NCERT Solutions for Class 6 Maths Chapter 03 Number Play

 

Question. 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.
Answer: Numbers play a role in many areas of daily life. We use them to track time on a clock by noting hours and minutes. When shopping, numbers help us check prices and work out the total cost of items. Sports rely on numbers to keep score, such as goals in football or runs in cricket. Height and weight measurements tell us how tall or heavy someone is. In school, numbers show marks earned in exams and assignments.
In simple words: Numbers are used to tell time, track prices, keep scores, measure size and weight, and record marks in school.

Exam Tip: When listing real-world uses of numbers, give concrete examples like specific activities (shopping, sports, school) rather than just general categories.

 

Question. Some children in a park are standing in a line. Each one says a number. What do you think these numbers mean?
Answer: The numbers each child says represent how many taller children are positioned nearby. A child surrounded by taller friends will say a higher number, while a child with fewer taller neighbours will say a lower number.
In simple words: The numbers tell how many taller children are standing next to each child in the line.

Exam Tip: Understand that a number spoken by a child depends on the relative heights of neighbouring children, not absolute height or position.

 

Question 1. Can the children rearrange themselves so that the children standing at the ends say '2'?
Answer: No, children at the ends of the line are unable to say '2'. This is because an end child has only one neighbour, so can say at most '1'. Since they lack two neighbours, they cannot reach '2'.
In simple words: Children at the ends have only one person next to them, so they can only say 1, never 2.

Exam Tip: Remember that position matters - children at the edges have fewer neighbours than those in the middle.

 

Question 2. Can we arrange the children in a line so that all would say only 0s?
Answer: No, this arrangement is not possible. For all children to say 0, they must each be taller than every neighbour. However, if everyone is taller than their neighbours, the heights would need to form an impossible pattern - it only works when they are all the same height.
In simple words: All children saying 0 is only possible if everyone has the exact same height.

Exam Tip: Think about what "0" means - a child says 0 when no neighbour is taller, which requires special conditions.

 

Question 3. Can two children standing next to each other say the same number?
Answer: Yes, it is possible. Two neighbouring children can say the same number if they both have matching numbers of taller neighbours around them. For example, two children of similar height standing between taller children can each say '1'.
In simple words: Two children next to each other can say the same number when they have the same number of taller friends around them.

Exam Tip: Look for symmetric arrangements where two adjacent children have identical neighbour patterns.

 

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?
Answer: No, this is not possible. If four children each say '1', they each have one taller neighbour. This means every one of them has exactly one person taller near them. But the fifth child, being different in height from all others, cannot simultaneously be taller than everyone (to get a '0') while also fitting into this arrangement. Since all children have different heights, achieving this specific pattern is impossible.
In simple words: With everyone at a different height, you cannot get four children saying 1 and one saying 0 at the same time.

Exam Tip: Test the logical constraints: count how many "taller neighbours" each position requires, then check if the unique heights can satisfy all demands.

 

Question 5. For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
Answer: No, the sequence 1, 1, 1, 1, 1 is not possible with five children. For everyone to say '1', each child must have exactly one taller neighbour. However, with an odd number of children and all different heights, at least one child must have no taller neighbours. This means they would need to say '0', not '1'.
In simple words: Each child saying 1 means one taller friend next to them. With five children of different heights, at least one person must have zero taller neighbours.

Exam Tip: Count total "neighbour slots" - an odd number of children makes certain symmetric patterns impossible.

 

Question 6. Is the sequence 0, 1, 2, 1, 0 possible? Why or why not?
Answer: Yes, the sequence 0, 1, 2, 1, 0 is possible. This represents a situation where the middle child is the tallest (and has two taller neighbours, which is zero - wait, this needs revision). Actually: the middle child is tallest and says '2' because two children on either side are shorter. The two children next to the middle child are shorter but each has one taller neighbour (the middle child), so they say '1'. The two end children are shortest and have no taller neighbours, so they say '0'.
In simple words: Put the tallest child in the middle with shorter children on both sides. The pattern works: tallest says 2, next to tallest say 1, shortest at ends say 0.

Exam Tip: Arrange tallest in the middle surrounded by progressively shorter children - this creates the required pattern naturally.

 

Question 7. How would you rearrange the five children so that the maximum number of children say '2'?
Answer: To get the most children saying '2', place the tallest child in the centre with the next two tallest children directly on either side of them. The two shortest children should go at the ends. This way, the three middle children can each say '2', because they will each have two taller neighbours. The two end children will say '1'.
In simple words: Put the three tallest children in the middle and shortest at the ends. The middle children will say 2, and end children will say 1.

Exam Tip: Maximise '2's by clustering the tallest children in the middle where they can have two taller neighbours each.

 

Figure it Out

 

Question 1. Colour or mark the supercells in the table below.
Answer: Supercells in the table are: 6828, 670, 9435, 6828, 3780, 7308, 8000, 5583, 52.
In simple words: A supercell is a cell where the number is larger than all its neighbours around it.

Exam Tip: Check each cell by comparing it to its immediate neighbours (up, down, left, right, and sometimes diagonals if the table is set up that way).

 

Question 2. Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells.
Answer: 4-digit numbers to make the coloured cells as supercells: 5346, 6000, 1000, 1258, 1100, 1200, 1300, 9635, 9800.
In simple words: Choose numbers so that each green cell is bigger than all its neighbours, and no other cell is.

Exam Tip: Plan before filling - make sure green cells get high numbers and white cells get lower numbers.

 

Question 3. Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.
Answer: Getting as many supercells as possible using numbers between 100 and 1000: 800, 400, 500, 300, 450, 200, 700, 470, 900.
In simple words: Fill cells with different numbers so that as many as possible are larger than all their neighbours.

Exam Tip: Place large numbers spaced apart in a checkerboard pattern to maximise the count of supercells.

 

Question 4. Out of the 9 numbers, how many supercells are there in the table above?
Answer: There are 5 supercells in the table above.
In simple words: Count the cells that are bigger than all their neighbours.

Exam Tip: Supercells tend to occur more often in the middle regions of the grid where each cell has the most neighbours to compare against.

 

Question 5. Find out how many supercells are possible for different numbers of cells.
Answer: The number of supercells depends on how many cells are in the table and how they are arranged. Generally, supercells tend to occur more frequently in the middle area than at the edges. The maximum number of supercells is roughly half the total number of cells, or slightly less than half (half + 1) of the total number of cells.
In simple words: The more cells you have, the more supercells you can make. Usually, supercells happen in about the middle region, not at the edges.

Exam Tip: Observe that corner and edge cells are less likely to be supercells since they have fewer neighbours to dominate.

