Let n 1 be an integer. Which of the followi | vedclass

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(B) We analyze the remainder of $n$ when divided by $3$. Let $n \equiv r \pmod{3}$,where $r \in \{0, 1, 2\}$.Case $1$: If $n \equiv 0 \pmod{3}$,then $

Vedclass · Accepted Answer

$n^{19}, n^{38}-1$

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(B) We analyze the remainder of $n$ when divided by $3$. Let $n \equiv r \pmod{3}$,where $r \in \{0, 1, 2\}$.Case $1$: If $n \equiv 0 \pmod{3}$,then $n$ is a multiple of $3$. Consequently,$n^{19}$ is a multiple of $3$.Case $2$: If $n \equiv 1 \pmod{3}$,then $n^{38} \equiv 1^{38} \equiv 1 \pmod{3}$. Thus,$n^{38} - 1 \equiv 0 \pmod{3}$,meaning $n^{38} - 1$ is a multiple of $3$.Case $3$: If $n \equiv 2 \pmod{3}$,then $n^{38} \equiv 2^{38} \equiv (-1)^{38} \equiv 1 \pmod{3}$. Thus,$n^{38} - 1 \equiv 0 \pmod{3}$,meaning $n^{38} - 1$ is a multiple of $3$.In all cases,at least one of the numbers in the set $\{n^{19}, n^{38}-1\}$ is a multiple of $3$. Therefore,the correct option is $B$.

Let $n > 1$ be an integer. Which of the following sets of numbers necessarily contains a multiple of $3$?

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