Question

Two numbers $a$ and $b$ are selected uniformly at random from the set $\{1,2,3,\cdots ,99\}.$ The probability that $3$ divides $a^3+b^3$ is

• A. $\frac{16}{147}$
• B. $\frac{11}{49}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

Two numbers $a$ and $b$ can be selected in ${}^{99}{C}_2$ ways.

Case $1:$
$a\in \{1,4,7,\cdots ,97\}$
$b\in \{2,5,8,\cdots ,98\}$
Probability, $p_1=\frac{{}^{33}{C}_1\times {}^{33}{C}_1}{{}^{99}{C}_2}$

Case $2:$
$a\in \{3,6,9,\cdots ,99\}$
$b\in \{3,6,9,\cdots ,99\}$
Probability, $p_2=\frac{{}^{33}{C}_2}{{}^{99}{C}_2}$

Total probability $=p_1+p_2$
$=\frac{{}^{33}{C}_1\times {}^{33}{C}_1+{}^{33}{C}_2}{{}^{99}{C}_2}$

$=\frac{33\times 33+33\times 16}{99\times 49}$
$=\frac{1}{3}$