Exercises
Q1. Write the locker numbers that remain open.
Answer:
The lockers that remain open are those whose numbers are perfect squares. The locker numbers from 1 to 100 that are perfect squares are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. These lockers are toggled an odd number of times because they have an odd number of factors.
Q2. Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?
Answer:
Between n² and (n+1)², there are exactly 2n numbers (excluding both squares). For example, between 1²=1 and 2²=4: numbers 2,3 (2 numbers = 2×1); between 2²=4 and 3²=9: numbers 5,6,7,8 (4 numbers = 2×2); between 3²=9 and 4²=16: numbers 10,11,12,13,14,15 (6 numbers = 2×3). So the pattern is that between consecutive squares n² and (n+1)² there are 2n numbers.
Q3. How many square numbers are there between 1 and 100? How many are between 101 and 200? Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of 100. What is the largest square less than 1000?
Answer:
Between 1 and 100: squares are 1,4,9,16,25,36,49,64,81,100 → 10 squares. Between 101 and 200: squares are 121,144,169,196 → 4 squares. Tabulation: 1-100: 10; 101-200: 4; 201-300: 225,256,289 → 3; 301-400: 324,361,400 → 3; 401-500: 441,484 → 2; 501-600: 529,576 → 2; 601-700: 625,676 → 2; 701-800: 729,784 → 2; 801-900: 841,900 → 2; 901-1000: 961 → 1. The largest square less than 1000 is 961 (31²).
Q4. Find whether 1156 and 2800 are perfect squares using prime factorisation.
Answer:
For 1156: prime factorisation = 2 × 578 = 2 × 2 × 289 = 2 × 2 × 17 × 17 = (2×17)×(2×17) = 34². So 1156 is a perfect square, √1156 = 34. For 2800: prime factorisation = 28×100 = 4×7×4×25 = 2²×7×2²×5² = 2⁴×5²×7. The factor 7 appears only once, so cannot be paired. Therefore 2800 is not a perfect square.
Q5. Akhil has a square piece of cloth of area 125 cm². He wants to know if he can cut out a square handkerchief of side 15 cm. If not, he wants to know the maximum size handkerchief that can be cut out from this piece of cloth with an integer side length.
Answer:
We need to check if 125 is a perfect square. 11²=121, 12²=144. 125 lies between 121 and 144, and 125≠(integer)². So 125 is not a perfect square. Since 125 < 225 (15²), a 15 cm side handkerchief cannot be cut. For integer side length, the largest square less than or equal to area 125 is 121 = 11². So maximum side length is 11 cm.
Q6. Using the pattern above, find 36², given that 35² = 1225.
Answer:
The nth odd number is 2n-1. To go from 35² to 36², add the 36th odd number. 36th odd number = 2×36-1 = 72-1 = 71. So 36² = 1225 + 71 = 1296.
Questions
Q1. What are the digits in the units places of these numbers? All these numbers end with 0, 1, 4, 5, 6 or 9. None of them end with 2, 3, 7 or 8.
Answer:
This is an observation, not a question. The text notes that perfect squares can only end in 0,1,4,5,6,9 and never in 2,3,7,8.
Q2. If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?
Answer:
No. For example, 26 ends with 6 but is not a square (since 5²=25, 6²=36). Similarly 14 ends with 4 but 14 is not a square. So the units digit condition is necessary but not sufficient.
Q3. Write 5 numbers such that you can determine by looking at their units digit that they are not squares.
Answer:
Any 5 numbers ending with 2,3,7, or 8. Example: 12, 23, 37, 48, 52.
Q4. Which of the following numbers have the digit 6 in the units place? (i) 38² (ii) 34² (iii) 46² (iv) 56² (v) 74² (vi) 82²
Answer:
Calculate squares: 38² = 1444 (ends in 4); 34² = 1156 (ends in 6); 46² = 2116 (ends in 6); 56² = 3136 (ends in 6); 74² = 5476 (ends in 6); 82² = 6724 (ends in 4). So numbers with units digit 6: (ii) 34², (iii) 46², (iv) 56², (v) 74².
Q5. If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Answer:
If a number ends with 3 zeros, it is divisible by 1000 = 2³×5³. Its square will be divisible by (2³)²×(5³)² = 2⁶×5⁶ = (2×5)⁶ = 10⁶, so it will end with 6 zeros.
Q6. What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?
Answer:
If a number ends with n zeros (i.e., is divisible by 10ⁿ), its square ends with 2n zeros. This is because (10ⁿ)² = 10^(2n). So squares always have an even number of zeros at the end (0,2,4,6,...). Yes, this always happens.
Q7. What can you say about the parity of a number and its square?
Answer:
If a number is even, its square is even. If a number is odd, its square is odd. More precisely, an even number squared is divisible by 4, and an odd number squared is odd.
Q8. What is the square root of 64?
Answer:
√64 = ±8, since 8×8=64 and (-8)×(-8)=64. Positive square root is 8.
Q9. What is the length of the side of a square with area 49 sq. cm?
Answer:
Side = √49 = 7 cm (positive root).
