Question

From a set of first $18$ natural number two number $\{x,y\}$ are selected randomly then the probability that sum of their cubes is divisible by $3$, is

• A. $\frac{5}{51}$
• B. $\frac{17}{51}$
• C. $\frac{3}{51}$
• D. $\frac{15}{51}$

Answer 1

The correct option is B $\frac{17}{51}$

Given set $\{x,y\}\in \{1,2,3,....18\}$

Now $x^3+y^3=(x+y)(x^2-xy+y^2)$ will be divisible by $3$ now number $x+y$ or $(x^2-xy+y^2)$ is divisible by $3$ now number $x+y$ will be divisible by $3$, if $x,y$ both are multiple of $3$ or among $x,y$ one leaves the remainder $1$ and other leaves the remainder $2$ when divided by $3$

Now $3,6,9,12,15,18$ are multiple of $3$
(these numbers are $6$)
and $1,4,7,10,13,16$ are those who leaves remainder $1$ when divided by $3$
and $2,5,8,11,14,17$ are those who leaves remainder

$2$ when divided by $3$

$n\left(A\right)={{}^{6}C}_1\times {{}^{6}C}_1+{{}^{6}C}_2=36+15=51$
$n\left(S\right)={{}^{18}C}_2=\frac{18}{2}.\frac{17}{1}=153$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{51}{153}=\frac{17}{51}=\frac{1}{3}$