Question 5

Mathematics Permutation and Combination Hard

Two numbers a and b are chosen ar random from a set of first 30 natural numbers, then the probability that a<sup>2</sup> - b<sup>2</sup> is divisible by 3 is

(A) <span class="math-tex">\(\frac {47}{87}\)</span>
(B) <span class="math-tex">\(\frac {15}{87}\)</span>
(C) <span class="math-tex">\(\frac {12}{87}\)</span>
(D) <span class="math-tex">\(\frac {9}{87}\)</span>
✓ Correct Answer: Option A Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

We need to find the probability that \( a^2 - b^2 \) is divisible by 3 when two numbers \( a \) and \( b \) are randomly chosen from the first 30 natural numbers. \[\] Step 1: Consider Remainders Modulo 3 \[\] The numbers from 1 to 30 can be classified based on their remainder when divided by 3: \[\] Numbers of the form \( 3k \) (i.e., divisible by 3) → \( 3, 6, 9, ..., 30 \) \[\] Numbers of the form \( 3k+1 \) → \( 1, 4, 7, ..., 28 \) \[\] Numbers of the form \( 3k+2 \) → \( 2, 5, 8, ..., 29 \) \[\] Each category has 10 numbers since there are 30 numbers in total. \[\] Step 2: Condition for \( a^2 - b^2 \) to be Divisible by 3 \[\] Since \( a^2 - b^2 = (a-b)(a+b) \), we need to check when it is divisible by 3. This happens if: \[\] 1. \( a \) and \( b \) have the same remainder modulo 3, because: \[\] If \( a \equiv b \pmod{3} \), then \( a^2 \equiv b^2 \pmod{3} \), so \( a^2 - b^2 \equiv 0 \pmod{3} \). \[\] 2. \( a \) and \( b \) have different remainders, but their squared difference is still 0 modulo 3. Checking cases: \[\] \( (3k+1)^2 = 9k^2 + 6k + 1 \equiv 1 \pmod{3} \) \[\] \( (3k+2)^2 = 9k^2 + 12k + 4 \equiv 1 \pmod{3} \) \[\] \( (3k)^2 = 9k^2 \equiv 0 \pmod{3} \) \[\] This means that when \( a \) and \( b \) belong to different nonzero remainder groups, their squared values give the same remainder (1), leading to \( a^2 - b^2 \equiv 0 \pmod{3} \). \[\] Step 3: Counting Favorable Cases \[\] Since we randomly pick \( a, b \), the total number of ways to pick any two numbers from 30 is: \[ \text{Total cases} = \binom{30}{2} = \frac{30 \times 29}{2} = 435 \] To count favorable cases: \[\] Picking two numbers from the same group: \[\] Ways to choose two numbers from the 10 numbers in each remainder category: \[ \binom{10}{2} = \frac{10 \times 9}{2} = 45 \] There are 3 groups, so total: \[ 3 \times 45 = 135 \] Picking one number from the (3k+1) group and one from the (3k+2) group: \[\] Ways to pick one from each group: \( 10 \times 10 = 100 \). \[\] Thus, the total number of favorable cases is: \[ 135 + 100 = 235 \] Step 4: Computing Probability \[ P = \frac{235}{435} \] Simplifying: \[ P = \frac{47}{87} \] Thus, the required probability is \( \frac{47}{87} \).

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