Number Play
Numbers aren't just for counting — they can be codes, puzzles, and even unsolved mysteries! Try the live Supercell grid and the Kaprekar calculator below.
Numbers can tell us things
Before we get to the puzzle: think about various situations where we use numbers. List five different situations in which numbers are used. See what your classmates have listed, share, and discuss. (Hint: think about time, calendars, counting objects or marks, measuring height & weight, phone numbers, prices… numbers are hiding everywhere!)
Each child looks only at the people standing right next to them (their neighbours) and counts how many of those neighbours are taller than them:
- Says '1' if exactly one neighbour is taller.
- Says '2' if both neighbours are taller.
- Says '0' if neither neighbour is taller.
A child standing at either end of the line only has ONE neighbour (nobody on the other side) — that's the trick to a lot of these puzzles!
Try answering these (cover and think first!)
No, never! A child at the end of the line only has ONE neighbour (there's nobody on the outside). To say '2' you need BOTH your neighbours to be taller — but end children only have one neighbour to compare with, so the biggest number they could ever say is '1'.
Yes! If every child is exactly the same height, nobody has a taller neighbour, so everybody says '0'.
Yes. For example, in a line arranged shortest-to-tallest, lots of middle children each have exactly one taller neighbour (the one further along) and say '1' — so neighbours can easily match.
Yes! Line them up in order from shortest to tallest. The shortest child (at one end) has only one neighbour, who is taller — says '1'. Every middle child has one taller neighbour (the next one along) — says '1'. The tallest child (at the other end) has only one neighbour, who is shorter — says '0'. That's four 1's and one 0!
No. Think about the TALLEST child in the whole group. Wherever they stand, none of their neighbours can be taller than them (they're the tallest!). So the tallest child MUST say '0' if they're in the middle, or... actually if they're at an end they'd need a taller neighbour to say '1', which is impossible. Either way, the tallest child can never say '1' — so all five saying '1' is impossible.
Yes, it's possible! Arrange 5 children by height like a little hill: short, taller, TALLEST (middle), taller, short — heights going up then down. The two shortest ends have one taller neighbour each ('1'... wait, check the ends first: the very end children have only 1 neighbour who is taller, so they'd say '1' not '0'). To make the true ends say '0', put the two tallest children at the two ends instead, and the shortest in the middle, going: tall, shorter, shortest (middle), shorter, tall. Then each end child's only neighbour is shorter than them (say '0'), the next-in children each have one taller neighbour outward... this shape (heights going DOWN then UP, like a valley) gives exactly 0, 1, 2, 1, 0 with the shortest child in the centre saying '2' (both neighbours taller).
At most 2 children can say '2' in a line of 5. Arrange heights so they go up-down-up-down (a zig-zag): short, TALL, shorter, TALL, shortest. The two "TALL" children (in positions 2 and 4) each have both neighbours taller... no — to say '2' a child needs BOTH neighbours taller than them, so put the two SHORT children in a zig-zag valley pattern instead: tall, SHORT, taller, SHORT, tall — both short children (positions 2 and 4) have two taller neighbours each, so both say '2'. That's the maximum possible.
Supercells
Look at each number's touching neighbours only (not diagonal). If it's bigger than ALL of them, it's a supercell — colour it in!
The biggest number in the whole grid is ALWAYS a supercell (nothing can beat it). The smallest number can NEVER be a supercell (something always beats it) — unless it's totally alone with no neighbours at all!
Figure It Out — more supercell puzzles
This is a "make your own" puzzle — lots of answers work! One filled row of 9 cells (three are coloured — the 1st, 4th and 9th):
Check: 5346 > 5347? No wait — just make sure each coloured number beats BOTH its neighbours, and each plain number loses to at least one neighbour. 5346 beats 5347? No — so instead make the coloured cells clearly bigger, e.g. 5346, 2000, 1258, 1100, 1200, 1300, 9635, 9600, 9200 — here 5346 > 2000 ✓, 1258 > 2000 and 1100 ✓, 9635 > 1300 and 9600 ✓. The exact numbers don't matter — what matters is coloured cells beat their neighbours and plain cells don't!
Alternate small-BIG-small-BIG all along the row, like a zig-zag:
Cells 1, 3, 5, 7, 9 (every OTHER cell, starting from the first) are supercells — that's 5 supercells out of 9 numbers!
