In chapter 3: Number play we know about Order of Number, Number Puzzle, Supercells, Patterns of numbers,Palindromic Patterns, Clock and Calendar Numbers, The Magic Number of Kaprekar, Simple Estimation and An Unsolved Mystery - the Collatz Conjecture
NCERT CLASS 6TH MATHS CHAPTER 3 NUMBER PLAY
- NCERT CLASS 6TH MATHS CHAPTER 1 PATTERNS IN MATHEMATICS
- NCERT CLASS 6TH MATHS CHAPTER 2 LINES AND ANGLES
- NCERT CLASS 6TH MATHS CHAPTER 4 DATA HANDLING AND PRESENTATION
- NCERT CLASS 6TH MATHS CHAPTER 5 PRIME TIME
- NCERT CLASS 6TH MATHS CHAPTER 6 PERIMETER AND AREA
- NCERT CLASS 6TH MATHS CHAPTER 7 FRACTION
- NCERT CLASS 6TH MATHS CHAPTER 8 PLATING WITH CONSTRUCTIONS
- NCERT CLASS 6TH MATHS CHAPTER 9 SYMMETRY
- NCERT CLASS 6TH MATHS CHAPTER 10 THE OTHER SIDE OF ZERO
Topic 3.1 Numbers can Tell us Things
Try answering the questions below and share your reasoning:
A child says ‘1’ if there is only one taller child standing next to them. A child says ‘2’ if both the children standing next to them are taller. A child says ‘0’, if neither of the children standing next to them are taller. That is each person says the number of taller neighbours they have.
Question 1
Can the children rearrange themselves so that the children standing at the ends say ‘2’?
Solution:
No,according to the rule children standing at the ends can say '2' if both the children standing next to them are taller than him.
Question 2
Can we arrange the children in a line so that all would say only 0s?
Solution:
We can not arrange the children in a line so that all would say only 0.
Question 3
Can two children standing next to each other say the same number?
Solution:
Yes,two children can standing next to each other say the same number 1.
Question 4
There are 5 children in a group, all of different heights. Can they stand such that four of them say ‘1’ and the last one says ‘0’? Why or why not?
Solution:
Yes, they can stand like this way in which four will say '1' and the last one will say '0'. If they will stand in ascending order of their height.
Question 5
For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
Solution:
No, 1, 1, 1, 1, 1 sequence is not posssble for this group of 5 children.
Question 6
Is the sequence 0, 1, 2, 1, 0 possible? Why or why not?
Solution:
Yes 0, 1, 2, 2, 0 sequence is possible.
Question 7
How would you rearrange the five children so that the maximum number of children say ‘2’?
Topic 3.2 Supercells
Figure It Out
Question 1
Colour or mark the supercells in the table below.
Question 2
Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells.
Solution:
Question 3
Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.
Solution:
(Note: You can use any number between 100 and 1000 here that satisfies the condition.)
Question 4
Out of the 9 numbers, how many supercells are there in the table above? ___________
Solution:
In the above table ,there are 5 supercells.
Question 5
Find out how many supercells are possible for different numbers of cells.
Do you notice any pattern? What is the method to fill a given table to get the maximum number of supercells? Explore and share your strategy.
Solution:
Make more cells and try by your self.
Question 6
Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
Solution :
No, Because in arrangement of number in any order (ascending or descending) at least one supercell we will get.
Question 7
Will the cell having the largest number in a table always be a supercell? Can the cell having the smallest number in a table be a supercell? Why or why not?
Solution: Yes, if the cell having the largest number in the table always be a supercell because value of this cell always greater than its neighbor cells.
NO, if the cell having the smallest number in the table cannot be a supercell because value of this cell always less than its neighbor cells.
Question 8
Fill a table such that the cell having the second largest number is not a supercell.
Solution:
Here ,800 is second largest number in the above table not a supercell.
(different answer may be possible)
Question 9
Fill a table such that the cell having the second largest number is not a supercell but the second smallest number is a supercell. Is it possible?
Solution: yes it is possible.
