Class 6 Mathematics Chapter 3: Number Play
- 4 topics
- 80 MCQs
- 24 short answers
- 12 long answers
- 60 flashcards
- 20 games
- ~137 min read
- English
What Numbers Can Tell Us
Numbers can describe an arrangement, not only a quantity. Children in a row report their taller neighbours, boxes in a table compete with the boxes beside them, and a number line turns size into position.
What you will be able to do
- I can work out what each child in a row would call out, once I know their heights.
- I can explain why the two children at the ends of a row can never call out 2.
- I can find the supercells in a row of boxes and say why the others miss out.
- I can work out the worth of one step on a number line and label the blank marks.
Key points
A number can report an arrangement. Each child calls out how many of their neighbours are taller than they are.
A child at the end of a row has one neighbour only, so 2 is never an answer there.
The tallest child in a row always calls out 0, so a row of all ones cannot happen.
The shortest child always has a taller neighbour, so a row of all zeros cannot happen either.
A box is a supercell when its number beats every box touching it, however many that is.
Two supercells can never sit side by side, because neither number can beat the other.
Colouring every second box gives the most supercells a row can hold.
On a number line the worth of one step is fixed by the labels, and it can be any size at all.
Full explanation
Meera lines up her friends
Meera plays a game in the school ground. Five friends stand in a row, shoulder to shoulder. Each one looks left and right, then calls out a single number. Meera writes down what she hears: 1, 1, 0, 1, 1. Her cousin sees the list and cannot make head or tail of it. The list looks like nonsense. It is not. Those five numbers describe the whole row.
The rule they are using
Every child counts the neighbours who are taller than they are. A child with a taller person on one side only calls out 1. A child with taller people on both sides calls out 2. A child who is taller than everyone beside them calls out 0. So the number is not the height. It is a count of taller neighbours.
Why the two ends are different
A child in the middle has a neighbour on the left and a neighbour on the right. A child at the end of the row has only one neighbour. That single fact decides a lot. An end child can call 0 or 1 and nothing else. Calling 2 would mean two taller neighbours, and there is nobody there to be the second one.
The tallest child always says nothing but zero
Somebody in the row is taller than everybody else. Nobody beside that child is taller. So the tallest child always calls out 0. This works however the children shuffle themselves. It also means a row can never be all ones. Wherever the tallest child stands, a 0 has to appear in the list.
Can every child call zero?
Try it and you will fail. Calling 0 means being taller than both your neighbours. Now think about the shortest child in the row. Somebody standing beside that child is taller, because everybody is. So the shortest child has to call 1 or 2, never 0. A row of all zeros is impossible.
Neighbours can call the same number
Meera's list begins 1, 1. That surprises people. It is easy to arrange. Put a short child, then a medium child, then a tall child. The first child sees a taller neighbour and says 1. The second child sees one shorter and one taller, so says 1 as well. Two neighbours saying the same thing is perfectly ordinary.
Reading the list backwards
The real fun starts when you are given the numbers and asked about the row. A 2 in the list marks a dip: that child is shorter than both neighbours. A 0 marks a peak. Peaks and dips cannot sit side by side, because a child cannot be taller and shorter than the same neighbour. So the list carries the shape of the row.
A table of numbers that compete
Now leave the ground and open your notebook. Write seven numbers in a row of boxes: 412, 806, 375, 940, 268, 719, 153. Colour a box when its number beats every number touching it. A coloured box is called a supercell. Three of these seven qualify. Find them before reading on.
Checking each box in turn
806 sits between 412 and 375, and beats both, so it is a supercell. 940 sits between 375 and 268, and beats both. 719 sits between 268 and 153, and beats both. The other four all lose to a neighbour. So the supercells here are 806, 940 and 719.
The end boxes count their own way
A box at the end of a row touches only one other box. So it has to beat one number, not two. That makes it easier for an end box to win, not harder. In our row 412 still loses, because 806 sits next to it. Swap 412 for 999 and the end box would win.
Two supercells never touch
Suppose two boxes sit side by side and both are coloured. The left one must beat the right one. The right one must beat the left one. Two numbers cannot each be bigger than the other. So that can never happen, and the coloured boxes always have a gap between them.
So how many can a row hold?
Colour one box, skip one, colour the next, and so on. That is the most you can manage. A row of seven boxes holds at most four supercells. A row of eight also holds four. A row of nine holds five. Work it out by filling a row with big and small numbers turn by turn.
Numbers have places, not just sizes
A number line does the same job in a different way. It turns size into position. The further right a number sits, the bigger it is. Two numbers you cannot tell apart in a list are obvious on a line, because one of them stands further along than the other.
One step is worth whatever the line says
Read the labels before you read the marks. On one line each step may be worth 200, so the marks run 3000, 3200, 3400 and onwards. On another each step may be worth 2, so the marks run 45670, 45672, 45674. The picture looks the same. Only the labels tell you the difference.
Finding a mark nobody has labelled
Very often a line labels two marks and leaves the rest blank. Count the gaps between the two labels. Share the difference out equally among them. That gives the worth of one step, and from there every blank mark can be filled in, going left as easily as right.
Concepts and terms
A number that describes a position rather than a quantity. In the lining-up game the number a child calls out counts their taller neighbours. It tells you nothing about the child's own height on its own.
Whoever stands immediately beside you, or whichever box touches yours. A child in the middle of a row has two neighbours and a child at the end has one. Counting neighbours first saves most of the mistakes.
A box whose number is greater than the number in every box touching it. An end box has to beat one neighbour and a middle box has to beat two, so the test is easier at the ends of a row.
The amount by which each mark rises above the mark before it. The step may be 1, or 200, or anything else. Two lines can look identical and carry completely different numbers.
Watch out!
Myth or reality
Case Study
Scenario:
A class is given two figures for a town: a population of 48,000 and a bus fare of 15 rupees. Meena is asked which number carries more information. She says the bigger one. Their teacher asks what each number would tell a visitor. The population says how many people live there, and that shapes how many schools and buses the town needs. The fare says what one journey costs. The class decides that neither is more informative, because they answer different questions. Meena rewrites her answer. A number means nothing on its own, and what it is counting is as much a part of it as its size.
Analysis:
Meena ranked the numbers by size. What a number counts, and the question it answers, decide what it can tell you.
Activity:
Find two numbers about your town. Write what question each one answers and who would need it.
Line up and call out
Title:
Line up and call out
What You Notice:
Every list holds at least one zero, and the ends never say 2. In the boxes, no two coloured cells ever end up touching, and four is the most you can squeeze out of seven boxes.
Did you know
A single row of numbers can hide a picture. The numbers 0, 2, 0, 2, 0 describe a row that goes tall, short, tall, short, tall.
Computers sort huge lists by asking the same small question again and again: is this item bigger than the one beside it?
Mapmakers mark hilltops and valley floors in much the same way, by comparing a place with the land touching it.
