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(B) For any integer $n$,$n^3 \equiv 0, 1, 	ext{ or } 8 \pmod{9}$,which is equivalent to $n^3 \equiv 0, 1, -1 \pmod{9}$.Statement $I$: We want to chec

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$I$ is true but $II$ is false

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(B) For any integer $n$,$n^3 \equiv 0, 1, 	ext{ or } 8 \pmod{9}$,which is equivalent to $n^3 \equiv 0, 1, -1 \pmod{9}$.Statement $I$: We want to check if $a^3+b^3+c^3 \equiv 0 \pmod{9}$ implies $abc \equiv 0 \pmod{3}$.If $abc$ is not divisible by $3$,then $a, b, c 
ot\equiv 0 \pmod{3}$. Thus $a, b, c \equiv 1, 2 \pmod{3}$.Then $a^3, b^3, c^3 \equiv 1, 8 \pmod{9}$ (since $1^3=1$ and $2^3=8 \equiv -1 \pmod{9}$).To have $a^3+b^3+c^3 \equiv 0 \pmod{9}$,we need the sum of three values from ${1, -1}$ to be $0 \pmod{9}$,which is impossible as the possible sums are $\{3, 1, -1, -3\}$.Thus,at least one of $a, b, c$ must be a multiple of $3$,so $abc$ is divisible by $3$. Statement $I$ is true.Statement $II$: Consider $a=1, b=2, c=1, d=2$.Then $a^3+b^3+c^3+d^3 = 1^3+2^3+1^3+2^3 = 1+8+1+8 = 18$,which is divisible by $9$.However,$abcd = 1 	imes 2 	imes 1 	imes 2 = 4$,which is not divisible by $3$.Thus,Statement $II$ is false.

Let $a, b, c, d$ be positive integers. Consider the following statements:
$I$. If $9$ divides $a^3+b^3+c^3$,then $3$ divides $abc$.
$II$. If $9$ divides $a^3+b^3+c^3+d^3$,then $3$ divides $abcd$.

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Let $a, b, c, d$ be positive integers. Consider the following statements:$I$. If $9$ divides $a^3+b^3+c^3$,then $3$ divides $abc$.$II$. If $9$ divides $a^3+b^3+c^3+d^3$,then $3$ divides $abcd$.