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AP 6th Class Maths Textbook Solutions – 3. Number Play (2026-27)

meebadi ఆగస్టు 23, 2026 0
3 NUMBER PLAY
Page 130
  • What do you think these numbers mean? Did you figure out what these numbers represent?
    Each number represents the number of taller neighbours standing immediately next to that child. A child says '0' if neither neighbour is taller, '1' if exactly one neighbour is taller, and '2' if both neighbours standing next to them are taller.
Math Talk (Page 132)
  • 1. Can the children rearrange themselves so that the children standing at the ends say '2'?
    No. A child standing at either end has only one neighbour next to them. Therefore, an end child can say at most '1' (or '0'), and can never say '2'.
  • 2. Can we arrange the children in a line so that all would say only 0s?
    No. If all children have distinct heights, the shortest child will have taller neighbours on both sides (or on one side if at an end), so that child cannot say '0'. Only if all children have the exact same height could everyone say '0'.
  • 3. Can two children standing next to each other say the same number?
    Yes. For example, in an arrangement in increasing order of height (shortest to tallest: C1 < C2 < C3 < C4 < C5), the sequence of numbers called out is 1, 1, 1, 1, 0. Here, adjacent children say 1.
  • 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?
    Yes. When 5 children of distinct heights stand in strictly increasing order of height from left to right (H1 < H2 < H3 < H4 < H5):
    - H1 has 1 taller neighbour on the right → says 1.
    - H2 has 1 taller neighbour on the right → says 1.
    - H3 has 1 taller neighbour on the right → says 1.
    - H4 has 1 taller neighbour on the right → says 1.
    - H5 is the tallest and at the end → says 0.
    The resulting sequence is 1, 1, 1, 1, 0.
  • 5. For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
    No. The tallest child in the group does not have any neighbour taller than them, so the tallest child must always say '0'. Thus, all five cannot say '1'.
  • 6. Is the sequence 0, 1, 2, 1, 0 possible? Why or why not?
    Yes. For example, arrange 5 children with heights in the order: 5, 2, 1, 3, 4 (where 5 is the tallest):
    - Person 1 (ht 5): Left end, taller than 2 → says 0.
    - Person 2 (ht 2): Between 5 and 1, only 5 is taller → says 1.
    - Person 3 (ht 1): Between 2 and 3, both are taller → says 2.
    - Person 4 (ht 3): Between 1 and 4, only 4 is taller → says 1.
    - Person 5 (ht 4): Right end, taller than 3 → says 0.
  • 7. How would you rearrange the five children so that the maximum number of children say '2'?
    Since the 2 end positions cannot say '2', at most the 3 middle positions can say '2'. For a child to say '2', both adjacent neighbours must be taller. Arranging in alternating tall and short order, for example with heights 4, 1, 5, 2, 3:
    - Ht 1 is between 4 and 5 → says 2.
    - Ht 2 is between 5 and 3 → says 2.
    Thus, a maximum of 2 children can say '2'.
Figure it Out - 3.1 (Page 134)
  • 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.
    a. Step size = 10:
    Positions: 1970, 1980, 1990, 2000, 2010, 2020, 2030, 2040, 2050
    - Smallest number: 1970
    - Largest number: 2050

    b. Step size = 1:
    Positions: 9993, 9994, 9995, 9996, 9997, 9998, 9999, 10000, 10001
    - Smallest number: 9993
    - Largest number: 10001

    c. Step size = 1:
    Positions: 15077, 15078, 15079, 15080, 15081, 15082, 15083, 15084, 15085
    - Smallest number: 15077
    - Largest number: 15085

    d. Step size = 1000:
    Positions: 83705, 84705, 85705, 86705, 87705, 88705, 89705, 90705, 91705
    - Smallest number: 83705
    - Largest number: 91705
Figure it Out - 3.2 (Page 136)
  • 1. Colour or mark the supercells in the table below.
    6828 670 9435 3780 3708 7308 8000 5583 52
    The supercells are: 6828 (greater than 670), 9435 (greater than 670 and 3780), 7308 (greater than 3708), and 8000 (greater than 7308 and 5583).
  • 2. Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells (where 5346 and 9635 are marked).
    A valid sequence of 4-digit numbers can be:
    5346, 2100, 3400, 1200, 9635, 4000, 5000, 2000, 1500.
    Here, only 5346 and 9635 are strictly greater than their respective adjacent neighbours.
  • 3. Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.
    (a) Out of the 9 cells, how many supercells are there in the above table?
    (b) Find out how many supercells are possible for different numbers of cells.
    (c) 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.
    (a) For a 9-cell row, the maximum number of supercells is 5.
    Example: 900, 100, 800, 200, 700, 300, 600, 400, 500.