Memory tricks
What a child at the end of a row can say — One neighbour, one chance — An end child compares with a single person. So the answer is 0 or 1. There is no second neighbour to make it 2.
Spotting supercells quickly — Beat both sides, or the one side you have — Count the boxes touching yours first. Then check your number against each of them. Nothing else matters.
Practice MCQs with answers (20)
- A How tall that child is in centimetres
- B Whether the child is standing on the left of the row
- C How many children are in the row altogether
- D How many neighbours that child has
Show answer
Why:
An end child compares with one person and a middle child with two, so the count of neighbours comes first.
- A Only 0
- B Only 1
- C 0 or 1
- D 0, 1 or 2
Show answer
Why:
One neighbour means at most one taller neighbour can be counted.
- A 0
- B 1
- C 2
- D It depends on the row length
Show answer
Why:
Both neighbours, 126 and 140, are shorter, so no taller neighbour is counted.
- A 2, because the row is uneven on both sides
- B 0, because 118 cm is shorter than she is
- C 1, because she stands next to the end of the row
- D 1, because only the 148 cm neighbour is taller
Show answer
Why:
Her neighbours are 118 and 148, and only one of those beats her.
- A It is fine, and the tallest child is at an end
- B It is fine, and two children must be the same height
- C It is impossible, because the tallest child has to call 0
- D It is impossible, because a list may not repeat a number
Show answer
Why:
Somebody in the row beats everybody, and that child has no taller neighbour.
- A taller than
- B standing between
- C the same height as
- D shorter than
Show answer
Why:
Two taller neighbours is exactly what a 2 reports.
- A A row of all zeros can be arranged if only the children are careful enough
- B The shortest child always calls out 0
- C A child at an end of the row often calls out 2
- D Two children standing next to each other can call out the same number
Show answer
Why:
Short, medium and tall in that order gives 1, 1 from the first two children.
- A (i) 0 and (ii) 2
- B (i) 1 and (ii) 0
- C (i) 2 and (ii) 1
- D (i) 0 and (ii) 1
Show answer
Why:
The tallest child counts nobody, and a child below both neighbours counts two.
- A the number in every box that touches it
- B every other number written in the table
- C the number on its left only
- D the average of the whole table
Show answer
Why:
Only the boxes actually touching it are compared.
- A I and II only
- B II and III only
- C I and III only
- D I, II and III
Show answer
Why:
I and III are correct. II is not: the tallest child must say 0, so a row of all ones cannot happen.
- A Because it is an odd number and odd numbers cannot win
- B Because 806 and 940 are both bigger than it
- C Because it sits at the end of the row and has one neighbour
- D Because a row may hold only two coloured boxes
Show answer
Why:
A box must beat every neighbour, and 375 beats neither of its own.
- A It is always a supercell, since it is easy to find
- B It is a supercell whenever it sits at an end of the row
- C It is never a supercell when the table has more than one box
- D It becomes a supercell after the table is rearranged
Show answer
Why:
Every neighbour it has is bigger than it, so it loses each comparison.
- A Yes, if both numbers are over a thousand
- B Yes, if the two numbers are close together
- C No, because each would have to beat the other
- D No, because a table allows only three coloured boxes
Show answer
Why:
No pair of numbers is each greater than the other.
- A Both true, and the reason explains the assertion
- B Both true, but the reason does not explain it
- C The assertion is true and the reason is false
- D The assertion is false and the reason is true
Show answer
Why:
Both are true, and the reason settles it. Whoever is shortest has to count at least one taller neighbour.
- A Write the numbers in order, smallest first, all the way along
- B Write a big number, then a small one, turn and turn about
- C Write every number close to every other number
- D Write all the big numbers together at one end of the row
Show answer
Why:
Alternating puts a winner in every second box, which is the most allowed.
- A 2
- B 200
- C 20
- D 2000
Show answer
Why:
Each label is 200 more than the label before it.
- A 1
- B 5
- C 10
- D 2
Show answer
Why:
The difference of 10 is shared among five equal gaps.
- A Because the digits of the two numbers are the same ones
- B Because the bigger number stands further to the right
- C Because a line always puts the smaller number first in the list
- D Because both numbers can be rounded to the same thousand
Show answer
Why:
A line turns size into position, so the eye does the comparing.
- A (i) says 2 and (ii) beats one neighbour only
- B (i) says 0 and (ii) beats every neighbour it has
- C (i) says 1 and (ii) holds the smallest number present
- D (i) says 0 and (ii) sits at an end of the row
Show answer
Why:
Both ideas describe something that beats everything immediately beside it.
- A No, the step is 1000, since that is the difference between the labels
- B Yes, because only the hundreds digit has changed between the two labels
- C Yes, because the numbers are close to each other
- D No, the step is 10, since the last digits match
Show answer
Why:
Subtract one label from the next and the gap is a thousand.
Important short questions (6)
It counts how many of the people standing right beside that child are taller than the child is.
Each of them has a neighbour on one side only. Counting two taller neighbours needs a person on both sides.
They call out 1, 1, 0 and 1. The 145 cm child beats both neighbours, and each of the others has exactly one taller neighbour.
Each box would need a number greater than the other box holds. That cannot happen with two different numbers.
Five. Put a large number in the first box and then in every second box after it, with small numbers in between.
One step is worth 1. The next marks read 15,079, then 15,080, then 15,081.
Long answer questions (3)
One of the five is taller than the other four. Nobody standing beside that child can be taller, so that child has to call out 0. A list of five ones therefore cannot happen, whoever stands where. A list that does work is 1, 1, 0, 1, 1. Arrange the children with the tallest in the middle, the next two on either side of her, and the two shortest at the ends. Each of the four outer children then has exactly one taller neighbour.
I would put a large number in the first box, a small number in the second, and carry on turn by turn. That colours the first, third, fifth and seventh boxes, so four supercells appear. I cannot do better, because two coloured boxes are never allowed to touch. Each one would have to beat the other, which no pair of numbers manages. Leaving a gap after every coloured box means at most four of seven can win.
The drawing carries no information by itself. Only the labels do. I would read the two labelled marks and take one away from the other. Then I would count the equal gaps lying between those marks and share the difference out among them. On the first line the difference is 1000 over five gaps, so each step is worth 200. On the second the difference is 10 over five gaps, so each step is worth 2. After that every blank mark can be filled by adding one step at a time.
Flashcards (15)
Play this topic
- Idea and meaning
- Possible or impossible?
- Fact or myth?
- Unscramble the word
- Label a blank number line
Playing with Digits
Digits can be added, reversed and rearranged. Digit sums group very different numbers together, palindromes read the same both ways, and one stubborn routine drags every four-digit number to 6174.
What you will be able to do
- I can work out a digit sum and find the smallest or largest number that gives one.
- I can test whether a number is a palindrome and count how many palindromes exist of a given length.