Activities
Q1. Find the squares of the first 30 natural numbers and fill in the table below.
Answer:
Table: 1²=1, 2²=4, 3²=9, 4²=16, 5²=25, 6²=36, 7²=49, 8²=64, 9²=81, 10²=100, 11²=121, 12²=144, 13²=169, 14²=196, 15²=225, 16²=256, 17²=289, 18²=324, 19²=361, 20²=400, 21²=441, 22²=484, 23²=529, 24²=576, 25²=625, 26²=676, 27²=729, 28²=784, 29²=841, 30²=900.
Q2. Using the pattern above (difference of consecutive squares equals successive odd numbers), find 36², given that 35²=1225.
Answer:
36² = 35² + (2×36-1) = 1225 + 71 = 1296.
Q3. Consider a number such as 38 that is not a square and subtract consecutive odd numbers starting from 1.
Answer:
38 - 1 = 37; 37 - 3 = 34; 34 - 5 = 29; 29 - 7 = 22; 22 - 9 = 13; 13 - 11 = 2; 2 - 13 = -11. Since we never reach 0, 38 is not a perfect square.
Figure it Out
Q1. Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
Answer:
A perfect square ends with 0, 1, 4, 5, 6, or 9. 2032 ends with 2, 2048 ends with 8, 1027 ends with 7 – these are not possible endings for a perfect square. 1089 ends with 9 and is a perfect square (33² = 1089). Therefore, 2032, 2048, and 1027 are not perfect squares.
Q2. Which one among 642, 1082, 2922, 362 has last digit 4?
Answer:
The last digit of a square depends on the last digit of the number. For a number ending in 2, its square ends with 4. Among the given numbers: 642 ends in 2 → square ends in 4; 1082 ends in 2 → square ends in 4; 2922 ends in 2 → square ends in 4; 362 ends in 6 → square ends in 6. All except 362 end in 4. The question likely asks which one among the numbers themselves has last digit 4. The numbers are 642, 1082, 2922, 362. Only 362 ends with 2, not 4. Actually, all numbers end with 2 except 362 ends with 2 as well. Wait – 642 ends with 2, 1082 ends with 2, 2922 ends with 2, 362 ends with 2. None of them have last digit 4. But the question asks about the square's last digit. So 642, 1082, 2922 all have squares ending in 4. The answer is 642, 1082, and 2922. Since it says "which one", possibly it expects 642 (or any one, but the text might intend to identify that 642, 1082, 2922 all have squares ending with 4).
Q3. Given 125² = 15625, what is the value of 126²? (i) 15625 + 126 (ii) 15625 + 262 (iii) 15625 + 253 (iv) 15625 + 251 (v) 15625 + 512
Answer:
126² = (125 + 1)² = 125² + 2×125×1 + 1² = 15625 + 250 + 1 = 15625 + 251. Therefore, the correct option is (iv) 15625 + 251.
Q4. Find the length of the side of a square whose area is 441 m².
Answer:
Area of square = side². So side = √441. We know 20² = 400, 21² = 441. So side = 21 m.
Q5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Answer:
Find LCM of 4, 9, 10. Prime factorisation: 4 = 2², 9 = 3², 10 = 2×5. LCM = 2² × 3² × 5 = 4×9×5 = 180. For a number to be a perfect square, each prime factor must have an even exponent. In 180, exponents: 2² (even), 3² (even), 5¹ (odd). Multiply by 5 to make exponent of 5 even: 180 × 5 = 900. 900 = 2²×3²×5² = 30². So the smallest square number divisible by 4, 9, 10 is 900.
Q6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Answer:
Prime factorisation of 9408: 9408 ÷ 2 = 4704, ÷2 = 2352, ÷2 = 1176, ÷2 = 588, ÷2 = 294, ÷2 = 147, ÷3 = 49, ÷7 = 7, ÷7 = 1. So 9408 = 2⁶ × 3 × 7². For a perfect square, all exponents must be even. Here 2⁶ (even), 3¹ (odd), 7² (even). Multiply by 3 to make exponent of 3 even: 9408 × 3 = 28224. 28224 = 2⁶ × 3² × 7² = (2³×3×7)² = (8×21)² = 168². Square root = 168.
Q7. How many numbers lie between the squares of the following numbers? (i) 16 and 17 (ii) 99 and 100
Answer:
Numbers between n² and (n+1)² = (n+1)² - n² - 1 = 2n. (i) Between 16² and 17²: 2×16 = 32 numbers. (ii) Between 99² and 100²: 2×99 = 198 numbers.
Q8. In the following pattern, fill in the missing numbers: 1² + 2² + 2² = 3², 2² + 3² + 6² = 7², 3² + 4² + 12² = 13², 4² + 5² + 20² = (___)², 9² + 10² + (___)² = (___)²
Answer:
Pattern: (n)² + (n+1)² + (n×(n+1))² = (n×(n+1)+1)². For n=4: 4² + 5² + (4×5)² = 16+25+400 = 441 = 21², so missing number is 21. For n=9: 9² + 10² + (9×10)² = 81+100+8100 = 8281 = 91². So missing numbers are 90 and 91.
