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About Squares and Square Roots — Class 8 Olympiad

Find squares and square roots using various methods; solve Olympiad problems involving perfect squares and patterns. This topic is part of the Olympiad Class 8 mathematics syllabus (chapter: Module 3). On this page you can practice 60 questions across three difficulty levels — 20 easy, 20 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 35-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Squares and Square Roots — solved examples for Class 8 Olympiad

Example 1easy

Which of the following numbers CANNOT be a perfect square?
  1. A)A) 12345676
  2. B)B) 98765432
  3. C)C) 56789001
  4. D)D) 23456789

Step-by-step solution

  1. The unit digit of a perfect square can only be 0, 1, 4, 5, 6, or 9.
  2. Numbers ending in 2, 3, 7, or 8 can never be perfect squares.
  3. Option B, 98765432, ends in 2. Therefore, it cannot be a perfect square.

Answer: B) 98765432

Example 2medium

A natural number N ends with the digit 5. If N is squared, its result (N²) has last two digits that are not both zero. What is the tens digit of N²?
  1. A)2
  2. B)0
  3. C)5
  4. D)7

Step-by-step solution

  1. A natural number N ending with 5 can be written in the form N = 10k + 5 for some integer k ≥ 0.
  2. Squaring N, we get N² = (10k + 5)² = (10k)² + 2(10k)(5) + 5² = 100k² + 100k + 25.
  3. Factoring out 100 from the first two terms: N² = 100(k² + k) + 25.
  4. This expression shows that N² will always end in 25. Therefore, the tens digit of N² is always 2.

Answer: 2

Example 3hard

If N is a perfect square and its last digit is 6, which of the following statements about its tens digit (T) must be true?
  1. A)A. T is an even digit.
  2. B)B. T is an odd digit.
  3. C)C. T is either 0 or 2.
  4. D)D. T is either 4 or 6.

Step-by-step solution

  1. A perfect square ending in 6 must be the square of a number whose unit digit is either 4 or 6.
  2. Case 1: Numbers ending in 4. Let the number be (10k + 4). Its square is (10k + 4)² = 100k² + 80k + 16. The tens digit is determined by (80k + 16), which means the tens digit is the unit digit of (8k + 1), which is always odd (e.g., 16, 196, 576, ...).
  3. Case 2: Numbers ending in 6. Let the number be (10k + 6). Its square is (10k + 6)² = 100k² + 120k + 36. The tens digit is determined by (120k + 36), which means the tens digit is the unit digit of (12k + 3), which is always odd (e.g., 36, 256, 676, ...).
  4. In both cases, the tens digit of a perfect square ending in 6 must be an odd digit.

Answer: B. T is an odd digit.

Practice questions on Squares and Square Roots

  1. Q1.easy

    If S = 1 + 3 + 5 + ... + (2n-1), and S = 1024, what is the value of n?
    1. A)A) 32
    2. B)B) 16
    3. C)C) 64
    4. D)D) 1024
    Show answer

    Answer: A) 32

    Hint: Remember the special property related to the sum of the first 'n' odd natural numbers.

  2. Q2.easy

    What is the smallest positive integer by which 1575 must be multiplied so that the product is a perfect square?
    1. A)A) 3
    2. B)B) 5
    3. C)C) 7
    4. D)D) 15
    Show answer

    Answer: C) 7

    Hint: Perform the prime factorization of 1575 and identify any prime factors that do not appear in pairs.

  3. Q3.easy

    Which of the following is the best approximation for the value of √(300 × 301 × 302 × 303 + 1)?
    1. A)A) 90900
    2. B)B) 90901
    3. C)C) 90902
    4. D)D) 90903
    Show answer

    Answer: B) 90901

    Hint: Consider the algebraic pattern for the product of four consecutive integers plus one. Let the smallest integer be 'n'.

  4. Q4.medium

    What is the smallest natural number by which the product of the first six even natural numbers must be multiplied to make it a perfect square?
    1. A)5
    2. B)3
    3. C)10
    4. D)15
    Show answer

    Answer: 5

    Hint: First, write down the product of the first six even numbers. Then, find the prime factorization of this product.

  5. Q5.medium

    Consider the sequence S_n = 1 + 3 + 5 + ... + (2n-1). If S_k + S_(k+1) = 225, what is the value of k?
    1. A)6
    2. B)7
    3. C)8
    4. D)9
    Show answer

    Answer: 7

    Hint: Recall the formula for the sum of the first 'n' odd natural numbers.

  6. Q6.medium

    If the square of a natural number has 'm' digits, and its square root has 'n' digits, which of the following statements is always true?
    1. A)m = 2n-1
    2. B)m = 2n
    3. C)2n-1 ≤ m ≤ 2n
    4. D)m ≥ 2n
    Show answer

    Answer: 2n-1 ≤ m ≤ 2n

    Hint: Consider the range of values for a number with 'n' digits and its square. Test with small examples like N=3, N=10, N=31, N=32.

  7. Q7.hard

    How many 4-digit perfect squares are there such that the sum of their digits is 19 and their units digit is 9?
    1. A)A. 1
    2. B)B. 2
    3. C)C. 3
    4. D)D. 4
    Show answer

    Answer: C. 3

    Hint: A 4-digit perfect square has a square root between 32 and 99. If its units digit is 9, its square root must end in 3 or 7. Systematically check numbers matching these criteria.

  8. Q8.hard

    Find the smallest natural number 'k' such that 1470 × k is a perfect square, and 1470 / k is also a perfect square.
    1. A)A. 30
    2. B)B. 70
    3. C)C. 105
    4. D)D. 210
    Show answer

    Answer: A. 30

    Hint: First, find the prime factorization of 1470. For a number to be a perfect square, all exponents in its prime factorization must be even. Consider how 'k' must affect the odd exponents.

  9. Q9.hard

    What is the value of √[ (1 + 3 + 5 + ... + 39) ]?
    1. A)A. 10
    2. B)B. 20
    3. C)C. 40
    4. D)D. 400
    Show answer

    Answer: B. 20

    Hint: Recall the fundamental property that the sum of the first 'n' consecutive odd natural numbers is n². Determine how many odd numbers are present in the sum 1 + 3 + 5 + ... + 39.

These are 9 of the 60 questions available for Squares and Square Roots. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.