- I can carry out the reverse-and-add game and say how many rounds it took.
- I can run Kaprekar's routine on a four-digit number and reach 6174.
Key points
There are 9 one-digit numbers, 90 two-digit numbers, 900 three-digit numbers and 9000 four-digit ones.
A digit sum adds the digits and ignores place value, so 89, 764 and 98,000 all come to 17.
A three-digit number whose digits run on has a digit sum of three times its middle digit.
Writing 1 to 100 uses the digit 4 twenty times, ten in the units place and ten in the tens.
A palindrome reads the same from either end, and there are 90 three-digit ones.
Reverse and add always reaches a palindrome from a two-digit start, though 89 needs 24 rounds.
Whether every three-digit start reaches a palindrome is still unknown, and 196 is the doubtful case.
Kaprekar's routine takes any four-digit number with two different digits to 6174.
Full explanation
Arjun counts the numbers
Arjun writes numbers in his book, starting at 1. He wonders how many numbers there are of each size. One-digit numbers run from 1 to 9, so there are nine of them. Two-digit numbers run from 10 to 99, and there are ninety. Three-digit numbers run from 100 to 999, and there are nine hundred. Each step up multiplies the count by ten. Five-digit numbers come to ninety thousand.
Adding the digits instead of using them
Here is a different thing to do with a number. Ignore what it is worth and just add its digits. The digits of 89 add to 17. The digits of 764 also add to 17. So do the digits of 98,000. These three numbers are miles apart in size, yet they share one digit sum.
Hunting for the smallest and the largest
Suppose you want a digit sum of 17 and the smallest number that gives it. Use as few digits as you can, and make the last digit as large as possible. That gives 89. Now ask for the largest five-digit number with the same digit sum. Load the front digits and leave the rest at zero, which gives 98,000.
A pattern in consecutive digits
Try three-digit numbers whose digits run on, like 234 or 567. Add the digits of 234 and you get 9. Add the digits of 567 and you get 18. Each answer is three times the middle digit. That is no accident. One digit is one less than the middle and one is one more, so the two changes cancel out.
Counting a digit rather than a number
Rekha writes out all the numbers from 1 to 100 and asks how many times she has written the digit 4. In the units place a 4 appears ten times, at 4, 14, 24 and so on. In the tens place it appears ten more times, all through the forties. So she has written it twenty times.
Numbers that read the same either way
Some numbers are unchanged when you read them backwards. 66 is one. So are 848 and 1111. These are called palindromes. Checking one is easy: pair the first digit with the last, the second with the second last, and so on. Every pair must match. In a number with an odd count of digits, the middle digit has no partner and can be anything.
How many palindromes are there?
Count the three-digit ones. The first digit may be any of 1 to 9, and the last is forced to copy it. The middle digit is free and may be any of 0 to 9. So there are ninety. Five-digit palindromes work the same way, with two free middle digits, and there are nine hundred of them.
Reverse and add
Now a game. Take a number, reverse it, and add the two together. If the answer is a palindrome, stop. If not, do the same thing again. Start with 23 and one round is enough: 23 plus 32 gives 55. Start with 57 and it takes two rounds, reaching 132 and then 363.
Some numbers keep you waiting
Start at 68 and you need three rounds, passing through 154 and 605 before landing on 1111. Start at 89 and the patience required is astonishing. It takes twenty-four rounds, and the palindrome at the end has thirteen digits. Nobody could guess that from looking at 89.
What nobody has settled
Every two-digit starting number does reach a palindrome in the end. Three-digit numbers are a different story, and this one is not finished. Mathematicians do not know whether every three-digit start reaches a palindrome. The number 196 is the famous doubtful case. Computers have run it for a very long time without a palindrome appearing, and people suspect it never will. Suspecting is not the same as knowing.
A teacher from Devlali
D.R. Kaprekar earned his living teaching mathematics at a state school in the town of Devlali, in Maharashtra. He spent his spare hours playing with numbers and found patterns nobody had noticed before. In 1949 he came across something very strange indeed while fiddling with four-digit numbers.
His routine, step by step
Pick a four-digit number with at least two different digits. Rearrange its digits to make the largest number you can, and call that A. Rearrange them again to make the smallest, and call that B. Work out A minus B. Then start over with the answer, and keep going.
Watching 2059 settle down
The digits of 2059 give a largest of 9520 and a smallest of 0259, which is just 259. Taking 259 from 9520 leaves 9261. Round two turns 9261 into 8352. Round three turns 8352 into 6174. Try a fourth round and nothing moves: the largest is 7641, the smallest is 1467, and the difference is 6174 all over again.
Why the routine has to stop somewhere
That number 6174 is called the Kaprekar constant. Every four-digit start with two different digits reaches it, and never in more than seven rounds. The warning about two different digits matters. A number like 5555 gives the same largest and smallest, so the difference is zero and there is nothing left to work with.
The same idea with three digits
Three-digit numbers have a constant of their own. Begin at 325 and the routine gives 297, then 693, then 594, then 495. From there it repeats 495 for ever. So the three-digit routine settles as well, just at a different resting place.
Concepts and terms
The total you get by adding the digits of a number, treating each one as a plain figure rather than a value. Place value is thrown away, which is why 89 and 98,000 share the digit sum 17.
A number whose digits are unchanged when the whole thing is read backwards, such as 848 or 1111. Pair the first digit with the last and work inwards. A middle digit with no partner is free.
A routine: reverse a number, add the reversal to the original, and repeat until a palindrome appears. From a two-digit start it always finishes. From a three-digit start nobody knows whether it always does.
The number 6174. Make the largest and smallest arrangements of a four-digit number, subtract, and repeat. Provided at least two of the digits differ, the working always arrives at 6174 and then sticks there.
Watch out!
Myth or reality
Case Study
Scenario:
A class notices a number plate reading 2059 and is asked to make the largest and smallest numbers from those digits. The largest comes quickly: 9520. The smallest causes an argument. Arjun writes 0259 and Nita says that is not a four-digit number at all. The class tests it by reading it aloud, and it says two hundred and fifty-nine. So the zero cannot lead. The smallest four-digit arrangement is 2059, the plate itself. Arjun sees why. A leading zero does not change the value, so it quietly turns a four-digit number into a three-digit one.
Analysis:
The argument is about what a digit in front actually does. A leading zero adds a symbol and no value, so the number shrinks a place.
Activity:
Take a four-digit number plate with a zero in it. Write the largest and smallest four-digit numbers you can make.
A digit workshop at home
Title:
A digit workshop at home
What You Notice:
Your birth year always reaches 6174, and so does everybody else's, though the number of rounds differs from person to person. The reverse-and-add game can finish at once or drag on for a surprising number of rounds.
Did you know
Kaprekar did all of his exploring while holding down a full timetable at a state-run school in the town of Devlali.
No four-digit number with two different digits needs more than seven rounds of the routine before it reaches 6174.