    (b) For $n$ cells in a line:
    - For 3 cells: 2 supercells.
    - For 4 cells: 2 supercells.
    - For 5 cells: 3 supercells.
    - For $n$ cells: $\frac{n+1}{2}$ supercells (for odd $n$) or $\frac{n}{2}$ supercells (for even $n$).

    (c) Strategy: Alternate high numbers and low numbers starting with a high number at the first cell (High, Low, High, Low, ...).
  • 4. Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
    No. In any finite collection of distinct numbers, there is always a unique maximum (largest) number. That maximum number will strictly be greater than all its adjacent neighbours, and therefore will always be a supercell.
  • 5. 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?
    - Largest number: Yes, it is always a supercell because no neighbouring number can be greater than or equal to it.
    - Smallest number: No, it can never be a supercell because its adjacent neighbours will always be larger than it.
  • 6. Fill a table such that the cell having the second largest number is not a supercell.
    Place the second largest number directly adjacent to the largest number.
    Example (1 to 5): 1, 2, 3, 4, 5. Here, 5 is the largest (supercell) and 4 is the second largest, but 4 is smaller than 5, so 4 is not a supercell.
  • 7. 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?
    Yes, it is possible.
    Example for numbers {1, 2, 3, 4, 5}: Arrange them as 2, 1, 3, 4, 5.
    - 2 (second smallest) is at the end and adjacent to 1 → 2 is a supercell.
    - 4 (second largest) is between 3 and 5 → 4 is adjacent to 5, so 4 is not a supercell.
Page 138
Table 2: 5-digit numbers using digits 1, 0, 6, 3, 9
96,301 36,109 93,610
13,609 60,319 19,306
91,036 60,193 90,631
10,963 63,901 16,039
  • Answer the following from the table:
    - The biggest number in the table is 96,301.
    - The smallest even number in the table is 19,306.
    - The smallest number greater than 50,000 in the table is 60,193.
  • Find out how many numbers have two digits, three digits, four digits, and five digits.
    1-digit numbers 2-digit numbers 3-digit numbers 4-digit numbers 5-digit numbers
    From 1-9 From 10-99 From 100-999 From 1000-9999 From 10000-99999
    9 90 900 9,000 90,000
Figure it Out - 3.3 (Page 140)
  • 1. Write other numbers whose digits add up to 14.
    Examples: 59, 77, 86, 95, 149, 257, 347, 554, 1148, 2345.
  • 2. What is the smallest number whose digit sum is 14?
    To make the smallest number, use the minimum number of digits with the largest possible last digit (9): 14 - 9 = 5.
    The smallest number is 59.
  • 3. What is the largest 5-digit number whose digit sum is 14?
    To make the largest 5-digit number, place the largest possible digits in the highest place values:
    Digits: 9, 5, 0, 0, 0 (9 + 5 + 0 + 0 + 0 = 14).
    The largest 5-digit number is 95,000.
  • 4. How big a number can you form having the digit sum of 14? Can you make an even bigger number?
    We can make the number as large as we want by adding as many '0's as we like at the end (e.g., 95000000000...). Furthermore, using smaller digits like '1's creates 14 digits: 11111111111111, and placing extra trailing zeros makes it arbitrarily large.
  • 5. Find out the digit sums of all the numbers from 40 to 70. Share your observations with the class.
    - Numbers 40 to 49: Digit sums are 4, 5, 6, 7, 8, 9, 10, 11, 12, 13.
    - Numbers 50 to 59: Digit sums are 5, 6, 7, 8, 9, 10, 11, 12, 13, 14.
    - Numbers 60 to 69: Digit sums are 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.
    - Number 70: Digit sum is 7.
    Observation: In each decade (group of ten), the digit sum increases by 1 for each step, and when transitioning to the next decade (like 49 to 50), the sum drops by 8 (13 → 5).
  • 6. Calculate the digit sums of 3-digit numbers whose digits are consecutive (for example 345). Do you see a pattern? Will this pattern continue?
    - 123: 1 + 2 + 3 = 6 = 3 × 2
    - 234: 2 + 3 + 4 = 9 = 3 × 3
    - 345: 3 + 4 + 5 = 12 = 3 × 4
    - 456: 4 + 5 + 6 = 15 = 3 × 5
    - 567: 5 + 6 + 7 = 18 = 3 × 6
    - 678: 6 + 7 + 8 = 21 = 3 × 7
    - 789: 7 + 8 + 9 = 24 = 3 × 8
    Pattern: The digit sum is always a multiple of 3, and equals exactly 3 times the middle digit. Yes, this pattern always continues because (n - 1) + n + (n + 1) = 3n.
  • 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?
    - From 1 to 100: The digit '7' occurs in units place (7, 17, 27, 37, 47, 57, 67, 77, 87, 97 → 10 times) and in tens place (70 to 79 → 10 times). Total = 10 + 10 = 20 times.

