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About Real Numbers — Class 10 Olympiad

Apply Euclid's division algorithm, fundamental theorem of arithmetic, and prove irrationality at Olympiad level. This topic is part of the Olympiad Class 10 mathematics syllabus (chapter: Module 1). On this page you can practice 59 questions across three difficulty levels — 20 easy, 20 medium, and 19 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.

Real Numbers — solved examples for Class 10 Olympiad

Example 1easy

When the square of any positive integer is divided by 4, what are the possible remainders?
  1. A)0, 1 or 2
  2. B)0 or 1
  3. C)1 or 2
  4. D)0, 1, 2 or 3

Step-by-step solution

  1. Any positive integer can be expressed in the form 2k or 2k+1 for some integer k.
  2. Case 1: If the integer is of the form 2k, its square is (2k)² = 4k². When 4k² is divided by 4, the remainder is 0.
  3. Case 2: If the integer is of the form 2k+1, its square is (2k+1)² = 4k² + 4k + 1 = 4(k² + k) + 1. When 4(k² + k) + 1 is divided by 4, the remainder is 1.
  4. Therefore, the possible remainders when the square of any positive integer is divided by 4 are 0 or 1.

Answer: 0 or 1

Example 2medium

Let `a` and `b` be two positive integers such that `a > b`. When `a` is divided by `b`, the quotient is `q` and the remainder is `r`. If `b = 2r`, and the HCF of `a` and `b` is 5, find the smallest possible value of `a`.
  1. A)10
  2. B)15
  3. C)20
  4. D)25

Step-by-step solution

  1. According to Euclid's Division Lemma, `a = bq + r` where `0 ≤ r < b`.
  2. Given `b = 2r`, we can substitute `r = b/2` into the equation: `a = bq + b/2 = b(q + 1/2)`.
  3. Since `a` and `b` are integers, `b` must be an even number. Also, `HCF(a, b) = 5`. This means both `a` and `b` must be multiples of 5.
  4. Combining `b` is even and `b` is a multiple of 5, `b` must be a multiple of `LCM(2, 5) = 10`. The smallest possible positive value for `b` is 10.
  5. If `b = 10`, then `r = b/2 = 5`. Substitute these into `a = bq + r`: `a = 10q + 5`.
  6. We are given `HCF(a, b) = 5`. So, `HCF(10q + 5, 10) = 5`. We can factor out 5: `5 × HCF(2q + 1, 2) = 5`. This implies `HCF(2q + 1, 2) = 1`. Since `2q + 1` is always an odd number, its HCF with 2 is always 1, so this condition holds for any integer `q`.
  7. We need the smallest possible value of `a`. Since `a > b`, `10q + 5 > 10`, which means `10q > 5`, or `q > 0.5`. The smallest integer value for `q` is 1.
  8. Substitute `q = 1` into `a = 10q + 5`: `a = 10(1) + 5 = 15`. Thus, the smallest possible value of `a` is 15.

Answer: 15

Example 3hard

Let N be a positive integer such that when N is divided by 5, the remainder is 3. What is the remainder when N² is divided by 5?
  1. A)A) 0
  2. B)B) 1
  3. C)C) 2
  4. D)D) 4

Step-by-step solution

  1. According to Euclid's Division Lemma, if N leaves a remainder of 3 when divided by 5, then N can be written as N = 5k + 3 for some integer k.
  2. Now, calculate N²: N² = (5k + 3)² = (5k)² + 2(5k)(3) + 3² = 25k² + 30k + 9.
  3. To find the remainder when N² is divided by 5, we can look at the terms modulo 5: 25k² ≡ 0 (mod 5), 30k ≡ 0 (mod 5).
  4. For the term 9: 9 = 5 × 1 + 4, so 9 ≡ 4 (mod 5). Therefore, N² ≡ 0 + 0 + 4 (mod 5), which means N² ≡ 4 (mod 5). The remainder is 4.