The trick is always the same: start the row with a supercell, then alternate small-big-small-big all the way along.
The pattern: for an even row of n cells, max supercells = n ÷ 2. For an odd row of n cells, max supercells = (n+1) ÷ 2.
No, never! Whatever numbers you choose, one of them has to be the biggest of the bunch. Wherever that biggest number sits in the table, nothing next to it can beat it — so it's always automatically a supercell. You can never avoid having at least one.
Yes, the largest number in a table is always a supercell — nothing can ever beat it. No, the smallest number can never be a supercell, because every neighbour it has will be bigger than it (that's what makes it the smallest!).
Here 9 is the largest and 8 is the second largest. Put 8 right next to 9 — since 9 > 8, the second-largest (8) loses to its neighbour and is not a supercell, even though it's the second-biggest number in the whole table!
Yes, it's possible!
Two fun ideas straight from the book: "Can you fill a row of 9 cells so that MORE than 5 are supercells?" (try it — you'll find it's impossible, since 5 is the max for 9 cells!) or "Can you fill a row of 9 cells with EXACTLY 4 supercells?"
Table 2 — five-digit supercells
Now let's use bigger numbers! Fill a 4×4 grid with 5-digit numbers made from the digits 1, 0, 6, 3, 9 (each digit used once per number, in any order) so that only the coloured cells beat all their neighbours.
Biggest digit-arrangement is 9,6,3,1,0 → 96,310. Smallest EVEN number: it must end in 0 or 6 — scanning the grid, 10,396 is the smallest one ending in an even digit. Smallest number greater than 50,000: scanning for numbers just over 50,000, 60,193 is the smallest that clears it.
Biggest: 96,310 · Smallest even: 10,396 · Smallest >50,000: 60,193. (Remember to put commas after the thousands digit, like 96,310 not 96310!)
Patterns of numbers on the number line
Place these on a line marked 1,000 to 10,000 in steps of 1,000: 2180, 2754, 1500, 3600, 9950, 9590, 1050, 3050, 5030, 5300, 8400.
Each number sits just after its "thousands" mark: 1050 and 1500 both between 1000-2000; 2180 and 2754 between 2000-3000; 3050 and 3600 between 3000-4000; 5030 and 5300 between 5000-6000; 8400 between 8000-9000; 9590 and 9950 between 9000-10,000.
Now try identifying the missing labels on these number lines (each tick is evenly spaced) and circle the smallest / box the largest in each row:
Steps of 5. Circle the smallest: 1990. Box the largest: 2035.
Steps of 1. Circle the smallest: 9993. Box the largest: 10,002. Notice how the line crosses over from 9999 straight into the 5-digit number 10,000!
Steps of 1. Circle the smallest: 15,077. Box the largest: 15,086.
Steps of 1,000 this time! Circle the smallest: 83,705. Box the largest: 92,705.
Playing with digits & palindromes
How many numbers of each length?
| Digits | Range | How many |
|---|---|---|
| 1-digit | 1 to 9 | 9 |
| 2-digit | 10 to 99 | 90 |
| 3-digit | 100 to 999 | 900 |
| 4-digit | 1000 to 9999 | 9,000 |
| 5-digit | 10000 to 99999 | 90,000 |
Each row is exactly 10× the row before it (9 → 90 → 900 → 9,000 → 90,000)! That's because adding one more digit slot multiplies your choices by 10.
Digit sums
Add up all the digits of a number to get its digit sum. For example 68 → 6+8=14. Fun fact: 176 and 545 both also have digit sum 14!
Loads of options! 248 (2+4+8=14), 653 (6+5+3=14), 356 (3+5+6=14), 815 (8+1+5=14), 833 (8+3+3=14), 12335 (1+2+3+3+5=14), 23351 (2+3+3+5+1=14) — there are endless more!
Inside each decade (40s, 50s, 60s) the digit sum goes up by 1 each time you count up by 1 — it's just the tens-digit plus the units-digit counting up. But watch the JUMP between decades: 49 (sum 13) to 50 (sum 5) — the sum suddenly drops! That's because rolling over to a new ten (like 49 → 50) resets the units digit back to 0.