 

Question 6. 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.
Answer: One key pattern that stands out is that supercells occur most often in the middle cells where there are more neighbouring cells to compare against. To get the most supercells, place the largest numbers in the central positions, ensuring they are surrounded by smaller numbers. Since a cell has more neighbours to compare to in the middle of the grid, it has a better chance of being a supercell if it has a larger number there. By strategically placing the largest numbers in central positions, you can maximise the number of supercells in the table.
In simple words: Put big numbers in the middle and small numbers at the edges. The middle cells are more likely to be supercells.

Exam Tip: Strategic placement beats random filling - central locations offer more neighbour positions to dominate.

 

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?
Answer: The largest number in a table will not always be a supercell. If the largest number is surrounded by other relatively large numbers, it may not be bigger than all of them. Conversely, the smallest number in a table can never be a supercell. A supercell must be larger than all its neighbours, but the smallest number in the entire table is smaller than at least one number somewhere, so it cannot satisfy the supercell condition.
In simple words: The biggest number might not be a supercell if large numbers sit next to it. The smallest number can never be a supercell since something is always bigger.

Exam Tip: Supercells depend on local neighbourhood, not global rank - context matters more than absolute size.

 

Question 8. Fill a table such that the cell having the second largest number is not a supercell.
Answer: One way to do this is to place the second-largest number next to the largest number (or surrounded by numbers larger than or equal to it). For example: 800, 750, 500, 300, 450, 200, 700, 470. Here, 750 is the second-largest, but it sits next to 800 (which is larger), so it is not a supercell.
In simple words: Put the second-biggest number next to the biggest number. That way it is not bigger than all its neighbours.

Exam Tip: Place high-ranked numbers adjacent to each other to prevent them from being supercells.

 

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?
Answer: Yes, this is possible. Place the second-smallest number in the middle of the table surrounded by the smallest number and other smaller numbers. At the same time, place the second-largest number next to the largest number so it is not a supercell. For example, in a grid: 800, 751, 500, 550, 450, 300, 200, 250. Here 751 is the second-largest but not a supercell (next to 800), while 550 is the second-smallest but can be a supercell if surrounded by smaller values.
In simple words: Yes, it works. Put the second-smallest in the middle with tiny numbers around it, and the second-biggest next to the biggest.

Exam Tip: Positioning matters - use the middle for the second-smallest (supercell candidate) and edges for the second-largest (non-supercell).

 

Question 10. Can a cell in the middle of the table ever be a supercell if all its neighbouring cells have consecutive numbers? Why or why not?
Answer: No, a middle cell cannot be a supercell if all its neighbouring cells have consecutive numbers. If a cell is surrounded by consecutive numbers and itself has a consecutive number in that sequence, it cannot be larger than all of them at once. For instance, if neighbours are 5, 6, 7, 8 and the cell itself is one of these, it fails the supercell test. The cell would need to be larger than all neighbours, which is impossible when they are all consecutive integers nearby in value.
In simple words: If all neighbours are consecutive numbers like 5, 6, 7, 8, the middle cell cannot beat them all.

Exam Tip: Consecutive numbers leave no room for a number to be strictly larger than all - use non-consecutive numbers to create supercells.

 

Question. Complete Table 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.
Answer:

The biggest number in the table is 96,310.
The smallest even number in the table is 10,936.
The smallest number greater than 50,000 in the table is 60,193.
In simple words: Use the five given digits to form 5-digit numbers. Place the largest numbers in the coloured cells so only they are bigger than their neighbours.

Exam Tip: For a supercell constraint, surround coloured (larger) cells with smaller numbers and ensure white cells are never the maximum in their region.

 

Question. 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.
Answer:
In simple words: Mark each number on the line from 1000 to 10,000 in the right place by measuring how far from the start each number should go.

Exam Tip: Order the numbers from smallest to largest first, then space them evenly based on the number line scale.

 

Question. Figure it Out: Identify the numbers marked on the number lines below, and label the remaining positions.
Answer:



In simple words: Look at where numbers sit on each line, then write in the numbers that belong at the empty spots.

Exam Tip: Count the intervals between marked numbers to find the step size, then use it to fill blank positions.

 

Question. Put a circle around the smallest number and a box around the largest number in each of the sequences above.
Answer:



In simple words: Mark the smallest with a circle and the biggest with a box on each number line.

Exam Tip: Always identify the extremes (smallest and largest) first before performing any other analysis.

 

Question. Find out how many numbers have two digits, three digits, four digits, and five digits:
Answer:

In simple words: Count how many numbers fit into each group by subtracting the smallest from the largest and adding 1.

Exam Tip: For any digit-length group, the count is always the difference between the upper and lower limits plus one.

 

Question. Figure it Out: 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?
Answer:
(a) Other numbers whose digits add up to 14:
59 (5 + 9 = 14)
86 (8 + 6 = 14)
167 (1 + 6 + 7 = 14)
275 (2 + 7 + 5 = 14)
593 (5 + 9 + 3 = 14)
(b) The smallest number whose digit sum is 14 is 59.
(c) To make the largest 5-digit number with a digit sum of 14, you want to place the highest possible digits in the higher place values (thousands, hundreds). The largest 5-digit number with a digit sum of 14 is 59000 (5 + 9 + 0 + 0 + 0 = 14).
(d) The largest number that can be formed with a digit sum of 14 is 59000. You cannot make a bigger number with the same digit sum constraint because adding larger digits in higher places would exceed a sum of 14.
In simple words: Put the biggest digits on the left. Put 5 and 9 first to make 59. Add zeros to make 59000. You cannot make it bigger.

Exam Tip: For maximising a number with a fixed digit sum, place larger digits leftmost and fill remaining places with 0.

 

Question 2. Find out the digit sums of all the numbers from 40 to 70. Share your observations with the class.
Answer:

Observation: The digit sums from 40 to 70 gradually increase from 4 to 15. The sums tend to repeat after every 10 numbers (for example, the sums for 40 and 50 are both 4 and 5, respectively). This pattern can help you quickly identify sums without manually calculating each one.
In simple words: As numbers get bigger from 40 to 70, their digit sums go up. The pattern repeats every 10 numbers, like 40 and 50 both start at 4 and 5.

Exam Tip: Look for repeating patterns in digit sums - knowing the pattern saves time in large number ranges.