Answer: D) 4

Practice questions on Real Numbers

  1. Q1.easy

    The HCF of two positive integers 'a' and 'b' is 18. If a = 2³ × 3² × 5 and b = 2ˣ × 3ʸ × 7, what is the value of x + y?
    1. A)3
    2. B)4
    3. C)5
    4. D)6
    Show answer

    Answer: 3

    Hint: Recall that the HCF of two numbers is found by taking the lowest power of each common prime factor.

  2. Q2.easy

    For which natural number 'n' can the number 12ⁿ end with the digit 0?
    1. A)For n = 1
    2. B)For n = 2
    3. C)For any natural number n
    4. D)Never
    Show answer

    Answer: Never

    Hint: A number ends with the digit 0 if and only if its prime factorization includes both 2 and 5.

  3. Q3.easy

    If 'a' and 'b' are two distinct prime numbers, what is HCF(a, b) + LCM(a, b)?
    1. A)a + b
    2. B)ab
    3. C)1 + ab
    4. D)a + b + ab
    Show answer

    Answer: 1 + ab

    Hint: Recall the definitions of HCF and LCM for prime numbers. What are the common factors and multiples?

  4. Q4.medium

    If `N` is a positive integer such that `N^2` has 15 divisors, and `N^3` has 28 divisors, what is the number of divisors of `N`?
    1. A)4
    2. B)6
    3. C)8
    4. D)12
    Show answer

    Answer: 6

    Hint: Recall the formula for the number of divisors of an integer from its prime factorization. Let `N = p_1^{a_1} p_2^{a_2} ... p_k^{a_k}` and determine the possible sets of exponents.

  5. Q5.medium

    Let `x` be a non-zero rational number and `y` be an irrational number. Which of the following statements is always true?
    1. A)x + y is rational
    2. B)x × y is rational
    3. C)x + y is irrational
    4. D)x^y is rational
    Show answer

    Answer: x + y is irrational

    Hint: Consider the definitions of rational and irrational numbers. For each statement, try to find a counterexample if you think it's not always true, or prove it by contradiction.

  6. Q6.medium

    The number `N = (2^k × 5^{k+1})` has its reciprocal `1/N` terminating in `k+1` decimal places. For which positive integer value of `k` does the fraction `(2k+1)/N` terminate in exactly `k` decimal places instead of `k+1`?
    1. A)1
    2. B)2
    3. C)3
    4. D)4
    Show answer

    Answer: 2

    Hint: The number of decimal places in a terminating decimal is determined by the maximum exponent of 2 or 5 in the denominator after the fraction is reduced to its simplest form. Consider when the numerator `(2k+1)` might simplify the denominator's factors.

  7. Q7.hard

    Find the smallest positive integer N such that 2N is a perfect square, 3N is a perfect cube, and 5N is a perfect fifth power.
    1. A)A) 2¹⁵ × 3²⁰ × 5²⁴
    2. B)B) 2¹⁵ × 3¹⁹ × 5²⁴
    3. C)C) 2¹⁴ × 3²⁰ × 5²⁴
    4. D)D) 2¹⁵ × 3²⁰ × 5²³
    Show answer

    Answer: A) 2¹⁵ × 3²⁰ × 5²⁴

    Hint: Let N = 2^a × 3^b × 5^c. For each condition, set up congruences for the exponents (a, b, c) based on the definitions of perfect squares, cubes, and fifth powers.

  8. Q8.hard

    Let p and q be distinct prime numbers. Which of the following expressions is always irrational?
    1. A)A) p + q
    2. B)B) √p × √q
    3. C)C) √p + √q
    4. D)D) (p+q)²
    Show answer

    Answer: C) √p + √q

    Hint: Consider proof by contradiction. Assume the expression is rational and derive a contradiction, possibly involving a rational square root.

  9. Q9.hard

    What is the smallest positive integer 'x' such that 7^x - 1 is divisible by 13?
    1. A)A) 3
    2. B)B) 4
    3. C)C) 6
    4. D)D) 12
    Show answer

    Answer: D) 12

    Hint: The problem asks for the order of 7 modulo 13. Fermat's Little Theorem can provide an upper bound for this order.

These are 9 of the 59 questions available for Real Numbers. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.