Palindromes turn up outside mathematics too. Malayalam is a language whose name reads the same both ways.
Memory tricks
Testing a palindrome — Outside in, pair by pair — Compare the outermost digits, then step inwards. One mismatch anywhere ends the test. A lone middle digit never needs checking.
The order of Kaprekar's routine — Biggest minus smallest, then go again — Arrange the digits high to make A, arrange them low to make B, and subtract. Feed the answer straight back in.
Practice MCQs with answers (20)
- A 900
- B 10,000
- C 90,000
- D 9,000
Show answer
Why:
They run from 1000 to 9999, which is nine thousand numbers.
- A 13
- B 16
- C 17
- D 20
Show answer
Why:
Adding 7, 6 and 4 gives seventeen.
- A Use as many digits as she can, all of them small
- B Use as few digits as she can, with the largest at the end
- C Start from 17 and keep adding ten to it
- D Pick any number and cross out digits until they add to 17
Show answer
Why:
Fewer digits means a smaller number, and 89 is the result.
- A 89
- B 764
- C 9,710
- D 5,432
Show answer
Why:
The digits of 5,432 come to 14, while the other three each come to 17.
- A The first
- B The middle
- C The last
- D None, the rule fails here
Show answer
Why:
One digit is one below the middle and the other is one above, so they cancel.
- A Ten
- B Nineteen
- C Twenty
- D Forty
Show answer
Why:
Ten appearances in the units place and ten through the forties.
- A constant
- B multiple
- C sequence
- D palindrome
Show answer
Why:
That is exactly what the word palindrome means.
- A 4,562
- B 7,337
- C 1,232
- D 8,910
Show answer
Why:
Its outer digits are both 7 and its inner digits are both 3.
- A Nine
- B Nine hundred
- C One thousand
- D Ninety
Show answer
Why:
Nine choices for the first digit and ten for the middle one.
- A Put it through the same steps again
- B Go back and start from a different number
- C Subtract the reversal from it instead
- D Stop, because the game has failed
Show answer
Why:
The routine simply repeats on whatever answer it produces.
- A 75
- B 112
- C 132
- D 363
Show answer
Why:
Adding 57 and 75 gives 132, which is not yet a palindrome.
- A Every two-digit start reaches a palindrome in one round
- B Every three-digit start has been proved to reach a palindrome
- C Every two-digit start reaches a palindrome in the end
- D The game has been proved to fail for 196
Show answer
Why:
Two-digit starts always finish, though 89 takes twenty-four rounds; the three-digit case is unsettled.
- A Nobody has settled it, and a palindrome is suspected never to appear
- B It has been proved that 196 does give a palindrome eventually
- C It has been proved that 196 never gives a palindrome
- D Nobody has settled it, and a palindrome is expected within a few rounds
Show answer
Why:
Long computer searches have found nothing, but a search is not a proof.
- A A state school in the town of Devlali, Maharashtra
- B A college of engineering in Pune, Maharashtra
- C A state school in the town of Nashik, Gujarat
- D A research institute for mathematics in Mumbai, Maharashtra
Show answer
Why:
He was a schoolteacher, not a research worker.
- A (i) 495 and (ii) 6174
- B (i) 6174 and (ii) 495
- C (i) 6174 and (ii) 1089
- D (i) 1089 and (ii) 495
Show answer
Why:
Four digits settle at 6174 and three digits settle at 495.
- A 2059
- B 0259, which is 259
- C 5902
- D 9250, putting the zero right at the end
Show answer
Why:
The zero goes to the front, so the number is really 259.
- A I and II only
- B II and III only
- C I and III only
- D I, II and III
Show answer
Why:
I and III are correct. II is not: there are 900 three-digit numbers, from 100 to 999.
- A Both true, and the reason explains the assertion
- B Both true, but the reason does not explain it
- C The assertion is true and the reason is false
- D The assertion is false and the reason is true
Show answer
Why:
Both are true, and the reason is the whole of it: (m-1) + m + (m+1) comes to 3m. So 456 adds to 15, which is 3 times 5.
- A It produces 6174 again
- B It produces 495
- C It produces zero
- D It produces a larger number each time
Show answer
Why:
Taking 1467 from 7641 leaves 6174 once more.
- A Every question about numbers has been answered by now
- B A question that is easy to ask is always easy to answer
- C Questions are only real once a computer has checked them
- D Some questions that are easy to ask are still open
Show answer
Why:
The 196 question can be explained in a minute and remains unsolved.
Important short questions (6)
A digit sum adds the digits and ignores what each is worth. The digits of both numbers are 8 and 9, plus zeros, so both come to 17.
Twenty times. It appears ten times in the units place and ten more times through the forties, and nothing else contributes.
Pair the outer digits and then the inner ones. Both outer digits are 7 and both inner digits are 3, so it reads the same either way.
Reversing 23 gives 32, and the two add to 55. That is a palindrome, so the game stops after a single round.
Nothing has been settled. Very long computer searches have produced no palindrome, so people suspect there is none, but that has never been proved.
If all four digits match, the largest and smallest arrangements are the same number. Their difference is zero, so there is nothing to carry forward.
Long answer questions (3)
The digits of 2059 make a largest of 9520 and a smallest of 0259, which is 259. Taking 259 from 9520 leaves 9261. In round two the largest is 9621 and the smallest is 1269, and the difference is 8352. In round three the largest is 8532 and the smallest is 2358, and the difference is 6174. A fourth round changes nothing at all. The routine shows that every four-digit number with two different digits is pulled to 6174, and never takes more than seven rounds to get there.
His sentence states something as a fact when nobody has shown it. Computers have carried the reverse-and-add game on 196 for an enormous number of rounds and no palindrome has turned up. That is strong evidence and it is not a proof. A search can only tell you what has not happened so far. A better sentence would be that 196 is suspected never to give a palindrome, and that the question is still open. Being honest about what is unknown is part of mathematics.
To make a five-digit number large I care most about the leading digit, so I give it the biggest share I can, which is 9. That leaves 8 to place, and the next digit takes all of it. The last three digits are then zeros, giving 98,000. To make a number small I want as few digits as possible instead, so I use two digits and put the larger one at the back. That gives 89. The two hunts pull in opposite directions, one loading the front and one shortening the number.
Flashcards (15)
Play this topic
- Term and meaning
- Settled or still open?
- Fact or myth?
- Finish the sentence
- One round of Kaprekar's routine
Numbers in Time, and Sums in Your Head
Clocks and calendars are full of number patterns, and large sums often come apart into friendly pieces. Both reward looking before calculating.
What you will be able to do
- I can test a clock reading or a date to see whether it is a palindrome.
- I can build the largest and smallest numbers from a set of digits and compare them.
- I can reach a target total by combining a few given pieces, repeating a piece where I need to.
- I can add a pattern of numbers quickly by grouping equal values instead of adding one by one.