Q9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
Answer:
The picture is not provided in the text. However, based on context, it likely shows a square grid or pattern. Without the picture, we cannot determine the number. [Note: In a typical textbook, this might be a 10×10 grid or similar. For a proper answer, the image is needed. But assuming a common pattern like a 5×5 grid gives 25 = 5². Since not given, solution cannot be provided accurately.]
Try This
Q1. How did Ramanujan know this? Well, he loved numbers. Find the two ways in which each of these can be expressed as the sum of two positive cubes: 4104 and 13832.
Answer:
4104 = 2³ + 16³ (2+4096=4098? Wait 2³=8, 16³=4096, sum=4104) and 9³ + 15³ (729+3375=4104). 13832 = 2³ + 24³ (8+13824=13832) and 18³ + 20³ (5832+8000=13832).
Q2. Can a cube end with exactly two zeroes (00)? Explain.
Answer:
A cube ending with two zeroes means it is divisible by 100 = 2²×5². For a cube, the exponent of each prime in its prime factorisation must be a multiple of 3. So if a cube ends with two zeroes, it must be divisible by 2² and 5², but since it's a cube, the exponents must be multiples of 3. The smallest such cube would be divisible by 2³ and 5³, giving at least three zeroes. So a cube cannot end with exactly two zeroes; it will end with either one zero (if only one factor of 10) or three or more zeroes.
Figure it Out (Cubes)
Q1. Find the cube roots of 27000 and 10648.
Answer:
27000 = 27 × 1000 = 3³ × 10³ = (3×10)³ = 30³. Cube root = 30. 10648: Check prime factors: 10648 ÷ 2 = 5324, ÷2 = 2662, ÷2 = 1331. 1331 = 11³. So 10648 = 2³ × 11³ = (2×11)³ = 22³. Cube root = 22.
Q2. What number will you multiply by 1323 to make it a cube number?
Answer:
Prime factorise 1323: 1323 ÷ 3 = 441, ÷3 = 147, ÷3 = 49, ÷7 = 7, ÷7 = 1. So 1323 = 3³ × 7². For a perfect cube, exponents must be multiples of 3. Here 3³ is fine, 7² needs one more 7. Multiply by 7: 1323 × 7 = 9261. 9261 = 3³ × 7³ = 21³.
Q3. State true or false. Explain your reasoning. (i) The cube of any odd number is even. (ii) There is no perfect cube that ends with 8. (iii) The cube of a 2-digit number may be a 3-digit number. (iv) The cube of a 2-digit number may have seven or more digits. (v) Cube numbers have an odd number of factors.
Answer:
(i) False. An odd number cubed is odd (odd×odd×odd = odd). (ii) False. For example, 2³ = 8, 12³ = 1728, 22³ = 10648 all end with 8. (iii) True. The smallest 2-digit number 10³ = 1000 (4 digits), but 4³=64 (2 digits), 5³=125 (3 digits). A 2-digit number like 4 is not 2-digit. Actually, smallest 2-digit is 10, its cube is 1000 (4 digits). So a 2-digit number's cube is at least 1000 (4 digits), so cannot be 3-digit. So false. Correction: The smallest 2-digit number is 10, cube is 1000 which is 4 digits. So no 2-digit number gives a 3-digit cube. Statement is false. (iv) True. The largest 2-digit number 99³ = 970299 which is 6 digits. So no 2-digit cube has 7 or more digits. So false. Actually 99³ = 970299 (6 digits). So false. (v) False. For example, 8 = 2³ has factors 1,2,4,8 (4 factors, even). However, 1³=1 has 1 factor (odd). So not always odd.
Q4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
Answer:
1331: Ends with 1, cube root ends with 1. 10³=1000, 15³=3375, so between 10 and 15. Try 11: 11³=1331. So cube root = 11. 4913: Ends with 3, cube root ends with 7 (since 7³=343 ends with 3). 15³=3375, 20³=8000, so between 15 and 20. 17³=4913. So cube root = 17. 12167: Ends with 7, cube root ends with 3 (3³=27 ends with 7). 20³=8000, 30³=27000, so between 20 and 30. 23³=12167. So cube root = 23. 32768: Ends with 8, cube root ends with 2 (2³=8). 30³=27000, 40³=64000, so between 30 and 40. 32³=32768. So cube root = 32.
Q5. Which of the following is the greatest? Explain your reasoning. (i) 67³ – 66³ (ii) 43³ – 42³ (iii) 67² – 66² (iv) 43² – 42²
Answer:
Use identities: a³ - b³ = (a-b)(a²+ab+b²) and a² - b² = (a-b)(a+b). (i) 67³-66³ = 1×(67²+67×66+66²) = 4489+4422+4356 = 13267. (ii) 43³-42³ = 1×(43²+43×42+42²) = 1849+1806+1764 = 5419. (iii) 67²-66² = (67-66)(67+66) = 1×133 = 133. (iv) 43²-42² = (43-42)(43+42) = 1×85 = 85. The greatest is (i) 67³-66³ = 13267.