(different answer may be possible)
Question 10
Make other variations of this puzzle and challenge your classmates.
Solution:
Activity based question do by your self.
* Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’, ‘3’, and ‘9’ in some order. Only a coloured cell should have a number greater than all its neighbours.
The biggest number in the table is ____________ .
The smallest even number in the table is ____________.
The smallest number greater than 50,000 in the table is ____________.
Solution:
The biggest number in the table is 96,310 .
The smallest even number in the table is 10396.
The smallest number greater than 50,000 in the table is 60193.
Topic 3.3 Patterns of Numbers on the Number Line
Figure It Out
We are quite familiar with number lines now. Let’s see if we can place some numbers in their appropriate positions on the number line. Here are the numbers: 2180, 2754, 1500, 3600, 9950, 9590, 1050, 3050, 5030, 5300 and 8400.
Solution:
Question 1
Identify the numbers marked on the number lines below, and label the remaining positions.
Put a circle around the smallest number and a box around the largest number in each of the sequences above.
Solution:
Topic 3.4 Playing with Digits
Find out how many numbers have two digits, three digits, four digits, and five digits:
Figure it Out
Question 1
Digit sum 14
a. Write other numbers whose digits add up to 14.
b. What is the smallest number whose digit sum is 14?
c. What is the largest 5-digit whose digit sum is 14?
d. How big a number can you form having the digit sum
14? Can you make an even bigger number?
Solution:(a) 59,68,77,86,95,149,158,167..............
(b) 59 is the smallest number and its digit sum is 14.
(c)95000 is the largest 5 digit number whose sum is 14.
(d) Not possible because numbers are infinite.
Question 2
Find out the digit sums of all the numbers from 40 to 70. Share your observations with the class.
Solution:
Numbers | Digit | Sum |
40 | 4+0 | 4 |
41 | 4+1 | 5 |
42 | 4+2 | 6 |
43 | 4+3 | 7 |
44 | 4+4 | 8 |
45 | 4+5 | 9 |
46 | 4+6 | 10 |
47 | 4+7 | 11 |
48 | 4+8 | 12 |
49 | 4+9 | 13 |
50 | 5+0 | 5 |
51 | 5+1 | 6 |
52 | 5+2 | 7 |
53 | 5+3 | 8 |
54 | 5+4 | 9 |
55 | 5+5 | 10 |
56 | 5+6 | 11 |
57 | 5+7 | 12 |
58 | 5+8 | 13 |
59 | 5+9 | 14 |
60 | 6+0 | 6 |
61 | 6+1 | 7 |
62 | 6+2 | 8 |
63 | 6+3 | 9 |
64 | 6+4 | 10 |
65 | 6+5 | 11 |
66 | 6+6 | 12 |
67 | 6+7 | 13 |
68 | 6+8 | 14 |
69 | 6+9 | 15 |
70 | 7+0 | 7 |
Question 3
Calculate the digit sums of 3-digit numbers whose digits are consecutive (for example, 345). Do you see a pattern? Will this pattern continue?
Solution:
Numbers | Digits | 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 |
When we find the digit sums of 3 digit number (whose digits are consecutive) We get multiples of 3 and this pattern stoped at 789.
Digit Detectives
Among the numbers 1–100, how many times will the digit ‘7’ occur? Among the numbers 1–1000, how many times will the digit ‘7’ occur?
Solution:
From 1 to 100 digit-7 will come up to 20 times.
(7, 17, 27, 37, 47, 57, 67, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 87, 97)
From 101 to 200 digit-7 will come up to 20 times.
(107, 117,127, 137, 147, 157,167, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 187,197)
From 201 to 300 digit-7 will come up to 20 times.
(207, 217, 227, 237, 247, 257, 267, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 287,2 97)
From 301 to 400 digit-7 will come up to 20 times.
(307, 317, 327, 337, 347, 357, 367, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 387, 397)
From 401 to 500 digit-7 will come up to 20 times.
(407, 417, 427, 437, 447, 457, 467, 470, 471, 472, 473, 474, 475, 476, 477, 478, 479, 487, 497)
From 501 to 600 digit-7 will come up to 20 times.