    - From 1 to 1000: The digit '7' occurs 100 times in units place, 100 times in tens place, and 100 times in hundreds place (700 to 799). Total = 100 + 100 + 100 = 300 times.
Page 142
  • Write all possible 3-digit palindromes using digits '1', '2', '3'.
    A 3-digit palindrome has the form ABA. Choosing A from {1, 2, 3} (3 choices) and B from {1, 2, 3} (3 choices) gives 3 × 3 = 9 palindromes:
    111, 121, 131, 212, 222, 232, 313, 323, 333.
Page 144
  • 1. Will reversing and adding numbers repeatedly, starting with a 2-digit number, always give a palindrome? Explore and find out.
    For almost all 2-digit numbers, it reaches a palindrome within a few steps (e.g., 89 takes 24 steps to reach 8813200023188). All 2-digit numbers eventually produce a palindrome.
  • 2. Will reversing and adding numbers repeatedly, starting with 196, always give a palindrome?
    196 is the most famous Lychrel number candidate. Even after millions of reverse-and-add iterations by computers, it has never formed a palindrome.
  • Puzzle time: I am a 5-digit palindrome. I am an odd number. My 'tens' digit is double of my 'unit' digit. My 'hundreds' digit is double of my 'tens' digit. Who am I?
    Let the 5-digit palindrome be represented by place values [tth, th, h, t, u]:
    1. Since the number is odd, the unit digit 'u' must be odd (1, 3, 5, 7, 9).
    2. Tens digit t = 2 × u. If u = 1, then t = 2. (If u ≥ 3, then t ≥ 6 and h ≥ 12, which is not a single digit).
    3. Hundreds digit h = 2 × t = 2 × 2 = 4.
    4. Since it is a palindrome, tth = u = 1 and th = t = 2.

    The number is 12421.
    In words: Twelve thousand four hundred and twenty-one.
  • On the usual 12-hour clock, find all possible palindromic times of the types 4:44, 10:10, 12:21.
    - 3-digit palindromic times (H:MM): 1:01, 1:11, 1:21, 1:31, 1:41, 1:51; 2:02, 2:12, 2:22, 2:32, 2:42, 2:52; 3:03, 3:13, ..., 3:53; up to 9:59 (6 times for each hour from 1 to 9 → 54 times).
    - 4-digit palindromic times (HH:MM): 10:01, 11:11, 12:21.
Let's Explore (Page 146)
  • If the date format is DD/MM/YYYY, what is the next palindromic date after 03/02/2030?
    In format DD/MM/YYYY, the year 2040 gives date 04/02/2040, and the year 2041 gives: 14/02/2041.
  • Are there any palindromic dates in the year 2031? Why or why not?
    No. Reversing the year 2031 gives the date and month as 13/02 (13th day of the 2nd month, February). But February has only 28 or 29 days. Therefore, 13/02/2031 is valid (13th February 2031).
    Note: In DD/MM/YYYY format, 2031 reverses to 13/02/2031, which is a valid date (13 Feb 2031).
  • Challenge: What is the last palindromic date of the 21st Century (up to the year 2100)?
    The highest valid year in the 21st century that yields a valid date is 2092 (giving 29/02/2092, since 2092 is a leap year).
    The last palindromic date is 29/02/2092.
Figure it Out - 3.4 (Page 150)
  • Carry out the Kaprekar steps with a few 3-digit numbers. What number will start repeating?
    For 3-digit numbers, repeating the routine always leads to the 3-digit Kaprekar constant: 495.
  • 1. Sarala uses the digits '4', '7', '3' and '2'. 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.
    a. Choose digits {1, 2, 3, 9}: Largest = 9321, Smallest = 1239. Difference = 9321 - 1239 = 8082 > 5085.

    b. Choose digits {1, 2, 3, 4}: Largest = 4321, Smallest = 1234. Difference = 4321 - 1234 = 3087 < 5085.

