Question

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

Answer 1

$(a^2-b^2)$ is divisible by 3

i.e $(a-b)\left(a+b\right)$ is divisible by 3.

Case(1) $\left(a+b\right)$ is divisible by 3

If a is chosen from (3k+1) set ,then b must be from (3k+2) set and viceversa.

If a is chosen from (3k) set, then b must be from (3k).

So, number of ways = 10*10

Note that when a is from 3k and b is from 3k,

then $\left(a-b\right)$ is also divisible by 3.

Case(2)$\left(a-b\right)$ but not $\left(a+b\right)$ is divisible by 3

If a is chosen from (3k+1) set ,then b must be from (3k+1) set .

Same for (3k+2) set.

So, number of ways = 2* 10*10

Number of ways of choosing such that a,b such that $(a^2-b^2)$ is divisible by 3

=30*30

Probability

$=\frac{5\times 10\times 10}{30\times 30}=\frac{5}{9}$