(507, 517, 527, 537, 547, 557, 567, 570, 571, 572, 573, 574, 575, 576, 577, 578, 579, 587, 597)
From 601 to 700 digit-7 will come up to 21 times.
(607, 617, 627, 637, 647, 657, 667, 670, 671, 672, 673, 674, 675, 676, 677, 678, 679, 687, 697,700)
From 701 to 800 digit-7 will come up to 120 times.
701 to 710 → 11 times.
711 to 720 → 11 times.
721 to 730 → 11 times.
731 to 740 → 11 times.
741 to 750 → 11 times.
751 to 760 → 11 times
761 to 770 → 12 times.
771 to 780 → 20 times.
781 to 790 → 11 times.
791 to 800 → 11 times.
Total → 120 times
From 801 to 900 digit-7 will come up to 20 times
(8087, 817, 827, 837, 847, 857, 867, 870, 871, 872, 873, 874, 875, 876, 877, 878, 879, 887, 897)
From 901 to 100 digit-7 will come up to 20 times.
(907, 917, 927, 937, 947, 957, 967, 970, 971, 972, 973, 974, 975, 976, 977, 978, 979, 987, 997)
⇒ From 1 to 100 digit-7 will come up to 20 times
⇒ From 1 to 1000 digit-7 will come =20+20+20+20+20+20+21+120+20+20 = 301 times
Topic 3.5 Pretty Palindromic Patterns
When arrangement of digits in number left to right and right to left are same are called palindromes or palindromic numbers.
Example: 2332, 123321, 525, 707 etc.
* All palindromes using 1, 2, 3 The numbers 121, 313, 222 are some examples of palindromes using the digits ‘1’, ‘2’, 3’. Write all possible 3-digit palindromes using these digits.
Solution:
111, 222, 333, 121, 212, 131, 313, 232, 323.
* Puzzle time
I am a 5-digit palindrome.
I am an odd number.
My ‘t’ digit is double of my ‘u’ digit.
My ‘h’ digit is double of my ‘t’ digit.
Who am I? _________________
Solution:
Condition 1: Number is odd ,so possible digit on 'u' is 1, 3, 5, 7 ,9.
Condition 2: 't' digit is double of 'u' digit so possible digit on 't' from 1, 3, 5, 7 ,9 are 2 (double of 1) and 6 (double of 3).
Condition 3:'h' digit is double of 't' digit so possible digit on 'h' is 4 (double of 2).
Now , digit on 'u' is 1, digit on 't' is 2 and digit on 'h' is 4.
Hence,the 5-digit palindrome is 12421
Topic 3.6 The Magic Number of Kaprekar
When we take any four digit number and formed the largest and the smallest number from digits of this number then find difference between the largest and smallest number. Continue this work forming of the largest and the smallest number and subtracting,we will always get the number 6174. This number 6174 is called The Magic Number of Kaprekar or Kaprekar constant.
Example:
4 digit number = 6823
The largest number = 8632
The smallest number= 2368
Difference= 8632 - 2368 = 6264
New 4 digit number =6264
The largest number =6642
The smallest number= 2466
Difference= 6642 - 2466 = 4176
New 4 digit number = 4176
The largest number = 7641
The smallest number= 1467
Difference= 7641 - 1467 = 6174
Topic 3.7 Clock and Calendar Numbers
Figure it Out
Question 1
Pratibha uses the digits ‘4’, ‘7’, ‘3’ and ‘2’, and makes the smallest and largest 4-digit numbers with them: 2347 and 7432. The difference between these two numbers is 7432 – 2347 = 5085. The sum of these two numbers is 9779. Choose 4–digits to make:
a. the difference between the largest and smallest numbers greater than 5085.
b. the difference between thanklargest and smallest numbers less than 5085.
c. the sum of the largest and smallest numbers greater than 9779.
d. the sum of the largest and smallest numbers less than 9779.