    c. Choose digits {5, 6, 7, 8}: Largest = 8765, Smallest = 5678. Sum = 8765 + 5678 = 14443 > 9779.

    d. Choose digits {1, 2, 3, 4}: Largest = 4321, Smallest = 1234. Sum = 4321 + 1234 = 5555 < 9779.
  • 2. What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
    - Smallest 5-digit palindrome = 10001
    - Largest 5-digit palindrome = 99999

    - Sum: 99999 + 10001 = 1,10,000.
    - Difference: 99999 - 10001 = 89,998.
  • 3. The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that?
    - Current time = 10:01.
    - Next palindromic time = 11:11. Difference from 10:01 to 11:11 = 1 hr 10 min = 70 minutes.
    - The one after that = 12:21. Difference from 10:01 to 12:21 = 2 hr 20 min = 140 minutes.
  • 4. How many rounds does the number 5683 take to reach the Kaprekar constant?
    - Round 1: Digits {8, 6, 5, 3} → 8653 - 3568 = 5085
    - Round 2: Digits {8, 5, 5, 0} → 8550 - 0558 = 7992
    - Round 3: Digits {9, 9, 7, 2} → 9972 - 2799 = 7173
    - Round 4: Digits {7, 7, 3, 1} → 7731 - 1377 = 6354
    - Round 5: Digits {6, 5, 4, 3} → 6543 - 3456 = 3087
    - Round 6: Digits {8, 7, 3, 0} → 8730 - 0378 = 8352
    - Round 7: Digits {8, 5, 3, 2} → 8532 - 2358 = 6174
    It takes 7 rounds.
Figure it Out - 3.5 (Page 152)
  • 1. Estimate the steps you would take to walk:
    a. From the place you are sitting to the classroom door: Around 10 to 15 steps.
    b. Across the school ground from start to end: Around 100 to 150 steps.
    c. From your classroom door to the school gate: Around 50 to 80 steps.
    d. From your school to your home: Around 1,500 to 3,000 steps (for 1 to 2 km).
  • 2. Number of times you blink your eyes or number of breaths you take:
    - Eye blinks (approx. 15 per minute):
    a. In a minute: 15 times.
    b. In an hour: 15 × 60 = 900 times.
    c. In a day (16 waking hours): 900 × 16 = 14,400 times.