Yes — every single digit sum is a multiple of 3 (6, 9, 12, 15, 18, 21, 24), and each one is exactly 3 more than the last! This pattern stops after 789 though, because 890 would need a 10th digit ("8,9,10") which isn't a single digit — so it does NOT continue forever.
From 1 to 100, digit 7 appears 20 times. From 1 to 1000, it appears 300 times! Counting is much faster if you think about which "place" (units, tens, hundreds) the 7 lands in, rather than checking every number one by one.
Palindromic patterns
A palindrome reads the same forwards and backwards, like 66, 848, or 1111.
Take any number, reverse its digits, and add. Keep going until you hit a palindrome!
For 3-digit numbers, nobody knows if reverse-and-add always works! Mathematicians strongly suspect that starting from 196, you NEVER reach a palindrome, no matter how many times you try — but nobody has proven it. It's an unsolved puzzle!
Kaprekar's magic number
Arrange your number's digits to make the biggest number and the smallest number, then subtract (smallest from biggest). Repeat with your answer. You will always land on 6174 — and once you're there, it just keeps making 6174 again forever!
The very same idea works for 3-digit numbers too — they always land on 495 instead! Try 123 → 198 → 792 → 693 → 594 → 495.
Clock and calendar numbers
All digits the same: 2:22, 3:33, 4:44. All pairs the same: 10:10, 11:11, 12:12, 9:09. Mirror/palindrome style: 12:21, 5:50, 10:01.
Lots more exist if you keep looking — try 1:11, 6:06, and think about what makes each one special!
Manish's date, written DD/MM/YYYY, is 20 12 2012 — the day and month together ("2012") exactly match the year ("2012")! So we need dates where the day and month digits, joined together, spell out the year.
In fact EVERY month from 01 to 12 gives one such date with day "19" (for years 1901-1912) and another with day "20" (for years 2001-2012) — as long as that day actually exists in that month!
Write the date as 8 digits DDMMYYYY and check if it reads the same backwards. Meghana's is 11 02 2011 → "11022011" reversed is "11022011" — a perfect match!
The trick: for an 8-digit palindrome DDMMYYYY, the LAST digit of the year must match the FIRST digit of the day, and so on working inward — try a few dates yourself with pen and paper, writing out all 8 digits and reading them backwards!
Yes! A calendar repeats itself after 6 years if exactly one leap year falls in that stretch. It repeats after just 5 years if two leap years fall in that stretch. That's because a normal year shifts the days-of-the-week by 1 (365 = 52 weeks + 1 day), but a leap year shifts them by 2 — so the shifts have to add up to a multiple of 7 to land back on the same pattern.
Figure It Out — Pratibha's digit puzzle
Pratibha uses the digits 4, 7, 3, 2 to make the largest (7432) and smallest (2347) 4-digit numbers. Their difference is 7432−2347=5085 and their sum is 7432+2347=9779.
Try digits 1, 3, 4, 7 → largest = 7431, smallest = 1347. 7431 − 1347 = 6084, and 6084 > 5085 ✓ — using a bigger spread of digits (especially a very small one like 1) widens the gap.
Try digits 3, 3, 4, 7 (repeating the 3 is fine) → largest = 7433, smallest = 3347. 7433 − 3347 = 4086, and 4086 < 5085 ✓ — digits close together keep the gap small.
Using 3, 3, 4, 7 → largest = 7433, smallest = 3347. 7433 + 3347 = 10,780, and 10,780 > 9779 ✓.
Using 1, 3, 4, 7 → largest = 7431, smallest = 1347. 7431 + 1347 = 8778, and 8778 < 9779 ✓.
Mental math puzzles
The arrow-sum puzzle
The middle column numbers (25,000 · 400 · 13,000 · 1,500 · 60,000) can be added together (using each as many times as needed) to reach the numbers on the sides.
1,000 is NOT possible — the only middle number smaller than 1,000 is 400, and 1,000 is not a multiple of 400 (400, 800, 1200... it jumps right past 1,000).
So 14,000, 15,000 and 16,000 are ALL possible — only 1,000 among these cannot be made!
The adding-and-subtracting box puzzle
This time we can BOTH add and subtract the box numbers (40,000 · 7,000 · 300 · 1,500 · 12,000 · 800) to reach the target.
