Out of the first 5 consecutive natural number | vedclass

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Question

(C) The set of the first $5$ natural numbers is $S = \{1, 2, 3, 4, 5\}$.The total number of ways to choose two different numbers $x$ and $y$ from $S$

Vedclass · Accepted Answer

$\frac{3}{5}$

Vedclass · Answer

(C) The set of the first $5$ natural numbers is $S = \{1, 2, 3, 4, 5\}$.The total number of ways to choose two different numbers $x$ and $y$ from $S$ is $^5C_2 = \frac{5 	imes 4}{2} = 10$.According to Fermat's Little Theorem,for any integer $a$ not divisible by $5$,$a^4 \equiv 1 \pmod{5}$.If $a$ is divisible by $5$,then $a^4 \equiv 0 \pmod{5}$.Let $x, y \in \{1, 2, 3, 4, 5\}$. We want $x^4 - y^4$ to be divisible by $5$,i.e.,$x^4 \equiv y^4 \pmod{5}$.Case $1$: Both $x$ and $y$ are not divisible by $5$. Then $x^4 \equiv 1$ and $y^4 \equiv 1$,so $x^4 - y^4 \equiv 0 \pmod{5}$.The numbers not divisible by $5$ are $\{1, 2, 3, 4\}$. The number of ways to choose $2$ distinct numbers from these $4$ is $^4C_2 = 6$.Case $2$: One of the numbers is $5$. If $x=5$,then $x^4 \equiv 0$. For $x^4 - y^4$ to be divisible by $5$,$y^4$ must also be $0$,which is impossible as there is only one multiple of $5$ in the set.Thus,the favorable outcomes are $6$.The probability is $\frac{6}{10} = \frac{3}{5}$.

Out of the first $5$ consecutive natural numbers,if two different numbers $x$ and $y$ are chosen at random,then the probability that $x^4 - y^4$ is divisible by $5$ is

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