Solution:
(a)
Four digits 3, 7, 1, 2 ( You can take any digit that fulfills the condition)
The largest number = 7321
The smallest number = 1237
Difference between numbers =7321 - 1237 =6084
Sum of numbers = 7321 + 1237 = 8558
(b)
Four digit =2, 3, 5, 1 ( You can take any digit that fulfills the condition)
The largest number = 5321
The smallest number =1235
Difference between numbers =5321 - 1235 = 4086.
(c)
Four digits 6, 4, 3, 4 ( You can take any digit that fulfills the condition)
The largest number = 6443
The smallest number = 3446
Sum of numbers = 6443 + 3446 = 9889
(d)
Four digits 3 ,5, 1, 3 ( You can take any digit that fulfills the condition)
The largest number =5331
The smallest number = 1335
Sum of numbers =5331 + 1335 = 8668
Question 2
What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Solution:
The largest 5-digit palindrome = 99999
The smallest 5-digit palindrome = 10001
Sum = 99999 + 10001 = 110000
Difference= 99999 -10001= 89998
Question 3
The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that?
Solution:
10:01 + 1 hour 10 minutes = 11:11
Hence, after 70 minutes (1 hour 10 minutes) the clock shows the next palindromic time11:11 the next and after 70 minutes next palindromic time is 12:21.
Question 4
How many rounds does the number 5683 take to reach the Kaprekar constant?
Solution:
Round:1
Given number = 5683
The largest number = 8653
The smallest number= 3568
Difference= 8653 - 3568 = 5085
Round:2
New number =5085
The largest number =8550
The smallest number= 5058
Difference= 8550 - 5058 = 3492
Round:3
New number = 3492
The largest number = 9432
The smallest number= 2349
Difference= 9432 - 2349 = 7083
Round:4
New number = 7083
The largest number =8730
The smallest number= 3078
Difference= 8730 - 3078 = 5652
Round:5
New number =5652
The largest number =6552
The smallest number= 2556
Difference= 6552 - 2556 = 3996
Round:6
New number =3996
The largest number =9963
The smallest number= 3699
Difference= 9963 - 3699 = 6264
Round:7
New number =6264
The largest number =6642
The smallest number= 2466
Difference= 6642 - 2466 = 4176
Round:8
New number = 4176
The largest number = 7641
The smallest number= 1467
Difference= 7641 - 1467 = 6174
Hence after 8 rounds the number 5683 takes to reach the Kaprekar constant.
Topi8c 3.8 Mental Math
Observe the figure below. What can you say about the numbers and the lines drawn?
Numbers in the middle column are added in different ways to get the numbers on the sides (1500 + 1500 + 400 = 3400). The numbers in the middle can be used as many times as needed to get the desired sum. Draw arrows from the middle to the numbers on the sides to obtain the desired sums.
Two examples are given. It is simpler to do it mentally!
Solution:
39,800 = 40,000 – 800 + 300 + 300
Question 1
Question 2
Solution:
* Can we make 1,000 using the numbers in the middle? Why not? What about 14,000, 15,000 and 16,000? Yes, it is possible. Explore how. What thousands cannot be made?
Solution:
We can not make 1000 because in the middle table there are no number whose sum is 1000.
For 14000 = 1500 ✖ 8 + 400 ✖ 5
For 15000 = 1500 ✖ 10
For 16000 = 13000 + 1500 ✖ 2
Adding and Subtracting
Here, using the numbers in the boxes, we are allowed to use both addition and subtraction to get the required number. An example is shown.
39,800 = 40,000 – 800 + 300 + 300
45,000 =
5,900 =
17,500 =
21,400 =
Solution:
45,000 = 40,00 + 800 - 300
5,900 = 7,000 -800 - 300
17,500 = 12,000 + 7,000 - 1,500
21,400 = 12,000 + 7,000 + 1,500 + 300 + 300 +300
Figure it Out
Question 1
Write an example for each of the below scenarios whenever possible.