These aren't the only ways to reach each target — there are lots of correct combinations! Grab the numbers 40,000 / 7,000 / 300 / 1,500 / 12,000 / 800 and see if you can find a different path to the same answers — that's great practice for mental addition and subtraction.
Digits and Operations
Adding two 5-digit numbers to get another 5-digit number:
Subtracting two 5-digit numbers to get another 5-digit number:
Figure It Out — digits and operations
Can you write an example for each of these 10 scenarios? (Some are IMPOSSIBLE — figuring out why is the real challenge!)
| # | Scenario | Example / why impossible |
|---|---|---|
| 1 | 5-digit + 5-digit > 90,250 | 45,000 + 45,400 = 90,400 ✓ > 90,250 |
| 2 | 5-digit + 3-digit = 6-digit | 99,999 + 999 = 100,998 ✓ a 6-digit sum |
| 3 | 4-digit + 4-digit = 6-digit | Impossible — even 9999+9999=19,998, only 5 digits |
| 4 | 5-digit + 5-digit = 6-digit | 60,000 + 40,000 = 1,00,000 ✓ a 6-digit sum |
| 5 | 5-digit + 5-digit = 18,500 | Impossible — smallest 5-digit + 5-digit is 10,000+10,000=20,000, already over 18,500 |
| 6 | 5-digit − 5-digit < 56,503 | 80,000 − 50,000 = 30,000 ✓ < 56,503 |
| 7 | 5-digit − 3-digit = 4-digit diff | 10,000 − 999 = 9,001 ✓ a 4-digit difference |
| 8 | 5-digit − 4-digit = 4-digit diff | 12,000 − 2,500 = 9,500 ✓ a 4-digit difference |
| 9 | 5-digit − 5-digit = 3-digit diff | 50,999 − 50,000 = 999 ✓ a 3-digit difference |
| 10 | 5-digit − 5-digit = 91,500 | Impossible — biggest possible difference is 99,999−10,000=89,999, less than 91,500 |
Always, Sometimes, Never?
Only sometimes true. 10,000+10,000=20,000 stays 5-digit, but 20,000+80,000=1,00,000 becomes a 6-digit number. Depends on the numbers!
Only sometimes true. 1,000 + 10 = 1,010 stays 4-digit, but the biggest possible case — 9,999 + 99 = 10,098 — jumps up to 5-digit. Depends on the numbers!
Only sometimes true. 90,000−10,000=80,000 stays 5-digit, but 12,000−10,000=2,000 drops to 4-digit. Depends on the numbers!
Never true. Even the smallest 5-digit number (10,000) minus the biggest 2-digit number (99) gives 10,000 − 99 = 9,901 — still 4 digits. Subtracting a tiny 2-digit number can never drop a 5-digit number all the way to 3 digits.
(Part "c. 4-digit + 2-digit gives a 6-digit number" is answered already down in Section 5!)
Patterns, estimation & games
🧩 Playing with number patterns
Instead of adding 21 boxes one by one, GROUP by value: there are 12 boxes showing "40" and 10 boxes showing "50".
The quick way: count how many boxes of EACH number there are, multiply, then add — way faster than adding 22 numbers one at a time!
Neat trick to notice: 32 boxes of "32" gives the SAME total as 16 boxes of "64" — because 16 is half of 32, and 64 is double 32! Multiplying and dividing by 2 cancel out.
Don't count every dot one by one — instead, sort the dice into two groups by what they show, count each group, then multiply:
Same trick every time: group the cells by their value, multiply each group, then add the group totals — no need to count 144 individual dots one by one!
The book also shows a snowflake-like hexagon pattern (with 15s, 25s and 35s) and a circle-of-circles pattern (with 125s, 250s, 500s and one 1000 right in the middle). These have a repeating 6-fold symmetry — the same little group of numbers repeats 6 times around the shape.
Instead of one official total, try Patto's method: count how many times ONE number appears (say, all the 125s), multiply, then move to the next number, and add all the totals together. That's the "quicker way" the book is really asking you to discover for yourself — grab the picture in your textbook and give it a go!
🔎 Estimation
Sometimes counting every single thing is impossible — so we estimate instead. Paromita worked out her school has about 500 students by noticing her class has about 100 students spread across 3 sections, times about 5 classes (6 to 10). She didn't count every single student — she reasoned it out!
