3.3 Number Line — Figure It Out
Page 59
Each number line’s tick spacing is found from the two given labels, then extended in both directions.
(a) Given 2010 and 2020 → gaps of 10:
| 1990 | 1995 | 2000 | 2005 | 2010 | 2015 | 2020 | 2025 | 2030 | 2035 |
circled (smallest) = 1990 · boxed (largest) = 2035
(b) Given 9996 and 9997 → gaps of 1:
| 9993 | 9994 | 9995 | 9996 | 9997 | 9998 | 9999 | 10000 | 10001 | 10002 |
circled = 9993 · boxed = 10002
(c) Given 15,077 / 15,078 / 15,083 → gaps of 1:
| 15077 | 15078 | 15079 | 15080 | 15081 | 15082 | 15083 | 15084 | 15085 | 15086 |
circled = 15077 · boxed = 15086
(d) Given 86,705 and 87,705 → gaps of 1,000:
| 83705 | 84705 | 85705 | 86705 | 87705 | 88705 | 89705 | 90705 | 91705 | 92705 |
circled = 83705 · boxed = 92705
3.4 Playing with Digits — Figure It Out
Page 60–61
There is no biggest number with digit sum 14 — you can always insert one more zero to make a bigger one, forever.
The digit sum climbs steadily as the units digit rises, then drops sharply and starts over each time the tens digit increases:
| Numbers | Digit sums |
|---|---|
| 40–49 | 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 |
| 50–59 | 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 |
| 60–69 | 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 |
| 70 | 7 |
Observation: within any block of ten (like 40–49), the digit sum simply increases by 1 each time. But crossing into a new ten (49→50) the digit sum drops — because the tens digit increases by 1 but the units digit resets from 9 back to 0.
| Number | Digit sum |
|---|---|
| 123 | 1+2+3 = 6 |
| 234 | 2+3+4 = 9 |
| 345 | 3+4+5 = 12 |
| 456 | 4+5+6 = 15 |
| 567 | 5+6+7 = 18 |
| 678 | 6+7+8 = 21 |
| 789 | 7+8+9 = 24 |
Yes — every digit sum is a multiple of 3, and each one is exactly 3 more than the last. This pattern cannot continue past 789, though, since there’s no 3‑digit number with consecutive digits after that (there’s no digit after 9).
3.7 Clock and Calendar Numbers — Figure It Out
Page 64–65
Choosing digits closer together (like 3,3,4,7) shrinks the difference; choosing digits farther apart (like 1,3,4,7) widens it.
Sum = $10{,}001+99{,}999=\mathbf{110{,}000}$ · Difference = $99{,}999-10{,}001=\mathbf{89{,}998}$
| Round | A − B = C |
|---|---|
| 1 | 8653 − 3568 = 5085 |
| 2 | 8550 − 0558 = 3492 |
| 3 | 9432 − 2349 = 7083 |
| 4 | 8730 − 0378 = 5652 |
| 5 | 6552 − 2556 = 3996 |
| 6 | 9963 − 3699 = 6264 |
| 7 | 6642 − 2466 = 4176 |
| 8 | 7641 − 1467 = 6174 |
It takes 8 rounds to reach the Kaprekar constant 6174.
3.8 Mental Math — Figure It Out
Page 66–67
| Scenario | Example / Reason |
|---|---|
| 5‑digit + 5‑digit > 90,250 | 45,000 + 45,400 = 90,400 |
| 5‑digit − 5‑digit < 56,503 | 80,000 − 50,000 = 30,000 |
| 4‑digit + 4‑digit → 6‑digit sum | Not possible. Even $9999+9999=19{,}998$ — only 5 digits. |
| 5‑digit − 4‑digit → 4‑digit difference | 12,000 − 2,500 = 9,500 |
| 5‑digit + 3‑digit → 6‑digit sum | 99,999 + 999 = 100,998 |
| 5‑digit − 3‑digit → 4‑digit difference | 10,000 − 999 = 9,001 |
| 5‑digit + 5‑digit → 6‑digit sum | 60,000 + 40,000 = 100,000 |
| 5‑digit − 5‑digit → 3‑digit difference | 50,999 − 50,000 = 999 |
| 5‑digit + 5‑digit = 18,500 | Not possible. Two 5‑digit numbers total at least $10{,}000+10{,}000=20{,}000 > 18{,}500$. |
| 5‑digit − 5‑digit = 91,500 | Not possible. The biggest possible difference is $99{,}999-10{,}000=89{,}999 < 91{,}500$. |
Reasoning through the “not possible” cases is the real learning here — it’s about knowing the smallest/largest values a digit‑count can produce.