 

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?
Answer: Examples of 3-digit numbers with consecutive digits:
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
Observation and Pattern: The digit sums of these consecutive 3-digit numbers increase by 3 for each next number in the sequence. Starting from 6 (for 123), it goes up in steps of 3 (9, 12, 15, 18, 21, 24, etc.). Yes, this pattern continues as long as the digits remain consecutive. Each step up in the number sequence increases the digit sum by 3.
In simple words: When digits are consecutive like 3-4-5, each new number jumps up by 3 in digit sum: 6, then 9, then 12, then 15, and so on.

Exam Tip: Consecutive digits always increase digit sum by the same amount - spot the pattern and use it to predict future sums.

 

Question. 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?
Answer: To solve this, break it into parts based on where '7' appears: in the tens place or in the ones place.

• In the tens place: The numbers 70 to 79 contain the digit '7' in the tens place. There are 10 numbers: 70, 71, 72, 73, 74, 75, 76, 77, 78, and 79.
• In the ones place: The numbers that have '7' in the ones place are 7, 17, 27, 37, 47, 57, 67, 87, and 97. So there are 10 numbers where '7' appears in the ones place.

Total occurrences of '7' between 1 and 100: 10 + 10 = 20 occurrences of the digit '7' in the numbers 1 to 100.
Among the numbers 1-1000, we break this into three parts: hundreds place, tens place, and one's place:

• In the hundreds place: The numbers 700 to 799 contain the digit '7' in the hundreds place. There are 100 numbers from 700 to 799.
• In the tens place: In each set of 100 numbers (i.e. 000-099, 100-199, 200-299, etc.), '7' will appear in the tens place 10 times (just like in the 1-100 range). Since we have 10 sets of 100 numbers (000-999, 100-199, ..., 900-999), '7' will appear in the tens place 10 × 10 = 100 times.
• In the ones place: Similarly, for each set of 100 numbers, '7' will appear in the ones place 10 times. Across 10 sets of 100 numbers, '7' will appear in the ones place 10 × 10 = 100 times.

Total occurrences of '7' between 1 and 1000: 100 + 100 + 100 = 300 occurrences of the digit '7' in the numbers 1 to 1000.
In simple words: Count how many times '7' shows up in each position (ones, tens, hundreds). Add them up. From 1 to 100, it appears 20 times. From 1 to 1000, it appears 300 times.

Exam Tip: Always break the counting by digit position (ones, tens, hundreds) rather than trying to count all instances at once.

 

Question. 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.
Answer: All possible 3-digit palindromes using the digits 1, 2, and 3 are: 111, 121, 131, 212, 222, 232, 313, 323 and 333. These are the 9 possible palindromes we can create with the digits 1, 2, and 3.
In simple words: A palindrome reads the same forwards and backwards. List all palindromes: 111, 121, 131, 212, 222, 232, 313, 323, 333.

Exam Tip: For a 3-digit palindrome, the first and last digits must be the same - choose the first digit (3 ways), then the middle can be any digit (3 ways), for 3 × 3 = 9 total.

 

Question. Explore: Will reversing and adding numbers repeatedly, starting with a 2-digit number, always give a palindrome? Explore and find out.
Answer: Yes, reversing and adding numbers repeatedly, starting with a 2-digit number, will eventually give a palindrome in most cases. The process involves the following steps:

• Take a 2-digit number.
• Reverse the digits of that number.
• Add the original number and the reversed number.
• If the result is a palindrome, stop. If not, repeat the process with the result by reversing its digits and adding again.

Example 1: Take 56 and the reverse of 56 = 65. Add: 56 + 65 = 121 (palindrome)
Example 2: Take 29 and the reverse of 29 = 92. Add: 29 + 92 = 121 (palindrome)
Example 3: Take 84 and the reverse of 84 = 48. Add: 84 + 48 = 132. Reverse 132: 231. Add: 132 + 231 = 363 (palindrome)
Exceptions: There are certain numbers for which this process does not lead to a palindrome or it takes a very large number of steps to determine if a palindrome will occur. A well-known example is 196, which, despite many attempts, has not been proven to result in a palindrome through this process. For most 2-digit numbers, however, this method will eventually produce a palindrome.
In simple words: Flip a number and add it to the original. Keep doing this until you get a palindrome. It works most of the time, but some numbers like 196 might take forever.

Exam Tip: The reverse-and-add method is a well-known algorithm - explain the steps clearly and mention that exceptions like 196 (Lychrel numbers) exist but are rare.

 

Question. 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?
Answer: 5-digit odd palindrome number: 12421
Number in words: Twelve thousand four hundred and twenty-one.
In simple words: The number reads the same forwards and backwards. It is odd. If 'u' = 1, then 't' = 2 and 'h' = 4. So the number is 12421.

Exam Tip: For palindromes, the first and last digits are the same, and the second and fourth are the same. Use the clues to figure out each position.

 

Question. Explore: Take different 4-digit numbers and try carrying out these steps. Find out what happens. Check with your friends what they got.
Answer: Starting with the example 4-digit number 3524:

• Arrange in descending order: 5432 (A)
• Arrange in ascending order: 2345 (B)
• Subtract: 5432 - 2345 = 3087 (C)

Now, repeating the steps with 3087:

• Arrange in descending order: 8730
• Arrange in ascending order: 0378
• Subtract: 8730 - 0378 = 8352

Repeating again with 8352:

• Arrange in descending order: 8532
• Arrange in ascending order: 2358
• Subtract: 8532 - 2358 = 6174

Repeating again with 6174:

• Arrange in descending order: 8532
• Arrange in ascending order: 2358
• Subtract: 8532 - 2358 = 6174

Once you reach 6174, you will always return to 6174 when you repeat the process. This number is known as the Kaprekar constant. If you try this process with any 4-digit number (where not all digits are the same), you will always eventually reach 6174.
In simple words: Arrange digits biggest to smallest, then smallest to biggest. Subtract one from the other. Repeat. You always reach 6174 and stay there.

Exam Tip: The Kaprekar constant (6174) is a famous result in recreational mathematics - mention it by name to show deeper understanding.

 

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 the largest 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.
Answer:
(a) Find digits such that the difference between the largest and smallest numbers is greater than 5085.
Let us use the digits 9, 8, 1, 0:
Largest number: 9810
Smallest number: 0189 (or simply 189)
Difference: 9810 - 189 = 9621
Since 9621 is greater than 5085, this works.
(b) Find digits such that the difference between the largest and smallest numbers is less than 5085.
Let us use the digits 5, 4, 2, 1:
Largest number: 5421
Smallest number: 1245
Difference: 5421 - 1245 = 4176
Since 4176 is less than 5085, this works.
(c) Find digits such that the sum of the largest and smallest numbers is greater than 9779.
Let us use the digits 9, 8, 7, 3:
Largest number: 9873
Smallest number: 3789
Sum: 9873 + 3789 = 13662
Since 13662 is greater than 9779, this works.
(d) Find digits such that the sum of the largest and smallest numbers is less than 9779.
Let us use the digits 4, 3, 2, 0:
Largest number: 4320
Smallest number: 0234 (or simply 234)
Sum: 4320 + 234 = 4554
Since 4554 is less than 9779, this works.
In simple words: Pick different digits. Arrange them to make the biggest and smallest 4-digit numbers. Find their difference or sum. Check if it meets the condition.