Key points
A clock reading is tested for a palindrome exactly like a number, once the colon is ignored.
A 12-hour clock shows 57 palindromic times, six in each single-digit hour and one each at 10, 11 and 12.
A date written as day, month and year can read the same both ways, as 21/02/2012 does.
A common year is one day longer than 52 weeks, so every date moves on a weekday the following year.
Sorting four digits downwards gives the largest number and upwards gives the smallest.
When building a total from given pieces, a piece may be used as many times as you like.
Allowing subtraction as well as addition makes many totals much quicker to reach.
A four-digit number plus a two-digit number can never reach six digits, whatever you choose.
Full explanation
A clock with something to say
Nisha glances at the wall clock and sees 5:55. A minute later the pattern is gone. She starts watching for readings that look special. Some repeat a digit, like 11:11. Others read the same backwards, like 3:03 or 12:21. A clock face is only a pair of numbers, and yet it throws up patterns all day long.
Testing a time for a palindrome
Ignore the colon and look at the digits alone. The reading 3:03 gives 303, which reads the same either way. The reading 12:21 gives 1221, which also passes. The reading 4:15 gives 415, and backwards that is 514, so it fails. The test is exactly the one you use on a plain number.
How many such times are there?
Take an hour with a single digit, say 3. The reading is 3 then two minute digits, so the last digit has to be a 3 as well. The first minute digit may be 0, 1, 2, 3, 4 or 5. That gives six times in every such hour, and there are nine of those hours. The hours 10, 11 and 12 contribute one each. The total is fifty-seven.
Dates can do it too
A date written as day, month and year makes an eight-digit string. The date 21/02/2012 gives 21022012, and that reads the same from either end. So does 10/02/2001. Such dates are rare, so people notice them and remember them.
Will a calendar ever come round again?
Jaspreet wonders why a new calendar is printed every year. The answer is that the days slide. A year of 365 days is fifty-two weeks and one day over, so every date shifts on by one weekday. A leap year shifts it by two. A calendar can be used again only by a year that starts on the same weekday and is of the same kind, leap or common.
Building the biggest and the smallest
Give yourself four digits, say 8, 1, 5 and 3. Sort them downwards for the largest number, 8531. Sort them upwards for the smallest, 1358. Their difference is 7173 and their sum is 9889. Change the digits you start from and both answers change, which makes a good puzzle for a friend.
Sums you can do without writing
A large addition is often easier in the head than on paper, provided you break it up sensibly. Suppose you have the pieces 30,000, 9,000, 1,200 and 200 to build with. You may take any piece as often as you like. To reach 40,600, take 30,000 and 9,000 and 1,200, then take 200 twice.
Why repeating a piece is allowed
Children often assume each piece is used once and then give up. Nothing in the rules says that. Reaching 40,600 needs the 200 twice. Once you see that a piece can come back, many more totals fall into place, and a short set of pieces goes a long way.
Taking away as well as adding
Now allow subtraction too. Some totals are much easier that way round. To reach 29,700 you could hunt for pieces that add up to it, or you could start from 30,000, take away 500 and put 200 back. Working down from above is often the quicker route.
How many digits does the answer have?
A five-digit number added to another five-digit number sometimes gives five digits and sometimes six. Try 10,000 and 10,000, which gives 20,000. Now try 99,999 twice, which gives 199,998. So the answer is sometimes five digits and sometimes six, and the only way to know is to look.
Some answers are simply out of reach
Other cases can be ruled out at once. A four-digit number plus a two-digit number can never have six digits. The largest you can manage is 9,999 plus 99, which is 10,098, and that has five. In the same way, a five-digit number take away a two-digit number never drops to three digits.
Examples worth keeping
It helps to have one clean example of each kind in your notebook. Adding 54,321 and 67,890 gives 122,211, which is two five-digit numbers making six digits. Taking 40,905 from 41,208 leaves 303, which is two five-digit numbers making three digits. Writing your own examples is better practice than reading mine.
Adding a figure the quick way
A pattern of numbers can be added one at a time, but that is slow and mistakes creep in. Look for equal numbers instead and count them. Four rows of six twenty-fives is twenty-four twenty-fives, and 24 x 25 = 600. One multiplication has replaced twenty-three additions.
Splitting a figure into groups
Some patterns hold two different numbers. Deal with each group on its own and add at the end. Twelve fifteens come to 180, because 12 x 15 = 180. Eight thirty-fives come to 280. Adding those two totals gives 460. The order in which you group them does not change the answer, so choose whichever grouping you find easiest to see.
Concepts and terms
A clock reading whose digits are unchanged when read backwards, such as 3:03 or 12:21. Drop the colon first and treat what is left as an ordinary number. A 12-hour clock shows fifty-seven of them.
Sorting a set of digits downwards builds the largest number they can make, and sorting them upwards builds the smallest. A zero among them moves to the front of the smallest and shortens it.
Reaching a target by combining a short list of given numbers. Each piece may be taken more than once, and allowing subtraction as well as addition opens up totals that addition alone cannot reach easily.
Counting how many times each value appears in a pattern and multiplying, instead of adding the numbers one after another. Twelve fifteens is one multiplication rather than eleven additions.
Watch out!
Myth or reality
Fatima plans the school bell
Scenario:
Fatima helps her teacher set the bell times for the school day in Bhopal. The first period begins at 9:00 in the morning. Each period runs for forty minutes and there is no gap between them. Fatima writes out the starting time of every period in a list. Partway down she notices that one of the times reads the same backwards. Her teacher then asks her a second question. The school wants to reuse the same printed timetable card next year. Fatima says that the times will be fine but the dates and weekdays printed beside them will not, because a year does not divide evenly into whole weeks. She works out both answers in her notebook.
Analysis:
Number patterns and calendar arithmetic both turn up in an ordinary school day, and both can be settled by careful counting rather than by guesswork.
Activity:
Write out the start times of six lessons in your own school and mark any that read the same backwards.
Did you know
A common year runs one day past fifty-two whole weeks, which is why your birthday lands on a new weekday each year.
Palindromic dates cluster in certain years and are missing altogether from others, depending on how the digits of the year fall.
Memory tricks
Testing a clock reading — Drop the colon, read it back — The colon is punctuation and carries no value. What is left is a plain number, and the usual palindrome test applies.
Building from pieces — Any piece, any number of times — Nothing limits you to one go with each piece. Remembering this is what unlocks totals that look impossible at first.
Practice MCQs with answers (20)
- A 5:35
- B 4:15
- C 7:20
- D 9:41
Show answer
Why:
Its digits are 535, which reads the same from either end.
- A ignore the colon and look at the digits
- B convert the time to the 24-hour system
- C round the minutes to the nearest ten
- D add the hour digits to the minute digits
Show answer
Why:
The colon is punctuation, so the digits are what matter.
- A Twelve
- B Twenty-four
- C Fifty-seven
- D Sixty
Show answer
Why:
Nine hours contribute six each, and 10, 11 and 12 contribute one each.
