Two numbers x and y are chosen at random (wit | vedclass

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(C) Let the set be $S = \{1, 2, \dots, 1000\}$. The total number of ways to choose two distinct numbers $x$ and $y$ is $\binom{1000}{2} = \frac{1000 \

Vedclass · Accepted Answer

$\frac{679}{999}$

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(C) Let the set be $S = \{1, 2, \dots, 1000\}$. The total number of ways to choose two distinct numbers $x$ and $y$ is $\binom{1000}{2} = \frac{1000 	imes 999}{2} = 500 	imes 999$.We want $|x^4 - y^4|$ to be divisible by $5$. By Fermat's Little Theorem,$n^4 \equiv 1 \pmod{5}$ if $n$ is not divisible by $5$,and $n^4 \equiv 0 \pmod{5}$ if $n$ is divisible by $5$.Let $S_0$ be the set of numbers in $S$ divisible by $5$ $(|S_0| = 200)$ and $S_1$ be the set of numbers not divisible by $5$ $(|S_1| = 800)$.$|x^4 - y^4| \equiv 0 \pmod{5}$ if:$1$. Both $x, y \in S_0$: Number of ways = $\binom{200}{2} = \frac{200 	imes 199}{2} = 100 	imes 199 = 19900$.$2$. Both $x, y \in S_1$: Number of ways = $\binom{800}{2} = \frac{800 	imes 799}{2} = 400 	imes 799 = 319600$.Total favorable outcomes = $19900 + 319600 = 339500$.Probability = $\frac{339500}{500 	imes 999} = \frac{3395}{5 	imes 999} = \frac{679}{999}$.

Two numbers $x$ and $y$ are chosen at random (without replacement) from the set $\{1, 2, 3, \dots, 1000\}$. The probability that $|x^4 - y^4|$ is divisible by $5$ is -

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