Exam Tip: Use extreme digit combinations (like 9, 8, 0, 1) to get larger differences, and milder combinations (like 4, 3, 2, 1) for smaller results.

 

Question 2. What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Answer: The smallest 5-digit palindrome is 10001.
The largest 5-digit palindrome is 99999.
Sum: 10001 + 99999 = 110000
Difference: 99999 - 10001 = 89998
In simple words: The smallest 5-digit palindrome that reads the same forwards and backwards is 10001. The biggest is 99999. Add them to get 110000. Subtract to get 89998.

Exam Tip: For 5-digit palindromes, the pattern is abcba - choose the smallest and largest valid combinations to find extremes.

 

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?
Answer: The next palindromic time after 10:01 is 11:11.
The difference between 10:01 and 11:11 is 1 hour and 10 minutes, which equals 70 minutes.
The next palindromic time after 11:11 is 12:21.
The difference between 11:11 and 12:21 is 1 hour and 10 minutes, which also equals 70 minutes.
In simple words: After 10:01, the next palindromic time is 11:11 (70 minutes later). After that is 12:21 (another 70 minutes).

Exam Tip: For palindromic times in 12-hour format, look for times like HH:HH or similar patterns that read the same backwards.

 

Question 4. How many rounds does the number 5683 take to reach the Kaprekar constant?
Answer: Let us apply the Kaprekar routine, starting with 5683:
Largest number: 8653
Smallest number: 3568
Subtract: 8653 - 3568 = 5085
Next step: Largest number: 8550, Smallest number: 0558 (or 558)
Subtract: 8550 - 558 = 7992
Next step: Largest number: 9972, Smallest number: 2799
Subtract: 9972 - 2799 = 7173
Next step: Largest number: 7731, Smallest number: 1377
Subtract: 7731 - 1377 = 6354
Next step: Largest number: 6543, Smallest number: 3456
Subtract: 6543 - 3456 = 3087
Next step: Largest number: 8730, Smallest number: 0378
Subtract: 8730 - 378 = 8352
Next step: Largest number: 8532, Smallest number: 2358
Subtract: 8532 - 2358 = 6174
It took 7 rounds for 5683 to reach the Kaprekar constant 6174.
In simple words: Keep arranging digits big to small, then small to big, and subtracting. Count how many times you do this until you reach 6174. The answer is 7 times.

Exam Tip: Show all steps clearly - each subtraction is one "round" until you reach 6174.

 

Question. 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?
Answer: If we are trying to make 1,000 using the numbers in the middle, the reason it is not possible is because the numbers given in the middle of the table are likely much larger than 1,000 and adding or subtracting these larger values will not allow us to reach exactly 1,000.
Yes, we can make 14,000, 15,000 and 16,000 by carefully adding or subtracting the given middle numbers.
Let us assume the middle numbers are 1,500, 3,000, 4,000 and 7,000. Here is how we can make each number:

• 14,000: 7,000 + 4,000 + 3,000 = 14,000
• 15,000: 7,000 + 4,000 + 3,000 + 1,000 = 15,000
• 16,000: 7,000 + 4,000 + 3,000 + 2,000 = 16,000

These are just examples based on possible middle numbers. We can explore different combinations depending on the exact numbers given.
Not all combinations of thousands can be made using simple additions or subtractions from the middle numbers. For example, if the differences between the middle numbers are too large or too small, we won't be able to reach exact amounts for some thousands.
In simple words: The middle numbers are big, so you cannot make small numbers like 1,000 from them. But you can make 14,000, 15,000, and 16,000 by adding and subtracting the middle numbers. Some numbers like 11,000 or 13,000 might not be possible.

Exam Tip: Think about which sums and differences are possible with the given numbers - some targets are unreachable due to the gaps between available values.

 

Question. Figure it Out: Write an example for each of the below scenarios whenever possible.
Answer: The following are examples for each scenario:

• 5-digit + 1-digit to give a 5-digit sum more than 90,250: 50,000 + 50,000 = 100,000 (but this is 6-digit). Try: 50,000 + 9 = 50,009 (still under 90,250). A working example: 85,000 + 8 = 85,008.
• 5-digit + 3-digit to give a 6-digit sum: 50,000 + 500 = 50,500 (5-digit). Try: 99,000 + 999 = 99,999 (still 5-digit). A working example: 99,500 + 999 = 100,499.
• 4-digit + 4-digit to give a 6-digit sum: 9,999 + 9,999 = 19,998 (5-digit). Try: 9,000 + 9,000 = 18,000 (still 5-digit). A working example: 7,000 + 4,000 + 3,000 = 14,000 (still doesn't work). Actually: 50,000 + 50,000 would work, but these are 5-digit. For 4-digit: 5,000 + 5,000 = 10,000 (5-digit). The maximum with 4-digit is 9,999 + 9,999 = 19,998 (5-digit). A 6-digit sum from 4-digit + 4-digit is not possible in the usual sense. But if we interpret this differently: it's impossible as stated.
• 5-digit + 5-digit to give a 6-digit sum: 50,000 + 50,000 = 100,000 (6-digit). ✓ This works.
• 5-digit - 5-digit to give a 3-digit difference: 55,000 - 54,500 = 500 (3-digit). ✓ This works.

In simple words: For each scenario, pick two numbers and add or subtract them. Check if the result has the right number of digits.

Exam Tip: Understand the digit count of sums and differences - practise estimating before calculating.

 

Question 1. Could you find examples for all the cases? If not, think and discuss what could be the reason. Make other such questions and challenge your classmates.
Answer: Yes, we can find examples for most situations. A 5-digit number plus a 5-digit number gives a 5-digit number when both numbers are small. For example, 10,000 + 45,000 = 55,000. However, when both numbers are large, like 90,000 + 90,000 = 180,000, we get a 6-digit result. A 4-digit number plus a 2-digit number always gives a 4-digit result because the biggest sum of two 4-digit numbers is 99,999 + 9,999 = 109,998, which is still a 5-digit number. We cannot get a 6-digit sum. For subtraction, a 5-digit number minus a 5-digit number gives different results depending on how close the numbers are. If the numbers are near each other, like 90,000 - 50,000 = 40,000, we get a 5-digit result. If they are far apart, like 10,000 - 9,999 = 1, we may get a 1-digit result. The reason certain cases seem impossible is that the size limits of numbers prevent certain outcomes.
In simple words: Some additions and subtractions give certain digit results, while others don't - it depends on how big the numbers are.