- A One
- B Three
- C Ten
- D Six
Show answer
Why:
The minutes must end in 6, and the first minute digit may be 0 to 5.
- A 12/11/2013
- B 03/04/2005
- C 19/08/2019
- D 21/02/2012
Show answer
Why:
Its eight digits are 21022012, unchanged when reversed.
- A Because months have different lengths
- B Because a common year is one day longer than fifty-two weeks
- C Because the calendar is reprinted every year with a new design
- D Because a week sometimes has eight days
Show answer
Why:
That extra day pushes every date on by one weekday.
- A next
- B previous
- C same
- D opposite
Show answer
Why:
Both the starting weekday and the leap or common status must match.
- A 8135
- B 8531
- C 8513
- D 1358
Show answer
Why:
Sorting the digits downwards puts the 8 in front, then 5, then 3, then 1.
- A 7173
- B 7371
- C 8531
- D 9889
Show answer
Why:
Taking 1358 from 8531 leaves 7173.
- A (i) 7173 and (ii) 9889
- B (i) 9889 and (ii) 7173
- C (i) 9889 and (ii) 8531
- D (i) 8531 and (ii) 1358
Show answer
Why:
They add to 9889 and differ by 7173.
- A By using each of the four pieces exactly once
- B By using 30,000 twice and 1,200 once
- C It cannot be reached from these pieces
- D By using 30,000, 9,000, 1,200 and then 200 twice
Show answer
Why:
The last four hundred needs the 200 piece taken a second time.
- A A piece may be used only once in any total
- B Subtraction is never allowed, only addition
- C A piece may be used as many times as you like
- D Every total between the pieces can always be reached
Show answer
Why:
Repeating a piece is what makes many targets reachable.
- A Add 500, then take away 200
- B Take away 500, then add 200
- C Take away 300, then take away 200
- D Add 700, then take away 200
Show answer
Why:
Thirty thousand less five hundred is 29,500, and two hundred more is 29,700.
- A Always
- B Sometimes
- C Never
- D Only when both are even
Show answer
Why:
The largest such sum is 9,999 plus 99, which is 10,098 and has five digits.
- A Always five
- B Five or six, depending on the numbers
- C Always six
- D Four, five or six, depending on the sum
Show answer
Why:
Two ten-thousands give five digits, while 99,999 twice gives six.
- A I and II only
- B II and III only
- C I and III only
- D I, II and III
Show answer
Why:
I and III are correct. II is not: a 12-hour clock shows 57, six in each single-digit hour and one each at 10, 11 and 12.
- A Add the twenty-fives one at a time down each row
- B Add the four row totals and then divide by six
- C Multiply 25 by 4 and stop there
- D Count twenty-four of them and multiply by 25
Show answer
Why:
One multiplication settles it, since 24 x 25 = 600.
- A Both true, and the reason explains the assertion
- B Both true, but the reason does not explain it
- C The assertion is true and the reason is false
- D The assertion is false and the reason is true
Show answer
Why:
Both are true, and the reason is the arithmetic: 52 weeks is 364 days, so 365 leaves one day over.
- A A five-digit number minus a two-digit number can have three digits
- B A four-digit number plus a two-digit number can have six digits
- C A one-digit number plus a one-digit number can have three digits
- D A five-digit number minus a five-digit number can have three digits
Show answer
Why:
Taking 40,905 from 41,208 leaves 303, which shows it happening.
- A (i) testing a time, (ii) adding a pattern quickly
- B (i) adding a pattern quickly, (ii) testing a time
- C (i) testing a date, (ii) reusing a calendar
- D (i) reusing a calendar, (ii) sorting digits
Show answer
Why:
One habit prepares a clock reading and the other speeds up a long addition.
Important short questions (6)
Drop the colon, which leaves the digits 1221. Read them from the other end and they are unchanged, so the time is a palindrome.
The final minute digit has to copy the hour digit. The first minute digit can only be 0, 1, 2, 3, 4 or 5, which gives six choices.
The largest is 9642 and the smallest is 2469. Their difference is 7173.
Take 30,000, then 9,000, then 1,200, and then the 200 piece twice over. Those five pieces together come to 40,600.
The largest four-digit number is 9,999 and the largest two-digit one is 99. Their sum is 10,098, which has only five digits, so nothing larger is possible.
Twelve fifteens come to 180 and eight thirty-fives come to 280. Adding those two group totals gives 460.
Long answer questions (3)
Take an hour with one digit, such as 4. The reading has three digits, so the last one has to match the hour and be a 4. The first minute digit may be 0, 1, 2, 3, 4 or 5, since minutes stop below sixty. That gives six times in each such hour, and there are nine hours with a single digit, which makes fifty-four. The hours 10, 11 and 12 each have four digits, and only one arrangement works in each, giving 10:01, 11:11 and 12:21. Fifty-four and three together make fifty-seven.
His rule is one he has invented. Nothing in the puzzle says a piece is used up once it has been taken. Working with his rule, the four pieces come to 40,400, which does indeed fall short. Once the rule is dropped the target is easy. Take 30,000, then 9,000, then 1,200, and then the 200 twice. That reaches 40,600 exactly. The lesson is worth remembering, because a rule you add for yourself can make a simple problem look impossible.
Some claims can be settled by testing the extreme case alone. A four-digit number plus a two-digit number can never reach six digits, because even 9,999 plus 99 gives 10,098, which has five. Nothing bigger is available, so the case is closed. Other claims depend on the numbers. A five-digit number added to another sometimes gives five digits and sometimes six. Ten thousand twice comes to 20,000, which has five. Adding 99,999 to itself gives 199,998, which has six. So one claim is never, and the other is only sometimes.
Flashcards (15)
Play this topic
- Idea and example
- Always, sometimes or never
- True or false?
- Unscramble the word
- Add a pattern the quick way
Estimating, and Winning Games
One simple rule leads to a problem nobody has solved, estimates answer questions that counting cannot, and a game of adding numbers turns out to have a rule hidden inside it.
What you will be able to do
- I can build a Collatz run from any starting number and say how many steps it took.
- I can explain why checking many numbers is not the same as proving something.
- I can build an estimate in stages and say why I rounded where I did.
- I can work out the winning strategy for an adding game by starting from the target.
Key points
The Collatz rule halves an even number and trebles an odd one before adding 1.
A power of two falls straight to 1, because halving it never produces an odd number.
The run from 27 climbs as high as 9,232 and takes 111 steps to finish.
Lothar Collatz suggested in 1937 that every whole number reaches 1, and nobody has proved it.
Checking billions of cases supports a belief but cannot settle it, since the whole numbers never end.
An estimate is built in stages, rounding on purpose so the next stage stays easy.
Counting for a minute and scaling up gives a usable estimate for a whole day.
In an adding game the safe numbers are spaced by one more than the largest jump allowed.