Solution:
5-digit + 5-digit to give a 5-digit sum more than 90,250
⇒ 64324 + 29123 = 93447
5-digit + 3-digit to give a 6-digit sum
⇒ 99876 + 523 =100399
4-digit + 4-digit to give a 6-digit sum
⇒ Not possible
5-digit + 5-digit to give a 6-digit sum
⇒98523 + 32561 = 131084
5-digit + 5-digit to give 18,500
⇒Not possible
5-digit – 5-digit to give a difference less than 56,503
⇒38126 - 23015 = 15111
5-digit – 3-digit to give a 4-digit difference
⇒10000 - 999 =9001
5-digit − 4-digit to give a 4-digit difference
⇒18456 - 9345 = 9111
5-digit − 5-digit to give a 3-digit difference
⇒82365 - 82124 = 241
5-digit − 5-digit to give 91,500
⇒Not possible
(You can use different number also)
Question 2
Always, Sometimes, Never?
Below are some statements. Think, explore and find out if each of the statement is ‘Always true’, ‘Only sometimes true’ or ‘Never true’. Why do you think so? Write your reasoning; discuss this with the class.
a. 5-digit number + 5-digit number gives a 5-digit number
b. 4-digit number + 2-digit number gives a 4-digit number
c. 4-digit number + 2-digit number gives a 6-digit number
d. 5-digit number – 5-digit number gives a 5-digit number
e. 5-digit number – 2-digit number gives a 3-digit number
Solution:(a) Only sometimes true
(b) Always true
(c) Never true
(d) Only sometimes true
(e) Never true
Topic 3.9 Playing with Number Patterns
Here are some numbers arranged in some patterns. Find out the
sum of the numbers in each of the below figures. Should we add
them one by one or can we use a quicker way?
Solution:
(a)
Total number of 50 = 10
Total number of 40 = 12
Sum = 50 ✖ 10 + 40 ✖ 12 = 500 + 480 = 980
(b)
Total number of 5 = 20
Total number of 1 = 44
Sum = 5 ✖ 20 + 1 ✖ 44 = 100 + 44 = 144
(c)
Total number of 32 = 32
Total number of 64 = 16
Sum = 32 ✖ 32 + 64 ✖ 16 = 1024 + 1024 = 2048
(d)
Total number of 4 = 18
Total number of 3 = 17
Sum = 4 ✖ 18 + 3 ✖ 17 = 72 + 51 = 123
(e)
Total number of 15 = 22
Total number of 25 = 22
Total number of 35 = 22
Sum = 15 ✖ 22 + 25✖ 22 + 35✖ 22 = 330 + 550 + 770 = 1650
(f)
Total number of 1000 = 1
Total number of 500 = 4
Total number of 250 = 8
Total number of 125 = 18
Sum = 1000 ✖ 1 + 500 ✖ 4 + 250 ✖ 8 + 125 ✖ 18 = 1000 + 2000 +1000 + 2250 = 6250
Topic 3.10 An Unsolved Mystery - the Collatz Conjecture!
Rule : Make a sequence starts with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat.The sequence will always reach 1.
Example
Make some more Collatz sequences like those above, starting with your favourite whole numbers. Do you always reach 1?
Solution:
- 40, 20, 10, 5, 16, 8, 4, 2, 1
- 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
- 76, 38, 19, 58, 29,88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
Yes, we always reach at 1.
Topic 3.11 Simple Estimation
Figure it Out
Question 1
Steps you would take to walk:
a. From the place you are sitting to the classroom door
b. Across the school ground from start to end
c. From your classroom door to the school gate
d. From your school to your home
Solution:
(a) 10 steps
(b) 500 steps
(c) 200 steps
(d) 5000 steps
(different answer may be possible)
Question 2
Number of times you blink your eyes or number of breaths you take:
a. In a minute
b. In an hour
c. In a day
Solution:
(a) 20 times
(b) 1000 times
(c) 20,000 times
(different answer may be possible)
Question 3
Name some objects around you that are:
a. a few thousand in number
b. more than ten thousand in number
Solution:
(a) Books in the library.
(b) Bricks used to construct school building.