Exam Tip: Always check the maximum and minimum possible values for each operation to understand why certain digit combinations are possible or impossible.

 

Question 2. 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) A 5-digit number + 5-digit number gives a 5-digit number
(b) A 4-digit number + 2-digit number gives a 4-digit number
(c) A 4-digit number + 2-digit number gives a 6-digit number
(d) A 5-digit number - 5-digit number gives a 5-digit number
(e) A 5-digit number - 2-digit number gives a 3-digit number
Answer:
(a) Only sometimes true. Two small 5-digit numbers, such as 10,000 + 10,000 = 20,000, result in a 5-digit sum. However, two large 5-digit numbers, like 90,000 + 90,000 = 180,000, give a 6-digit number.
(b) Only sometimes true. A small 4-digit number like 1,000 + 2-digit number 50 = 1,050 gives a 4-digit result. However, a large 4-digit number like 9,999 + 2-digit number 99 = 10,098 gives a 5-digit result.
(c) Never true. The biggest sum of a 4-digit number and a 2-digit number would be 9,999 + 99 = 10,098, which is a 5-digit number, not a 6-digit number. A 4-digit number and a 2-digit number can never add up to a 6-digit number.
(d) Only sometimes true. If two 5-digit numbers are near in value, like 90,000 - 50,000 = 40,000, the result is a 5-digit number. If they are far apart, like 10,000 - 9,999 = 1, the result is a 1-digit number.
(e) Only sometimes true. If the 5-digit number is small, like 10,000 - 9 = 9,991, the result is still a 5-digit number. If the 5-digit number is small enough, like 1,001 - 9 = 992, the result becomes a 3-digit number.
In simple words: When you add or subtract numbers, the number of digits in the result changes based on the actual values of the numbers, not just their digit counts.

Exam Tip: Test each statement with extreme cases - the smallest and largest possible values for each digit category - to determine if a rule is always, sometimes, or never true.

 

Question 3. Additional challenges for classmates:
(i) Find a 5-digit number and subtract a 3-digit number to get a 5-digit difference.
(ii) Find a 4-digit number that, when added to a 3-digit number, gives a sum less than 2,000.
(iii) Find two 4-digit numbers that sum to exactly 15,000.
Answer:
(i) Consider 50,000 as the 5-digit number and 100 as the 3-digit number. When we subtract: 50,000 - 100 = 49,900, we get a 5-digit difference.
(ii) Take 1,000 as the 4-digit number and 500 as the 3-digit number. Adding them: 1,000 + 500 = 1,500, which is less than 2,000.
(iii) The two 4-digit numbers could be 7,000 and 8,000. Adding them: 7,000 + 8,000 = 15,000.
In simple words: By choosing appropriate numbers based on the limits of addition and subtraction, we can satisfy all these conditions.

Exam Tip: When faced with these challenges, identify the constraints first, then work backwards from the target result to find suitable numbers.

 

Question 4. Checking whether always, sometimes or never:
(a) A 5-digit number + 5-digit number gives a 5-digit number
Answer: Only sometimes true. Two small 5-digit numbers, like 10,000 + 10,000 = 20,000, result in a 5-digit number. However, two large 5-digit numbers, like 90,000 + 90,000 = 180,000, would give a 6-digit number.
In simple words: Adding two 5-digit numbers gives a 5-digit result only when the numbers are small enough that they don't create a carry to the hundred-thousands place.

Exam Tip: Always test boundary cases - the smallest and largest values in each digit range - to verify whether a statement holds always, sometimes, or never.

 

Question 5. Checking whether always, sometimes or never:
(b) A 4-digit number + 2-digit number gives a 4-digit number
Answer: Only sometimes true. A small 4-digit number like 1,000 + a 2-digit number like 50 = 1,050 gives a 4-digit result. However, a large 4-digit number like 9,999 + a 2-digit number like 99 = 10,098 gives a 5-digit result.
In simple words: Whether the sum stays 4-digit or becomes 5-digit depends on whether the addition causes a carry to the ten-thousands place.

Exam Tip: The maximum sum of a 4-digit and 2-digit number is 9,999 + 99 = 10,098 (5-digit), and the minimum is 1,000 + 10 = 1,010 (4-digit), showing why this rule is sometimes, not always, true.

 

Question 6. Checking whether always, sometimes or never:
(c) A 4-digit number + 2-digit number gives a 6-digit number
Answer: Never true. The maximum possible sum of a 4-digit number and a 2-digit number would be 9,999 + 99 = 10,098, which is a 5-digit number. A 4-digit number and a 2-digit number can never add up to a 6-digit number.
In simple words: No matter how large the 4-digit and 2-digit numbers are, their sum cannot reach 100,000, so they can never produce a 6-digit result.

Exam Tip: When determining if something is impossible, find the maximum possible value for that operation - if it still doesn't reach the target range, the statement is never true.

 

Question 7. Checking whether always, sometimes or never:
(d) A 5-digit number - 5-digit number gives a 5-digit number
Answer: Only sometimes true. When two 5-digit numbers are close in value, like 90,000 - 50,000 = 40,000, the result is a 5-digit number. However, when the 5-digit numbers are far apart, like 10,000 - 9,999 = 1, the result is a 1-digit number.
In simple words: Subtracting two 5-digit numbers gives a 5-digit result only when the numbers are large and close enough to each other in value.

Exam Tip: In subtraction, the number of digits in the result depends on the gap between the two numbers, not just their initial digit counts.

 

Question 8. Checking whether always, sometimes or never:
(e) A 5-digit number - 2-digit number gives a 3-digit number
Answer: Only sometimes true. If the 5-digit number is small, like 10,000 - 9 = 9,991, the result is still a 5-digit number. However, if the 5-digit number is small enough, like 1,001 - 9 = 992, the result becomes a 3-digit number.
In simple words: Subtracting a 2-digit number from a 5-digit number usually leaves a 5-digit result, unless the 5-digit number is very close to 1,000.

Exam Tip: Understand the range of 5-digit numbers (10,000 to 99,999) to see when subtracting a 2-digit number would drop the result below 10,000.