Full explanation
A rule anyone can follow
Here is a rule for making a run of numbers. Start anywhere you like. An even number gets cut in half. An odd number gets trebled, with one more added on. Then do the same to the answer. Nothing in that rule is hard. A child can follow it, and so can a computer.
Where the rule takes you
Start at 15. It is odd, so trebling and adding one gives 46. That is even, so halving gives 23. Carry on and the run climbs to 106, falls away, climbs again, and after seventeen steps arrives at 1. Start at 7 instead and it takes sixteen steps. Start at 32 and the whole run is a slide: 32, 16, 8, 4, 2, 1.
Powers of two fall straight down
A power of two is the easiest start of all. Halve it and you get another power of two. There is never an odd number to send you climbing again. So the run drops step by step and lands on 1 without a single detour.
Some runs wander a very long way
Not every start behaves so nicely. Begin at 27 and the run rises as high as 9,232 before it comes down. It takes a hundred and eleven steps to reach 1. Nothing about the number 27 warns you of that. Two starts that look alike can behave completely differently.
The part nobody has settled
In 1937 the German mathematician Lothar Collatz suggested that every whole number, followed by this rule, ends at 1. Mathematicians have worked on it ever since. Computers have checked an enormous stretch of numbers and every single one has reached 1. Even so, nobody has proved it for all numbers. It is still open, and among the best known questions of its kind anywhere.
Checking is not proving
The gap here is worth understanding. A computer can test the first many billions of starting numbers. That leaves infinitely many untested, because the whole numbers never run out. So a search can support a belief and can never finish the job. The honest sentence is that the conjecture is unproved.
When an exact number is not needed
Now put the puzzles aside for something practical. Your headteacher knows exactly how many children are on the roll. You do not, and you do not need to. Is it about two hundred, or about a thousand? That is an estimate, and for most purposes it is all anyone wants.
How an estimate is built
Maryam counts her own section: 38 children. The other two sections have 41 and 36. That comes to 115, so she calls her whole class about 120. Her school has five classes of a similar size. Five lots of 120 gives about 600. She has not counted a single child beyond her own section.
Rounding on purpose
Notice that Maryam rounded 115 up to 120 deliberately. An estimate is meant to be easy to carry in the head and easy to multiply. Carrying 115 through the next step would have given a false air of accuracy. An estimate that looks exact is worse than one that is plainly rough.
Estimating things you cannot count
Some quantities cannot be counted directly at all. How many breaths do you take in a day? Count them for one minute and suppose you get twenty. An hour has sixty minutes, and 20 x 60 = 1200. A day has twenty-four hours, and 1200 x 24 = 28800. The answer is rough, and it is far better than a shrug.
A game with a hidden rule
Now a game for two players. Each of you in turn puts 1, 2 or 3 on to a running total that begins at nothing. Whoever says 21 wins. Play it a few times and you will begin to feel that the first player has an advantage. There is a reason, and it is worth working out.
Working backwards from the win
Whoever says 21 wins, so 21 is a safe number. What about 17? From 17 your opponent can only reach 18, 19 or 20, and from any of those you can say 21. So 17 is safe too. Keep going down in fours: 13, 9, 5 and 1 are all safe. The first player says 1 and never lets go.
Why the gap is four
The largest jump allowed is 3. So whatever your opponent adds, you can add something that brings the pair of moves to exactly 4. Your rival adds 1 and you add 3. Your rival adds 3 and you add 1. Either way you move four along and stay on the safe list.
Changing the game changes the winner
Try a version where players add anything from 1 to 10 and the winner says 99. Now the gap is eleven, and the safe numbers are 11, 22, 33 and onwards to 99. The first player cannot reach 11 in one move, so the second player takes control. Here the second player wins with correct play.
Making the rule your own
The pattern is easy to state. Add one to the largest jump allowed, and that is the gap between safe numbers. If the target sits exactly on that gap, the second player wins. If it does not, the first player wins by opening on the remainder. Invent your own target and jump size and test it on a friend.
Concepts and terms
Halve the number if it is even; treble it and add one if it is odd; then repeat on the answer. Every run tried so far ends at 1, though some wander upwards a long way before they come down.
A statement somebody believes is true but nobody has proved. Collatz made his in 1937. A conjecture stays a conjecture however many examples agree with it, because examples cannot cover every case.
A rough answer built on purpose, good enough for the question being asked. It is made in stages, with deliberate rounding at each stage so the working stays easy to carry in the head.
A running total that hands your opponent a losing position in an adding game. The safe numbers are spaced by one more than the largest jump, and the target itself is always one of them.
Watch out!
Myth or reality
Estimating the school water tank
Scenario:
The tank on the roof of a school in Nagpur runs dry by the middle of the afternoon, and the headmaster asks the Class 6 children to work out why. Nobody can measure the tank, so the children estimate instead. Ritu counts the pupils at one water point during a single break and finds that about forty fill a bottle. There are three such points and three breaks in the day. A bottle holds about one litre. The children also allow roughly two hundred litres for the washrooms and the kitchen together. They write their working on the board in stages, rounding at each stage, and mark every figure with the word about so that nobody mistakes it for a measurement.
Analysis:
An estimate built in clear stages can answer a practical question that no available instrument can settle, provided each rounding is deliberate and openly recorded.
Activity:
Estimate how many litres your own household uses in a day by counting one bucket, then scaling up.
Invent a game and crack it
Title:
Invent a game and crack it
What You Notice:
The first five rounds look like luck and the last five do not. Whether the first or the second player wins is settled before anyone moves, by whether the target lands exactly on the gap.
Did you know
Lothar Collatz put his question forward in 1937, and it has resisted every attempt since.
The run that begins at 27 climbs to 9,232 before it turns, and needs 111 steps to reach 1.
The rule is often called the 3n plus 1 problem, after the step that every odd number has to take.
Memory tricks
Applying the Collatz rule — Even halve, odd climb — An even number is cut in two. An odd number is trebled and given one more. Those are the only two moves there are.
Finding a winning strategy — Start at the finish and walk back — Mark the target as safe, then step back by one more than the largest jump, again and again, until you reach the beginning.
Practice MCQs with answers (20)
- A It is halved
- B It is trebled and given one more
- C It is doubled and given one less
- D It is left as it is
Show answer
Why:
Odd numbers are the ones that climb before they fall.
- A 7
- B 16
- C 30
- D 46
Show answer
Why:
Fifteen is odd, so trebling it gives 45 and one more gives 46.
- A 23
- B 92
- C 139
- D 45
Show answer
Why:
Forty-six is even, so it is halved.
- A Both true, and the reason explains the assertion
- B Both true, but the reason does not explain it
- C The assertion is true and the reason is false
- D The assertion is false and the reason is true
Show answer
Why:
Both are true, and the reason is why a proof is needed. The run from 27 alone climbs to 9,232 and takes 111 steps.