    - Breaths taken (approx. 18 per minute):
    a. In a minute: 18 breaths.
    b. In an hour: 18 × 60 = 1,080 breaths.
    c. In a day: 1080 × 24 = 25,920 breaths.
Page 154
  • 3. Name some objects around you that are:
    a. A few thousand in number: Leaves on a large tree, pages in school library books, tiles on the school floor.
    b. More than ten thousand in number: Grains in a bowl of rice, hairs on our head, stars visible on a clear night.
  • 4. Number of words in your maths textbook:
    Answer: a. More than 5000 (A textbook contains around 30,000 to 50,000 words).
  • 5. Number of students in your school who travel to school by bus:
    Answer: a. More than 200 (or depending on your school size).
  • 6. Achyuth 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?
    No, I do not agree.
    Reason: 1 litre of milk costs around ₹30–₹40, and 3 types of fruit (like apples, bananas, grapes) cost at least ₹30–₹50 each, plus custard powder. The total realistic cost will be around ₹150 to ₹200.
  • 7. Estimate the distance between Amaravati (in Andhra Pradesh) to Hyderabad (in Telangana).
    The distance is approximately 275 km to 300 km.
  • 8. Amara 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?
    No, I do not agree.
    Reason: A school year has about 220 working days with 6 hours per day = 220 × 6 = 1,320 hours/year. Over 6 years (Grades 1 to 6), total time = 6 × 1320 = 7,920 hours, which is much less than 13,000 hours.
  • 9. Approximately, how long would it take you to walk at normal pace (approx. 4 km/h) from:
    a. Current location to a favourite place nearby (2 km): About 30 minutes.
    b. Current location to neighbouring state capital (e.g., to Chennai/Hyderabad, 300 km): 300 ÷ 4 = 75 hours of continuous walking (about 7 to 8 days).
    c. Southernmost point (Kanyakumari) to northernmost point (Kashmir) (approx. 3,200 km): 3200 ÷ 4 = 800 hours of walking (about 2 to 3 months).
Page 156
  • Mental Math: Obtain the target numbers using middle column values {25000, 400, 13000, 1500, 60000}.
    - 28,000 = 25,000 + 1,500 + 1,500
    - 61,600 = 60,000 + 400 × 4
    - 31,000 = 25,000 + 6,000
    - 63,000 = 60,000 + 1,500 + 1,500
    - 19,500 = 13,000 + 1,500 × 4 + 400 + 100
    - 20,900 = 13,000 + 1,500 × 5 + 400
  • Can we make 1,000 using the numbers in the middle? What about 14,000, 15,000 and 16,000?
    - 1,000: Cannot be made by only adding whole units of 400, 1500, etc. without subtraction.
    - 14,500: 13,000 + 1,500
    - 16,000: 13,000 + 1,500 × 2 = 16,000.
Page 158
  • Adding and Subtracting: Using {40000, 7000, 300, 1500, 12000, 800}, represent the numbers:
    - 45,000 = 40,000 + 7,000 - 1,500 - 500
    - 5,900 = 7,000 - 1,500 + 800 - 400
    - 17,500 = 12,000 + 7,000 - 1,500
    - 21,400 = 12,000 + 7,000 + 1,500 + 900
  • Figure it Out - 3.6 (Question 1): Write an example for each of the scenarios.
    - 5-digit + 5-digit to give 5-digit: 10,000 + 20,000 = 30,000
    - 4-digit + 3-digit to give 5-digit: 9,500 + 600 = 10,100
    - 5-digit + 4-digit to give 6-digit: 95,000 + 6,000 = 1,01,000
    - 5-digit + 5-digit to give 6-digit: 60,000 + 50,000 = 1,10,000
    - 5-digit + 5-digit to give 18,500: Impossible (the minimum sum of two 5-digit numbers is 10,000 + 10,000 = 20,000).
    - 5-digit - 5-digit to give 5-digit: 50,000 - 20,000 = 30,000
    - 5-digit - 3-digit to give 4-digit: 10,500 - 600 = 9,900
    - 5-digit - 4-digit to give 4-digit: 12,000 - 3,000 = 9,000
    - 5-digit - 5-digit to give 3-digit: 10,500 - 10,200 = 300
    - 5-digit - 5-digit to give 91,500: Impossible (the maximum difference between two 5-digit numbers is 99,999 - 10,000 = 89,999).
  • 2. Always, Sometimes, Never?
    a. 5-digit number + 5-digit number gives a 5-digit number: Only sometimes true.

    b. 4-digit number + 2-digit number gives a 4-digit number: Only sometimes true.

    c. 4-digit number + 2-digit number gives a 6-digit number: Never true.

    d. 5-digit number - 5-digit number gives a 5-digit number: Only sometimes true.

    e. 5-digit number - 2-digit number gives a 3-digit number: Never true.
Page 160
  • Find out the sum of the numbers in each of the figures:
    a. Twelve 40s arranged in a plus shape: 12 × 40 = 480.

    b. Ten 50s in a grid: 10 × 50 = 500.

    c. Forty 32s (4 × 10 grid): 40 × 32 = 1,280.

    d. Twenty 64s (4 × 5 grid): 20 × 64 = 1,280.
Page 162
  • Sum of figure (e) and figure (f):
    - Figure e (Hexagon pattern): Group complementary pairs: (15 + 35 = 50) and (25 + 25 = 50). Adding all pairs gives total sum = 1,200.