 

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Question 9. 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? Share and discuss in class the different methods each of you used to solve these questions.
Answer:
Pattern a: The pattern alternates between the numbers 40 and 50. By observing the rows and columns which show symmetry, we can count each number separately and multiply: Total count of 40s: 12, Total count of 50s: 9. Sum = (12 × 40) + (9 × 50) = 480 + 450 = 930
Pattern b: This pattern displays dice with dots, and each die has a specific value. We count the total dots (where each white dot equals 1), then multiply based on the number of clusters or repetitions. Total count of dots: 44. Total count of dice with 5 dots: 20. Sum = (44 × 1) + (20 × 5) = 44 + 100 = 144
In simple words: Instead of adding each number separately, group the numbers and count how many times each repeats, then multiply and add the products together.

Exam Tip: Look for patterns and groupings in figures - recognizing that numbers repeat allows you to use multiplication as a shortcut instead of tedious one-by-one addition.

 

Question 10. Pattern c - Rows and columns of 32s and 64s
Answer: Here we observe rows of 32s and 64s. Count the number of 32s and 64s and multiply accordingly: Total 32s: 32 numbers. Total 64s: 16 numbers. Sum = (32 × 32) + (16 × 64) = 1,024 + 1,024 = 2,048
In simple words: Count how many times each number appears, multiply each number by its count, then add these products together to get the total.

Exam Tip: Multiplication saves time when you have repeated numbers - always look for how many times a number appears before adding.

 

Question 11. Pattern d - Dice grid arrangement
Answer: This appears similar to a dice grid. We can count the dots as groups of 1s to work out the total sum quickly. Total count of dice with 3 dots: 17. Total count of dice with 4 dots: 18. Sum = (17 × 3) + (18 × 4) = 51 + 72 = 123
In simple words: Group the dice by the number of dots they have, multiply each dot value by how many dice show that value, then add these products.

Exam Tip: Organizing numbers into groups before multiplying is far quicker than adding them one at a time.

 

Question 12. Pattern e - Hexagonal arrangement with alternating numbers
Answer: This is a hexagonal shape with alternating numbers like 15, 25 and 35. Count how many times each number appears and multiply: 15s: 12 occurrences, 25s: 6 occurrences, 35s: 6 occurrences. Sum = (12 × 15) + (6 × 25) + (6 × 35) = 180 + 150 + 210 = 540
In simple words: Count each different number, multiply by how often it appears, then add all the products to find the total.

Exam Tip: When patterns involve multiple different numbers, organize them by value and count occurrences separately for accuracy.

 

Question 13. Pattern f - Circular arrangement of numbers
Answer: This is a circular pattern with values like 125, 250, 500 and 1,000. Count each occurrence and add them: 125s: 12 occurrences, 250s: 8 occurrences, 500s: 5 occurrences, 1,000s: 1 occurrence. Sum = (12 × 125) + (8 × 250) + (5 × 500) + (1 × 1,000) = 1,500 + 2,000 + 2,500 + 1,000 = 7,000
In simple words: Count how many times each value appears in the circle, multiply each by its count, then add all the products together.

Exam Tip: Using multiplication with counts is much faster and reduces the chance of errors compared to adding individual numbers.

 

Question 14. Methods Discussion
Answer: There are two main approaches to solving these pattern problems. Adding one by one is straightforward but time-consuming, especially when there are many numbers. Using multiplication is quicker because you identify repeated values and count how often they appear, then multiply each value by its count. This method works best when numbers repeat clearly in a pattern. This can form a good class discussion about how grouping and multiplication save time compared to adding each number individually.
In simple words: Multiplication is faster than adding when numbers repeat - count the repetitions, multiply, then add the products.

Exam Tip: In exam situations where speed matters, always ask yourself: "Does this pattern repeat?" - if yes, use multiplication instead of addition.

 

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Question 15. Make some more Collatz sequences like those above, starting with your favourite whole numbers. Do you always reach 1? Do you believe the conjecture of Collatz that all such sequences will eventually reach 1? Why or why not?
Answer: Collatz Sequence Examples: Starting with 10: 10 - 5 - 16 - 8 - 4 - 2 - 1. Starting with 15: 15 - 46 - 23 - 70 - 35 - 106 - 53 - 160 - 80 - 40 - 20 - 10 - 5 - 16 - 8 - 4 - 2 - 1. Starting with 19: 19 - 58 - 29 - 88 - 44 - 22 - 11 - 34 - 17 - 52 - 26 - 13 - 40 - 20 - 10 - 5 - 16 - 8 - 4 - 2 - 1. In all these examples and many others, the sequence always eventually reaches 1. This pattern has been observed for a vast range of starting numbers, though it has not been formally proven for all possible numbers. The Collatz conjecture appears to be true based on empirical evidence - every tested sequence eventually reaches 1, even for very large numbers. This makes it seem likely that the conjecture holds for all numbers. The reasoning behind this is rooted in how the rules work: dividing even numbers and multiplying odd numbers by 3 and adding 1 creates a tendency for most sequences to get reduced quickly. However, since the problem is still unsolved, there is always the possibility of an exception for some very large, unknown number.
In simple words: When you follow the Collatz rules (halve if even, triple and add 1 if odd), every starting number tested so far eventually reaches 1, but we can't be completely sure this is true for all numbers.

Exam Tip: When discussing unsolved conjectures, distinguish between empirical evidence (many examples tested) and formal proof (proven for all cases) - the Collatz conjecture has strong evidence but no complete proof yet.

 

Question 16. Figure it Out - We shall do some simple estimates. It is a fun exercise, and you may find it amusing to know the various numbers around us. Remember, we are not interested in the exact numbers for the following questions. Share your methods of estimation with the class.
Answer: Estimation questions encourage us to make reasonable guesses about large quantities without needing exact values. The goal is to develop number sense and understand magnitudes. For example, estimating distances involves knowing scale and geography. Estimating time involves understanding daily routines and durations. For object quantities, we think about density and groupings. These estimation skills help us make quick judgments in real life and develop intuition about the world around us.
In simple words: Estimation means making a smart guess about a quantity without counting or measuring everything - it's about developing a feel for how big or small something is.

Exam Tip: In estimation questions, showing your reasoning method is often more important than arriving at an exact number - examiners want to see that your thinking process is logical.

 

Question 17. 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
Answer:
(a) From where I am sitting to the classroom door: 15 - 20 steps
(b) Across the school ground from start to end: 100 - 150 steps
(c) From my classroom door to the school gate: 50 - 100 steps
(d) From my school to my home: 1,000 - 2,000 steps
In simple words: Short distances like within a classroom take just a few steps, while longer distances like from school to home take many hundreds or thousands of steps.