| Statement | Verdict | Example |
|---|---|---|
| (a) 5‑digit + 5‑digit = 5‑digit | Sometimes | 20,000+80,000=1,00,000 (6‑digit, so not always) |
| (b) 4‑digit + 2‑digit = 4‑digit | Sometimes | 9,999+99=10,098 (5‑digit here) |
| (c) 4‑digit + 2‑digit = 6‑digit | Never | Max possible: 9,999+99=10,098 — only 5 digits, never 6 |
| (d) 5‑digit − 5‑digit = 5‑digit | Sometimes | 12,000−10,000=2,000 (only 4‑digit here) |
| (e) 5‑digit − 2‑digit = 3‑digit | Never | Smallest possible: 10,000−99=9,901 — always 4‑digit or more |
3.11 Simple Estimation — Figure It Out
Page 69–71
Steps to walk (rough estimates, will vary by person):
- Seat to classroom door: ~10–15 steps
- Across the school ground: ~150–300 steps
- Classroom door to school gate: ~50–100 steps
- School to home: varies hugely — could be 500 steps or several thousand, depending on distance
Blinks / breaths:
Objects around us:
- A few thousand: pages printed in a school library, seats in a large auditorium, a 4‑digit PIN’s possible combinations.
- More than ten thousand: hairs on your head, grains of rice in a kilogram bag, students’ mobile phone numbers (10 digits).
Estimate the Answer
Page 70–71
More than 5000. A single page of a textbook typically has 150–300 words, and a textbook runs to well over 100 pages — so the total easily crosses several thousand.
This depends entirely on your specific school — a small school might have fewer than 200 bus riders, while a large school could easily have more. (Estimate based on your own school and compare with classmates.)
It depends on the quantity and fruit chosen. ₹100 is realistic if he buys small quantities of budget-friendly fruit (like banana, apple, papaya) plus a small milk packet. But it would not be enough if he chooses more expensive fruits (like grapes, kiwi, or pomegranate) or buys larger quantities for generous servings.
Gandhinagar sits in the west of India and Kohima in the far northeast — roughly 2,500 km apart as the crow flies, one of the longer cross‑country spans within India.
No — 13,000 hours is far too high.
$\dfrac{13{,}000}{6\times200}\approx 10.8$ years of schooling — but she’s only been in school about 8 years, so her estimate is noticeably too high.
- (a) Nearby favourite place (say 1–2 km): about 15–25 minutes of walking.
- (b) Neighbouring state capital (say 200–400 km): at a normal walking pace of ~4–5 km/h for 8 hours a day, that’s roughly 6–12 days of continuous walking.
- (c) Kanyakumari to the northern tip of India (~3,700 km): at the same pace, that’s around 90–115 days of non‑stop daily walking — clearly why long‑distance transport was so valuable historically!
- How many students study in your school?
- On average, how many hours does a person sleep across their entire lifetime?
- How many times does your heart beat in a single day?
3.12 Games and Winning Strategies — Figure It Out
Page 72–73
Original grid (only 62,871 in the centre is a supercell):
| 16,200 | 39,344 | 29,765 |
| 23,609 | 62,871 | 45,306 |
| 19,381 | 50,319 | 38,408 |
Swap the digits ‘6’ and ‘1’ in 62,871 to get 12,876:
| 16,200 | 39,344 | 29,765 |
| 23,609 | 12,876 | 45,306 |
| 19,381 | 50,319 | 38,408 |
Shrinking the centre number to 12,876 lets 16,200, 39,344, 29,765, and 50,319 all become supercells — four in total.
Worked example for the birth year 1980 (written as 4‑digit 1980):
| Round | A − B = C |
|---|---|
| 1 | 9810 − 0189 = 8721 |
| 2 | 8721 − 1278 = 7443 |
| 3 | 7443 − 3447 = 3996 |
| 4 | 9963 − 3699 = 6264 |
| 5 | 6642 − 2466 = 4176 |
| 6 | 7641 − 1467 = 6174 |
1980 takes 6 rounds to reach 6174. (Try it with your own birth year — the number of rounds varies!)
$18{,}000 + 300 + 370 = 18{,}670$ ✓
(Many other combinations work — as long as the three numbers add up exactly to 18,670.)
Chosen number: 250
Every power of 2 is always even until it finally becomes 1. Since the Collatz rule simply halves even numbers, starting from $2^k$ you just keep dividing by 2 — $2^k \to 2^{k-1} \to 2^{k-2} \to \cdots \to 4 \to 2 \to 1$ — with no odd numbers ever appearing along the way. So the sequence is guaranteed to reach 1 directly, with no detours.
$100,50,25,76,38,19,58,29,88,44,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1$
Yes — it reaches 1 after 25 steps, confirming the conjecture for this starting value.
Same idea as Game #1 from earlier in this section — since each turn adds 1, 2, or 3, a full round between both players can always be forced to add exactly 4.
The first player should claim 2, 6, 10, 14, 18, 22 (each 4 apart) to guarantee the win.