Question

Two numbers $a$ and $b$ are chosen at random from the first $30$ natural numbers. The probability that $a^2-b^2$ is divisible by $3$ is

• A. $\frac{47}{87}$
• B. $\frac{37}{87}$
• C. $\frac{57}{87}$
• D. $\frac{33}{87}$

Answer 1

The correct option is A $\frac{47}{87}$

The total number of ways of choosing two numbers out of $1,2,3,...,30$ is ${}^{30}{C}_2=435$
So, exhaustive number of cases is $435.$

$a^2-b^2=(a+b)(a-b)$
If $a$ and $b$ both are divisible by $3,$ then $a+b$ is divisible by $3.$ Consequently, $a^2-b^2$ is divisible by $3.$

If neither of $a$ and $b$ are divisible by $3:$
Case $1.$ If $a$ and $b$ leave the same remainder, then $a-b$ is divisible by $3.$
Case $2.$ If one leaves remainder $1$ and the other leaves remainder $2,$ then $a+b$ is divisible by $3.$
Consequently, $a^2-b^2$ is divisible by $3$ in both cases.

Thus, favourable number of cases is ${}^{10}{C}_2+{}^{20}{C}_2=235$
Hence, the required probability is $\frac{235}{435}=\frac{47}{87}$