- A Because it is always larger than 1
- B Because halving it gives another power of two, which is even
- C Because the rule leaves even numbers alone
- D Because powers of two were left out of the conjecture entirely
Show answer
Why:
No odd number ever appears, and only odd numbers make the run climb.
- A never reaches 1 at all
- B reaches 1 in only four steps
- C repeats the number 27 for ever
- D climbs to 9,232 and needs 111 steps
Show answer
Why:
A small start can take a very long and very high journey.
- A Lothar Collatz, in 1937
- B D.R. Kaprekar, in 1949
- C Lothar Collatz, in 1949
- D Srinivasa Ramanujan, in 1913
Show answer
Why:
He was a German mathematician and the conjecture carries his name.
- A It has been proved true for every whole number
- B It has been proved false by a counter-example
- C It is unproved, and no exception has ever been found
- D It was withdrawn once computers began testing it
Show answer
Why:
Enormous searches agree with it, and agreement is not proof.
- A Because computers make arithmetic mistakes too often to be relied on
- B Because the whole numbers never run out, so infinitely many stay untested
- C Because a search can only test the even numbers and skips every odd one
- D Because the results of a search are always kept secret by the people running it
Show answer
Why:
However far a search reaches, endlessly many starts lie beyond it.
- A a failed attempt at an exact answer, made when the exact one proves too hard
- B a rough answer that is close enough for the question
- C an answer that has been rounded by accident
- D a guess made without any working at all
Show answer
Why:
Being rough is the point, not a weakness.
- A About 90
- B About 120
- C About 200
- D About 400
Show answer
Why:
They come to 115, which rounds comfortably to 120.
- A About 250
- B About 400
- C About 600
- D About 1,200
Show answer
Why:
Five lots of about 120 come to about 600.
- A Because 115 is the wrong total for the three sections
- B Because a total that ends in a five may not be multiplied by a whole number
- C Because a round figure keeps the working easy and stays honest about accuracy
- D Because estimates must always be rounded upwards
Show answer
Why:
An estimate that looks exact misleads the reader.
- A I and II only
- B II and III only
- C I and III only
- D I, II and III
Show answer
Why:
I and III are correct. II is the opposite: a power of two falls straight to 1, because halving it never produces an odd number.
- A 2, 6, 10, 14, 18
- B 4, 8, 12, 16, 20
- C 3, 6, 9, 12, 15, 18, 21
- D 1, 5, 9, 13, 17, 21
Show answer
Why:
Step back from 21 in fours and these are the totals you reach.
- A the target 21 can be divided by four
- B there are four players taking part
- C each player is allowed to add any amount from one right up to four
- D the largest jump is three, and one more than that is four
Show answer
Why:
Whatever your rival adds, you can make the pair of moves come to four.
- A The first player, by opening with 1
- B The second player, by answering every move
- C Whichever player reaches 17 first, whoever that is
- D Neither, because the game always ends in a draw
Show answer
Why:
Twenty-one is not a multiple of four, so the opening move can claim the first safe number.
- A The first player, by opening with 1
- B The first player, by opening with 10
- C The second player, because 99 sits exactly on a gap of eleven
- D It depends entirely on luck, because neither player is able to plan ahead
Show answer
Why:
The safe totals are the multiples of eleven, and the first player cannot reach 11 in one move.
- A (i) first player and (ii) second player
- B (i) second player and (ii) first player
- C (i) first player and (ii) first player
- D (i) second player and (ii) second player
Show answer
Why:
Twenty-one is not a multiple of four, while ninety-nine is a multiple of eleven.
- A The first player wins by opening with 5
- B The first player wins by opening with 1
- C The game cannot be won by either player, since 30 is never landed on exactly
- D The second player wins, since 30 is a multiple of the gap of five
Show answer
Why:
The gap is five, the safe totals are the multiples of five, and the opener cannot reach 5.
Important short questions (6)
Halve an even number, and treble an odd number before adding one. Starting at 7 the run reads 7, 22, 11, 34.
Every number in it is a power of two. Halving one gives another power of two, so no odd number ever turns up to send it upwards.
Collatz put it forward in 1937 and it has never been proved. Huge computer searches have found no exception, but that is evidence rather than a proof.
The three sections come to 115, so a class is about 120. Five classes of that size give an estimate of about 600 children.
She should say 1. That is the lowest safe total, and from there she can answer every move to reach 5, 9, 13, 17 and 21.
Add one to the largest jump the rules allow. That gap lets you answer any move your opponent makes and still land on the next safe total.
Long answer questions (3)
Fifteen is odd, so it becomes 46, and 46 is even, so it becomes 23. The run climbs to 106, falls back, climbs once more and finally arrives at 1 after seventeen steps. Its shape is a series of climbs and drops rather than a smooth slide. What it shows is that this one starting number behaves as Collatz suggested. What it does not show is anything about the numbers nobody has tried. One run, or a billion runs, cannot settle a claim made about every whole number there is.
I would count the words in three or four lines and take a typical figure, say ten. Then I would count the lines on a full page, perhaps thirty, which gives about 300 words a page. Multiplying by the number of pages gives my estimate. It could go wrong in several ways. Pages holding pictures or exercises carry far fewer words than a page of prose. My sample lines might be unusually long or short. So I would treat the answer as a rough size rather than a count, and say so clearly.
In the game of 21 the largest jump is three, so safe totals are spaced four apart. Working back from 21 gives 17, 13, 9, 5 and 1. The first player says 1 and afterwards answers each move so that the pair of moves comes to four. She lands on every safe total and wins. Change the rules so that players add up to ten and the winner says 99. The gap becomes eleven and the safe totals are the multiples of eleven. The first player cannot reach 11 in a single move, so the second player takes 11 and keeps control from there.
Flashcards (15)
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- Term and meaning
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Class 6 Mathematics Chapter 3 — frequently asked questions
It counts how many of the people standing right beside that child are taller than the child is.
Each of them has a neighbour on one side only. Counting two taller neighbours needs a person on both sides.
They call out 1, 1, 0 and 1. The 145 cm child beats both neighbours, and each of the others has exactly one taller neighbour.
A digit sum adds the digits and ignores what each is worth. The digits of both numbers are 8 and 9, plus zeros, so both come to 17.
Twenty times. It appears ten times in the units place and ten more times through the forties, and nothing else contributes.
Pair the outer digits and then the inner ones. Both outer digits are 7 and both inner digits are 3, so it reads the same either way.
Drop the colon, which leaves the digits 1221. Read them from the other end and they are unchanged, so the time is a palindrome.
The final minute digit has to copy the hour digit. The first minute digit can only be 0, 1, 2, 3, 4 or 5, which gives six choices.
The largest is 9642 and the smallest is 2469. Their difference is 7173.
Halve an even number, and treble an odd number before adding one. Starting at 7 the run reads 7, 22, 11, 34.
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