Figure It Out — walking & body estimates
a. From where you're sitting to the classroom door — maybe 10–20 steps. b. Across the whole school ground — maybe 100–300 steps. c. Classroom door to the school gate — maybe 50–150 steps. d. School to home — could be hundreds to thousands of steps depending how far you live! These are all estimates — yours will depend on your own school and home. The idea is to reason sensibly, not to be exactly right.
a. In a minute: roughly 15–20 blinks or breaths. b. In an hour: about 60 × that, so roughly 900–1,200. c. In a day: about 24 × the hourly estimate, so tens of thousands — around 20,000–25,000 blinks or breaths in a day!
A few thousand: things like car number plates in a big parking area, or a 4-digit PIN's possible codes (0000 to 9999 is exactly 10,000, close to "a few thousand" territory). More than ten thousand: things like a person's monthly salary in rupees, or the range of mobile phone numbers possible.
Estimate the answer (guess within 30 seconds!)
A full-length textbook usually has more than 5,000 words — just one page can have 150-300 words, and a textbook has well over 100 pages!
This truly depends on YOUR school — a small school might have fewer than 200 bus-riders, a big school with many bus routes could easily have more. Estimate based on how many buses you see and roughly how many students fit on each one!
It depends! ₹100 IS enough if you buy a small quantity of each fruit (say, 1 of each type) and a modest amount of milk. But it would NOT be enough if you bought a large serving size or picked costly fruits like berries. So ₹100 is a reasonable rough estimate, but not guaranteed.
Roughly 2,500 kilometres — Gandhinagar is in the far west of India and Kohima is in the far north-east, so this is one of the longest distances you could measure within the country!
No, 13,000 hours is too high! Let's check with real numbers: about 6 school hours a day, about 200 school days a year.
A Grade 6 student has been in school (Nursery, KG, then Classes 1-6) for about 8 years, not almost 11. So 13,000 hours is an over-estimate — the real number is closer to 8 × 1,200 ≈ 9,600 hours.
a. A nearby favourite spot: maybe 10–30 minutes of walking. b. A neighbouring state's capital, often hundreds of km away: walking non-stop at ~5 km/h, that's days to weeks of nonstop walking! c. India's southernmost point (Kanyakumari) to northernmost point is roughly 3,000+ km — walking that whole way at ~5 km/h, non-stop, would take roughly 600+ hours, which is about 25 days of NONSTOP walking (in real life, with rest, it would take months)!
Try these: "How many students are there in your whole school?" or "About how many hours does a person spend sleeping over their entire lifetime?" Challenge a friend to guess in 30 seconds!
♾️ The Collatz Conjecture — an unsolved mystery
Rule: pick any number. If it's even, halve it. If it's odd, multiply by 3 and add 1. Repeat forever. German mathematician Lothar Collatz guessed in 1937 that you always, eventually, land on 1 — no matter which number you start with. Nobody has ever found a number that breaks the rule, but nobody has proven it always works either! It's still unsolved today.
🎲 Winning strategy games
Two players take turns adding 1, 2, or 3 to a running total (starting at 0). Whoever says 21 wins!
The first player says a number from 1 to 10. Then players take turns adding a number from 1 to 10 to the previous total. Whoever reaches 99 first wins!
If the biggest move allowed is m, the "safe gap" between control numbers is always m+1. In Game #1 the biggest move is 3, so the gap is 4 (1,5,9,13,17,21). In Game #2 the biggest move is 10, so the gap is 11 (1,12,23,...,89). Make your own version by picking any biggest move and any winning number — then work out the control numbers the same way!
Every question from the book
9435 and 3780. 9435 is the biggest number in the whole row, so it automatically wins. 3780 beats both its neighbours (670 and 3708). Every other number loses to at least one neighbour.
96310 — put the biggest digits first! Sort 9,6,3,1,0 from biggest to smallest and line them up.
Smallest: try to use as FEW digits as possible, and make the first digit as small as possible. 59 works (5+9=14) and nothing smaller does.
Largest 5-digit: put the biggest digit up front, then zeros: 95000 (9+5+0+0+0=14). You could even make it bigger with more digits and zeros — there's no single "biggest" number ever, since you can always add more zeros!
