Question

Out of the first 5 consecutive natural numbers, if two different numbers x and y are chosen at random, then the probability that x⁴- y⁴ is divisible by 5 is:

• A. \(\frac{2}{5}\)
• B. \(\frac{4}{5}\)
• C. \(\frac{3}{5}\)
• D. \(\frac{1}{5}\)

Hint:

For selecting r items out of n, use the combination formula \binom{n}{r} = \frac{n!}{r!(n-r)!}.

Answer 1

The Correct Option is C

We are given 5 consecutive natural numbers: \( 1, 2, 3, 4, 5 \). We need to find the probability that for two randomly chosen distinct numbers \(x\) and \(y\), the expression \(x^4 - y^4\) is divisible by 5. 

Step 1: Understanding the Condition for Divisibility 
From the identity: \[ x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x-y)(x+y) \] Since 5 consecutive natural numbers cover all residues modulo 5 (i.e., 0, 1, 2, 3, 4), we will compute the values of \(x^4 \mod 5\). 

Step 2: Values of \(x^4 \mod 5\) 
By Fermat’s Little Theorem: \[ x^4 \equiv 1 \pmod{5} \quad \text{for} \; x = 1, 2, 3, 4 \] \[ x^4 \equiv 0 \pmod{5} \quad \text{for} \; x = 5 \] 

Step 3: Condition for \(x^4 - y^4 \equiv 0 \pmod{5} \) 
- If \(x^4 \equiv 1\) and \(y^4 \equiv 1\), then \(x^4 - y^4 = 0\).
- If \(x^4 \equiv 0\) and \(y^4 \equiv 0\), then \(x^4 - y^4 = 0\).
- If \(x^4 \equiv 1\) and \(y^4 \equiv 0\) (or vice versa), then \(x^4 - y^4 \equiv 1\).
Step 4: Probability Calculation 
- Total number of ways to choose 2 distinct numbers out of 5: \[ \binom{5}{2} = 10 \] - Number of valid pairs that satisfy \(x^4 - y^4 \equiv 0\) (when both residues are equal or both are divisible by 5): \[ \text{Valid pairs:} \quad (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) = 6 \text{ pairs} \] 

Step 5: Probability Calculation 
\[ \text{Probability} = \frac{6}{10} = \frac{3}{5} \] 

Final Answer: (C) \( \frac{3}{5} \)