    - Figure f (Target concentric circles):
    * Center = 1000
    * Inner ring = 4 × 500 = 2000
    * Middle ring = 8 × 250 = 2000
    * Outer ring = 16 × 125 = 2000
    * Total Sum = 1000 + 2000 + 2000 + 2000 = 7,000.
  • Collatz sequences: Do you always reach 1?
    Yes, every tested whole number eventually falls into the loop 4 → 2 → 1. For example, starting with 6: 6 → 3 → 10 → 5 → 16 → 8 → 4 → 2 → 1.
Page 164
  • Game #1: Half-Century Chase (Reach 50, adding 1 to 4 runs each turn). Which player can always win? What is the winning pattern?
    - Winning Strategy: The sum of minimum and maximum additions is 1 + 4 = 5.
    - The key milestone numbers are multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.
    - Player B (the second player) can always win by choosing whatever number makes the pair sum to 5 (e.g., if Player A adds 3, Player B adds 2).
  • Game #2: Century Clash (Reach 100, adding 1 to 10 runs each turn). Which player can always win? What is the winning pattern?
    - Step size = 1 + 10 = 11.
    - Milestones counting backwards from 100: 100, 89, 78, 67, 56, 45, 34, 23, 12, 1.
    - Player A (first player) can always win by scoring 1 on the first turn, and then whenever Player B adds k runs, Player A adds 11 - k runs to land on each consecutive milestone: 1, 12, 23, 34, 45, 56, 67, 78, 89, 100.
Figure it Out - 3.7 (Page 168)
  • 1. There is only one supercell in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap.
    Swap the first two digits of the center number 62,871 to make it 26,871.
    Now, numbers 39,344, 23,609, 45,306, and 50,319 all become greater than their adjacent neighbours, creating 4 supercells.
  • 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?
    Using only odd digits {1, 3, 5, 7, 9}:
    - Largest number (< 75,000): Starts with 73... → 73,999.
    - Smallest number (> 35,000): Starts with 35... → 35,111.
    - Closest to 50,000: Comparing 49,999 (has even digit 4, not allowed) → the closest below is 39,999 (distance 10,001) and closest above is 51,111 (distance 1,111). The closest is 51,111.
  • 6. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.
    Example:
    18,000 + 350 + 320 = 18,670.
  • 8. Recall the sequence of Powers of 2 from Chapter 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence?
    Every power of 2 (2n) is an even number. Under the Collatz rule, dividing by 2 repeatedly gives 2n → 2n-1 → … → 8 → 4 → 2 → 1 directly without ever branching to odd multiplication.
  • 9. Check if the Collatz Conjecture holds for the starting number 100.
    Sequence: 100 → 50 → 25 → 76 → 38 → 19 → 58 → 29 → 88 → 44 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1.
    Yes, it holds.
  • 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?
    - Step size = 1 + 3 = 4.
    - Milestones counting back from 22: 22, 18, 14, 10, 6, 2.
    - Player 1 (first player) wins by choosing 2 on the first turn, and whenever Player 2 adds k, Player 1 adds 4 - k to hit the target numbers 2, 6, 10, 14, 18, 22.
CHAPTER MASTERY (Page 170)
  • A. True/False
    Statement Ans
    1. 575 is a palindromic number. True
    2. 6174 is called the Kaprekar constant for 4-digit numbers. True
    3. In a supercell table, the smallest number can be a supercell. False
    4. Every Collatz sequence eventually reaches 1. True
  • B. Multiple Choice Questions

    5. What is a palindromic number?
    a) A number divisible by 10
    b) A number read the same forwards and backwards
    c) A number with repeated digits
    d) A prime number
    Answer: b) A number read the same forwards and backwards
  • 6. The final digit sum of 1729 is:
    a) 19
    b) 10
    c) 1
    d) 7
    1 + 7 + 2 + 9 = 19 → 1 + 9 = 10 → 1 + 0 = 1.
    Answer: c) 1
  • 7. The 3-digit Kaprekar constant is:
    a) 495
    b) 999
    c) 174
    d) 100
    Answer: a) 495
  • 8. In the Collatz rule, if the number is even, we divide it by 2.
  • 9. Assertion (A): 1221 is a palindromic number.
    Reason (R): It reads the same from left to right and right to left.

    Choose the correct option:
    a) Both A and R are true and R is the correct explanation of A
    b) Both A and R are true but R is not the correct explanation of A
    c) A is true but R is false
    d) A is false but R is true
    Answer: a) Both A and R are true and R is the correct explanation of A
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  • 10. In the "Supercells" activity, what specific condition must a number meet to be considered a Supercell?
    A number in a cell is called a Supercell if its value is strictly greater than the numbers in all of its immediately adjacent neighbouring cells (left, right, top, bottom).
  • 11. Explain the two rules used to generate a sequence in the Collatz Conjecture.
    1. If the number is even: Divide it by 2 (i.e., take n/2).
    2. If the number is odd: Multiply it by 3 and add 1 (i.e., calculate 3n + 1).
  • 12. Write an example for the following statement: (4-digit number) - (4-digit number) = 3-digit number.
    Example:
    1,250 - 1,000 = 250.
  • 13. Compare the "Half-Century Chase" (reaching 50) and the "Century Clash" (reaching 100). Identify the winning strategy for the first player in the "Half-Century Chase" if they can add 1, 2, 3, or 4 runs each turn.
    - In "Half-Century Chase", each pair of turns can total 1 + 4 = 5 runs. Since 50 is an exact multiple of 5 (50 = 10 × 5), the second player has the winning strategy by always completing the multiple of 5.
    - If the target were not a multiple of 5, the first player would win by hitting the remainder first. For reaching 50, the second player controls the game by maintaining multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.
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AP TEXT BOOK SOLUTIONS

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