Exam Tip: When estimating step counts, think about your normal walking pace and the approximate distance in metres, then divide by the length of your stride (usually about 0.5 - 1 metre per step).

 

Question 18. 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
Answer:
(a) In a minute: 15 - 20 blinks and 12 - 16 breaths
(b) In an hour: 900 - 1,200 blinks and 720 - 960 breaths
(c) In a day: 14,000 - 19,000 blinks and 17,000 - 23,000 breaths
In simple words: Your body does many automatic actions every second - blinking and breathing happen constantly, adding up to huge numbers over hours and days.

Exam Tip: For these questions, count your actual blinks and breaths in one minute, then multiply: ×60 for an hour, ×1,440 for a day (60 × 24).

 

Question 19. Name some objects around you that are:
(a) A few thousand in number
(b) More than ten thousand in number
Answer:
(a) Objects that are a few thousand in number:
- Grains of sand or soil particles in a small patch of ground
- Blades of grass in a small lawn
- Pieces of chalk in a large box
(b) Objects that are more than ten thousand in number:
- Books in a large library
- Leaves on trees in a school yard
- Pins or needles in a factory or workshop
In simple words: Small scattered items like grains of sand or grass blades add up to thousands quickly, while larger collections like books or leaves can easily reach tens of thousands.

Exam Tip: When naming objects in a quantity range, think about density - how closely packed items are - which helps you estimate total counts.

 

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Question 20. Estimate the answer. Try to guess within 30 seconds. Check your guess with your friends.
Answer: This is a timed estimation exercise designed to develop quick number sense and intuition. The goal is not accuracy but speed and reasonable thinking. By comparing your estimates with classmates, you learn different approaches to estimation and see which strategies work best.
In simple words: Quick estimation means making a fast, reasonable guess based on what you know - you don't need the exact answer, just something close enough.

Exam Tip: In timed estimation problems, it's better to use a simple rough method and finish in time than to aim for perfect accuracy and run out of time.

 

Question 21. Number of words in your maths textbook:
(a) More than 5000
(b) Less than 5000
Answer: (a) More than 5000. Because most textbooks in Class 6 contain a large number of explanations, exercises and examples. It's highly likely that the total word count exceeds 5000 words.
In simple words: Textbooks have many pages with lots of text, so they almost always contain thousands of words.

Exam Tip: When estimating word counts in books, estimate words per page (usually 200-300 words), then multiply by the number of pages.

 

Question 22. Number of students in your school who travel to school by bus:
(a) More than 200
(b) Less than 200
Answer: (a) More than 200. Because my school is very large and in a large school, a significant number of students typically rely on buses for transportation, especially if the school covers a wide area.
In simple words: Large schools usually have many students who need to travel far, so most rely on buses.

Exam Tip: For school-based estimates, consider your school's size and location - urban schools with wide catchment areas typically have more bus users.

 

Question 23. Roshan wants to buy milk and 3 types of fruit to make fruit custard for 5 people. He estimates the cost to be Rs 100. Do you agree with him? Why or why not?
Answer: No, because the cost of milk and 3 types of fruit can exceed Rs 100 depending on the quantity and type of fruits. Fruits like apples, grapes or bananas could add up in cost, making Rs 100 a bit low for 5 people. Making Rs 100 a bit low for 5 people.
In simple words: The cost of milk and multiple types of fruit for five people usually comes to more than Rs 100.

Exam Tip: When evaluating cost estimates, consider current prices of items and the quantities needed - Rs 100 is quite tight for five servings of fruit custard.

 

Question 24. Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland). [Hint: Look at the map of India to locate these cities.]
Answer: The distance between Gandhinagar (in Gujarat) and Kohima (in Nagaland): Approximately 2,500 - 3,000 kilometres, because Gujarat and Nagaland are on opposite sides of India, with Gandhinagar in the west and Kohima in the northeast, making this a long-distance journey.
In simple words: These two cities are on opposite sides of India, so the distance between them is very large - several thousand kilometres.

Exam Tip: For geography-based distance estimates, visualize India's map and note the cardinal directions - opposite sides of the country means thousands of kilometres.

 

Question 25. 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?
Answer: No, because on average, a school day lasts around 6 hours and with about 200 school days in a year, she would spend approximately 1,200 hours in school per year. Given that she has been in school for around 6 years, her total would be closer to 7,200 hours, much less than 13,000 hours.
In simple words: School days are about 6 hours long and there are about 200 school days per year, so in 6 years, the total is far less than 13,000 hours.

Exam Tip: When checking time-based claims, work backwards: hours per day × school days per year × number of years - this simple calculation quickly shows if an estimate is reasonable.

 

Question 26. Walking at a 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 (Kanyakumari to Kashmir)
Answer:
(a) Your current location to one of your favourite places nearby: 10 - 30 minutes
(b) Your current location to any neighbouring state's capital city: Several days to weeks
(c) From the southernmost point to the northernmost point of India (Kanyakumari to Kashmir): It would take months to walk this distance
In simple words: Walking to nearby places takes minutes, to another state takes days, and across the entire country takes months.

Exam Tip: For walking-time estimates, assume an average walking speed of 1 metre per second or about 5 kilometres per hour, then calculate based on estimated distances.

 

Question 27. Make some estimation questions and challenge your classmates! Examples include:
- How many pencils do you think a class of 30 students uses in a year?
- Estimate the number of steps it would take to walk around your entire school building
- Guess the number of pages in all your textbooks combined
- Estimate how many litres of water your family uses in a week
Answer: Creating your own estimation questions helps develop deeper understanding. For a class of 30 students using pencils in a year, consider that each student might use 1-2 pencils per week, giving roughly 30-60 pencils per week. Over 40 school weeks, this would be 1,200 - 2,400 pencils. Estimating steps around the school building involves knowing the perimeter and your stride length. For combined textbook pages, add up the approximate pages in each subject's textbooks. For weekly water usage, think about daily activities like drinking, cooking, washing and bathing for each family member. These questions teach that estimation is everywhere in daily life and develops practical number sense.
In simple words: Estimation questions help you think about real numbers in your own life - pencils used, distances walked, pages read, water consumed.

Exam Tip: When creating estimation questions, pick topics from everyday life that everyone can relate to - this makes the exercise meaningful and engaging.

 

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Question 28. Figure it Out - 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.
Answer: With exchange of two digits of one of the numbers, there will be 4 supercells:

In simple words: By swapping specific digits in one number, you create more numbers that are larger than all their neighbouring numbers in the grid.

Exam Tip: To find supercells, compare each number with its neighbours (up, down, left, right) - a supercell must be greater than all of them.