(different answer may be possible)
Estimate the answer
Question 1
Number of words in your maths textbook:
a. More than 5000
b. Less than 5000
Solution:
Number of words in our maths textbook: More than 5000
Question 2
Number of students in your school who travel to school by bus:
a. More than 200
b. Less than 200
Solution
Number of students in our school who travel to school by bus: More than 200
Question 3
Roshan wants to buy milk and 3 types of fruit to make fruit custard for 5 people. He estimates the cost to be ₹ 100. Do you agree with him? Why or why not?
Solution:
Yes,we agree with him.
(different answer may be possible)
Question 4
Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland).
[Hint: Look at the map of India to locate these cities.]
Solution:
Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland) is 3000km.
(different answer may be possible)
Question 5
Sheetal is in Grade 6 and says she has spent around 13,000 hours in school till date. Do you agree with her? Why or why not?
Solution:
If she is studying from nursery and count this time from nursery to Grade 6 then we agree with her.
Question 6
Earlier, people used to walk long distances as they had no other means of transport. Suppose you walk at your normal pace. Approximately how long would it take you to go from:
a. Your current location to one of your favourite places nearby.
b. Your current location to any neighbouring state’s capital city.
c. The southernmost point in India to the northernmost point in India.
Solution:
(a) 20 km
(b) 400 km
(c) 3200 km
(different answer may be possible)
Question 7
Make some estimation questions and challenge your classmates!
Solution:
Do by your self.
Topic 3.12 Games and Winning Strategies
Figure it Out
Question 1
There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap.
Change middle number unit place digit to ten thousand place digit 62871 →12876
Question 2
How many rounds does your year of birth take to reach the Kaprekar constant?
Solution:
Birth year =1992
Round 1 : 9921 -1299 = 8622
Round 2 : 8622 - 2268 = 6354
Round 3 : 6543 - 3456 = 3087
Round 4 : 8730 - 3078 = 5652
Round 5 : 6552 -2556 = 3996
Round 7 : 9963 -3699 = 6264
Round 8 : 6642 - 2466 = 4176
Round 9 : 7641 - 1467 = 6174
Hence, year of birth take 9 rounds to reach the Kaprekar constant 6174.
Question 3
We are the group of 5-digit numbers between 35,000 and 75,000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50,000?
Solution:
Odd digits : 1, 3, 5, 7, 9.
the group of 5-digit numbers between 35,000 and 75,000 Such that all digits are odd
35179, 35917, 37159, 37915, 39157 , 51379, 51739, 51937, 53179 ,53917, 57139, 57319, 59137, 59317 , 71539, 71935, 73159, 73519.
The lagest number = 73519
The smallest number =35179
Number closest to 50000 =51379.
Question 4
Estimate the number of holidays you get in a year including weekends, festivals and vacation. Then try to get an exact number and see how close your estimate is.
Solution: See the calendar and do by your self
Question 5
Estimate the number of liters a mug, a bucket and an overhead tank can hold.
Solution:
A mug = 250 milliliter
A bucket = 10 liter
An overhead tank = 3000 liter
(different answer may be possible)
Question 6
Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.
Solution:
5 digit number = 18000
3 digit number =670
Hence, sum of both of the numbers is 18670
(different numbers may be possible)
Question 7
Choose a number between 210 and 390. Create a number pattern similar to those shown in Section 3.9 that will sum up to this number
Solution:
We choose 305
Question 8
Recall the sequence of Powers of 2 from Chapter 1, Table 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence?
Solution: Sequence of power of 2 : 1, 2, 4, 8, 16,.......
Now check the Collatz conjecture for the sequence,
- 1
- 2, 1
- 4, 2, 1
- 8, 4, 2 , 1
- 16, 8, 4, 2, 1
Hence ,the Collatz conjecture correct for all the starting numbers in this sequence.
Question 9
Check if the Collatz Conjecture holds for the starting number 100.
Solution:
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.
Hemce ,the Collatz Conjecture holds for the starting number 100.
Question 10
Starting with 0, players alternate adding numbers between 1 and 3. The first person to reach 22 wins. What is the winning strategy now?
Solution:
Play this game in the class and find the answer.
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