10th Maths Real Numbers Exercise 1.1 Solutions

Exercise 1.1 Solutions – Class X Mathematics

Exercise 1.1 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on Euclid’s algorithm and the division algorithm. Mathematical expressions are rendered using MathJax.

1. Use Euclid’s algorithm to find the HCF of:

Euclid’s algorithm: For two positive integers \(a\) and \(b\) (where \(a > b\)), divide \(a\) by \(b\), take the remainder, and repeat until the remainder is 0. The last non-zero remainder is the HCF.

(i) 900 and 270

Step 1: \( 900 = 270 \times 3 + 90 \) (Quotient = 3, Remainder = 90)
Step 2: \( 270 = 90 \times 3 + 0 \) (Quotient = 3, Remainder = 0)
Since remainder is 0, HCF is the last divisor: 90.

HCF(900, 270) = 90

(ii) 196 and 38220

Step 1: \( 38220 = 196 \times 194 + 164 \) (Quotient = 194, Remainder = 164)
Step 2: \( 196 = 164 \times 1 + 32 \) (Quotient = 1, Remainder = 32)
Step 3: \( 164 = 32 \times 5 + 4 \) (Quotient = 5, Remainder = 4)
Step 4: \( 32 = 4 \times 8 + 0 \) (Quotient = 8, Remainder = 0)
Since remainder is 0, HCF is the last divisor: 4.

HCF(196, 38220) = 4

(iii) 1651 and 2032

Step 1: \( 2032 = 1651 \times 1 + 381 \) (Quotient = 1, Remainder = 381)
Step 2: \( 1651 = 381 \times 4 + 127 \) (Quotient = 4, Remainder = 127)
Step 3: \( 381 = 127 \times 3 + 0 \) (Quotient = 3, Remainder = 0)
Since remainder is 0, HCF is the last divisor: 127.

HCF(1651, 2032) = 127

2. Use division algorithm to show that any positive odd integer is of the form \(6q + 1\), \(6q + 3\), or \(6q + 5\), where \(q\) is some integer.

Division algorithm: \( a = bq + r \), where \( 0 \leq r < b \). Let \( a \) be a positive odd integer, \( b = 6 \), so \( r = 0, 1, 2, 3, 4, 5 \).
Check remainders:
  • \( r = 0 \): \( a = 6q \) (even, e.g., 6, 12)
  • \( r = 1 \): \( a = 6q + 1 \) (odd, e.g., 1, 7)
  • \( r = 2 \): \( a = 6q + 2 \) (even, e.g., 2, 8)
  • \( r = 3 \): \( a = 6q + 3 \) (odd, e.g., 3, 9)
  • \( r = 4 \): \( a = 6q + 4 \) (even, e.g., 4, 10)
  • \( r = 5 \): \( a = 6q + 5 \) (odd, e.g., 5, 11)
Since \( a \) is odd, \( r = 1, 3, 5 \). Thus, \( a = 6q + 1 \), \( 6q + 3 \), or \( 6q + 5 \).

Conclusion: Every positive odd integer is of the form \(6q + 1\), \(6q + 3\), or \(6q + 5\).

3. Use division algorithm to show that the square of any positive integer is of the form \(3p\) or \(3p + 1\).

Let \( a \) be a positive integer, \( a = 3q + r \), where \( r = 0, 1, 2 \).
Compute \( a^2 \):
  • \( r = 0 \): \( a = 3q \), \( a^2 = (3q)^2 = 9q^2 = 3(3q^2) \), so \( a^2 = 3p \) (\( p = 3q^2 \)).
  • \( r = 1 \): \( a = 3q + 1 \), \( a^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 \), so \( a^2 = 3p + 1 \) (\( p = 3q^2 + 2q \)).
  • \( r = 2 \): \( a = 3q + 2 \), \( a^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 \), so \( a^2 = 3p + 1 \) (\( p = 3q^2 + 4q + 1 \)).
Thus, \( a^2 = 3p \) or \( 3p + 1 \).

Conclusion: The square of any positive integer is of the form \(3p\) or \(3p + 1\).

4. Use division algorithm to show that the cube of any positive integer is of the form \(9m\), \(9m + 1\), or \(9m + 8\).

Let \( a = 3q + r \), where \( r = 0, 1, 2 \).
Compute \( a^3 \):
  • \( r = 0 \): \( a = 3q \), \( a^3 = (3q)^3 = 27q^3 = 9(3q^3) \), so \( a^3 = 9m \) (\( m = 3q^3 \)).
  • \( r = 1 \): \( a = 3q + 1 \), \( a^3 = (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 \), so \( a^3 = 9m + 1 \) (\( m = 3q^3 + 3q^2 + q \)).
  • \( r = 2 \): \( a = 3q + 2 \), \( a^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 \), so \( a^3 = 9m + 8 \) (\( m = 3q^3 + 6q^2 + 4q \)).
Thus, \( a^3 = 9m \), \( 9m + 1 \), or \( 9m + 8 \).

Conclusion: The cube of any positive integer is of the form \(9m\), \(9m + 1\), or \(9m + 8\).

5. Show that one and only one out of \(n\), \(n + 2\), or \(n + 4\) is divisible by 3, where \(n\) is any positive integer.

Let \( n = 3q + r \), where \( r = 0, 1, 2 \).
Check remainders:
  • \( r = 0 \):
    • \( n = 3q \), remainder = 0 (divisible by 3).
    • \( n + 2 = 3q + 2 \), remainder = 2.
    • \( n + 4 = 3q + 4 = 3(q + 1) + 1 \), remainder = 1.
    • Only \( n \) is divisible by 3.
  • \( r = 1 \):
    • \( n = 3q + 1 \), remainder = 1.
    • \( n + 2 = 3q + 3 = 3(q + 1) \), remainder = 0 (divisible by 3).
    • \( n + 4 = 3q + 5 = 3(q + 1) + 2 \), remainder = 2.
    • Only \( n + 2 \) is divisible by 3.
  • \( r = 2 \):
    • \( n = 3q + 2 \), remainder = 2.
    • \( n + 2 = 3q + 4 = 3(q + 1) + 1 \), remainder = 1.
    • \( n + 4 = 3q + 6 = 3(q + 2) \), remainder = 0 (divisible by 3).
    • Only \( n + 4 \) is divisible by 3.
In each case, exactly one number is divisible by 3.

Conclusion: One and only one of \(n\), \(n + 2\), or \(n + 4\) is divisible by 3.

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