 

Question 29. How many rounds does your year of birth take to reach the Kaprekar constant?
Answer: My year of birth is 2013.
Round 1:
- Descending order: 3210
- Ascending order: 0123
- Subtract: 3210 - 0123 = 3087
Round 2:
- Descending order: 8730
- Ascending order: 0378
- Subtract: 8730 - 0378 = 8352
Round 3:
- Descending order: 8532
- Ascending order: 2358
- Subtract: 8532 - 2358 = 6174
After 3 rounds, I reach the Kaprekar constant 6174.
In simple words: Take your birth year, arrange its digits in descending order, then ascending order, subtract, and repeat - you'll eventually always reach 6174.

Exam Tip: The Kaprekar constant (6174) is a fascinating number where any 4-digit number will reach it after applying this rearrange-and-subtract process repeatedly.

 

Question 30. 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?
Answer: The possible odd digits are: 1, 3, 5, 7 and 9.
- Largest number: 73,999
- Smallest number: 35,111
- Closest to 50,000: 51,111
In simple words: When all digits must be odd and the number falls between 35,000 and 75,000, the largest uses 7 in ten-thousands place with 9s elsewhere, the smallest uses 3 and 5 for the first two digits with 1s elsewhere, and the closest to 50,000 uses 5 and 1 carefully.

Exam Tip: When constructing numbers with digit constraints, work systematically from left to right (most significant digit first) to find maximum and minimum values.

 

Question 31. 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.
Answer: Holidays in a year:
- Weekends: 104 days
- Summer vacation: 50 days
- Winter vacation: 12 days
- Festival holidays: 12 days
- Public holidays: 4 days
Total estimated holidays = 104 + 50 + 12 + 12 + 4 = 182 days.
In simple words: By counting Saturdays and Sundays (about 104 days), plus vacation periods and festival days, you get roughly 180-190 days of holidays per year.

Exam Tip: When estimating holidays, break them into categories (weekends, vacation, festivals, public holidays) and add them separately for accuracy.

 

Question 32. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.
Answer: One possible combination is:
5-digit number: 12,345
3-digit number 1: 789
3-digit number 2: 536
Sum: 12,345 + 789 + 536 = 18,670
In simple words: Choose a 5-digit number less than 18,670, then find two 3-digit numbers that add up to the remaining total.

Exam Tip: To solve these number combination problems, start with the largest number (5-digit), then adjust the 3-digit numbers to reach the target sum.

 

Question 33. 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.
Answer: We can create a simple addition pattern using a series of repeated numbers. For example, to reach 300: 40 + 40 + 40 + 50 + 50 + 50 + 30 = 300. The pattern uses 40 repeated 3 times, 50 repeated 3 times and then a 30 to complete the sum. Another method: we can use a simple repetitive number pattern like 50 + 50 + 50 + 50 + 50 + 50 = 300. This pattern repeats the number 50 six times to reach the sum of 300.
In simple words: Create a pattern where the same numbers repeat multiple times, and when added together, they reach your target number.

Exam Tip: When creating addition patterns, use repeated numbers and simple multiplication as a shortcut - it's faster than random combinations.

 

Question 34. 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?
Answer: The Collatz conjecture suggests that starting with any positive integer, repeatedly applying the following steps will eventually lead to the number 1: If the number is even, divide it by 2. If the number is odd, multiply it by 3 and add 1. For powers of 2 (like 1, 2, 4, 8, 16, 32, etc.), the conjecture holds because every power of 2 is even, so you always divide it by 2. Dividing powers of 2 by 2 continuously leads to smaller powers of 2, eventually reaching 1. This explains why the Collatz conjecture is correct for all the starting numbers in the sequence of powers of 2.
In simple words: Powers of 2 are all even numbers, so the Collatz rule always divides them by 2. Dividing by 2 keeps giving you smaller powers of 2 until you reach 1.

Exam Tip: Understanding why the Collatz conjecture works for powers of 2 (they're all even, so division is the only operation used) gives insight into the conjecture's broader validity.

 

Question 35. Checking if the Collatz Conjecture holds for the starting number 100.
Answer: Applying the steps of the Collatz Conjecture to 100:
100 is even: 100/2 = 50
50 is even: 50/2 = 25
25 is odd: 25 × 3 + 1 = 76
76 is even: 76/2 = 38
38 is even: 38/2 = 19
19 is odd: 19 × 3 + 1 = 58
58 is even: 58/2 = 29
29 is odd: 29 × 3 + 1 = 88
88 is even: 88/2 = 44
44 is even: 44/2 = 22
22 is even: 22/2 = 11
11 is odd: 11 × 3 + 1 = 34
34 is even: 34/2 = 17
17 is odd: 17 × 3 + 1 = 52
52 is even: 52/2 = 26
26 is even: 26/2 = 13
13 is odd: 13 × 3 + 1 = 40
40 is even: 40/2 = 20
20 is even: 20/2 = 10
10 is even: 10/2 = 5
5 is odd: 5 × 3 + 1 = 16
16 is even: 16/2 = 8
8 is even: 8/2 = 4
4 is even: 4/2 = 2
2 is even: 2/2 = 1
After these steps, we reach 1, confirming that the Collatz Conjecture holds for the starting number 100.
In simple words: Starting from 100 and following the Collatz rules, the sequence eventually reaches 1, proving the conjecture works in this case.

Exam Tip: When verifying the Collatz Conjecture, work step-by-step without skipping - each calculation must follow the rules precisely to avoid errors.

 

Question 36. Starting with 0, players alternate adding numbers between 1 and 3. The first person to reach 22 wins. What is the winning strategy now?
Answer: To ensure a win, I will aim to force my opponent to land on 18. Here's why: From 18, no matter what the opponent adds (1, 2 or 3), I can always add the right amount to reach 22:
If they add 1 (making it 19), I add 3 (making it 22).
If they add 2 (making it 20), I add 2 (making it 22).
If they add 3 (making it 21), I add 1 (making it 22).
The Sequence of Winning Moves: If I want to ensure that I can win, I must always aim to get to the following numbers: 4, 8, 12, 16 and 18. If I land on one of these numbers, I can always maintain control and eventually force my opponent to lose by making them reach 18.
Start by aiming for 4 and then 8 and continue until I reach 18. From there, I'll be able to make the final move to 22 and win.
This is the optimal strategy to win this game if I play first or are able to steer the game into this sequence.
In simple words: The winning strategy is to control certain key numbers (4, 8, 12, 16, 18) - if you land on these, your opponent can never escape, and you'll force them to lose.

Exam Tip: In games involving number sequences and moves, identify the "safe positions" where you have control - work backwards from the winning position to find them.