Question

Two numbers x and y are chosen at random (without replacement) from amongst the numbers $1,2,3,.....2004$. The probability that $x^3+y^3$ is divisible by 3 is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

We divide the number in three groups
$3k+1$ type $\left\{1,4,7,..................,2005\right\}$
$3k+2$ type $\left\{2,5,8,..................,2006\right\}$
$3k+3$ type $\left\{3,6,9,..................,2007\right\}$
$x^3+y^3$ is divisible by 3 if x and y both belong to $3^{rd}$ group or one of them belongs to the first group and the other to the second group. So favourable number of cases
${=}^{669}{C}_2+669\times 669$
Total number of cases ${}^{2007}{C}_2$
$\therefore$ Desired probability $=\frac{\frac{669\times 668}{2}+669\times 668}{\frac{2007\times 2006}{2}}$
$=\frac{669\times 2006}{2007\times 2006}=\frac{1}{3}$