9 palindromes total — the outer two digits must match (that's what makes it a palindrome), so you just pick the outer digit (3 choices) and the middle digit (3 choices) = 3×3=9.
Units digit must be odd (1,3,5,7,9) for the number to be odd. Try units=1: tens=2×1=2, hundreds=2×2=4. That fits (all single digits)! Any bigger units digit makes hundreds too big (10 or more, impossible for one digit).
The palindrome is 12421 (hundreds-tens-units-tens-hundreds = 4-2-1-2-4). In words: twelve thousand four hundred twenty one.
7 rounds! Try typing 5683 into the Kaprekar calculator in Section 3 to watch it happen live.
Largest: sort digits big to small → 7432. Smallest: sort small to big → 2347.
Smallest 5-digit palindrome: 10001 (can't start with 0). Largest: 99999.
Next is 11:11 ("1111"), which is 70 minutes after 10:01. The one after THAT is 12:21 ("1221"), another 70 minutes later!
Powers of 2 are always even, all the way down — so you just keep halving with no surprises:
Yes — it's obviously true for powers of 2, because halving an even number always gives another even number (until you hit 1). The REAL mystery is odd numbers, which jump around unpredictably.
It reaches 1 in 25 steps (26 numbers total, including the starting 100 and the final 1)!
Control the numbers 2, 6, 10, 14, 18, 22 (each 4 apart, since the biggest single move is 3, so 4 is the "safe gap"). Be the one to say "2" first, then always steer the total back onto that list!
Never! The biggest possible answer is 9999 + 99 = 10098, which only has 5 digits. You can never jump all the way to 6 digits just by adding a 2-digit number.
Never! Even the smallest 5-digit number (10000) minus the biggest 2-digit number (99) gives 9901, still a 4-digit number. It can never drop all the way to 3 digits.
More from Section 3.12's Figure It Out
Swap the digits 1 and 6 inside 62,871 to get 12,876:
By shrinking the center number from 62,871 down to 12,876, all FOUR of its neighbours suddenly beat it — and since none of them were beaten by anything else already, all four become supercells at once!
Just 3 rounds for 1980! Try typing YOUR OWN birth year into the Kaprekar calculator back in Section 3 — every 4-digit year (with at least 2 different digits) will land on 6174 eventually.
Every digit must come from {1, 3, 5, 7, 9}, and the number must be between 35,000 and 75,000.
If you're not allowed to repeat digits: largest becomes 73,951, smallest becomes 35,179, and closest to 50,000 becomes 51,379.
A rough estimate: about 104 weekend days (52 weeks × 2) + maybe 15-20 festival/national holidays + 30-60 vacation days (summer + winter breaks) — totalling somewhere around 150-180 days a year. Now go check your own school calendar for the EXACT number and see how close your estimate was!
Rough everyday estimates: a mug holds about 0.3-0.5 litres, a bucket holds about 15-20 litres, and a household overhead tank holds anywhere from 500 to 2,000 litres depending on its size. Notice how each one is roughly 40-100 times bigger than the last!
Lots of other combinations work too — try making up your own set of three numbers that adds to 18,670!
Chosen number: 250. One way — a small hexagon-ish ring of six 25s, plus a middle strip of three 50s:
Or even simpler: 25 boxes of "10" arranged in a 5×5 grid also sums to 25 × 10 = 250! Try picking your OWN number between 210 and 390 and designing a shape for it.
Practice like the real exam
Section A · MCQ (1 mark each)
(b) 6174
(c) forwards and backwards
Section B · short answer (2 marks each)
Section C · longer answer (3 marks each)
Largest: sort big to small = 8510. Smallest: sort small to big, but the first digit can't be 0, so swap 0 and 1: 1058.
Section D · 4 marks each
Reaches 6174 in 3 rounds! You can check this yourself in the Section 3 calculator.
Section E · case study (4 marks)
The winning "control numbers" are 1, 5, 9, 13, 17, 21 (each 4 apart).
(a) Aditi should say 1 first (going first, she can grab the first control number).
(b) The total is at 13 (a control number) before her opponent's turn. Whatever the opponent adds (1, 2, or 3), Aditi should add enough to land back on 17 — the next control number.
You did it! 🎉
🏁 Chapter 3 of 5 · Term 1 Maths · Niyati, Class 6