Level
1

Determine the number of 8 - tuples $£_1, £_2, \cdots , £_8$ such that $£_1, £_2, \cdots £_8$ belongs to { $1, -1$ } and

$£_1 + 2£_2 + 3£_3 + \cdots + 8£_8$ Is a multiple of 3

84
94
85
88

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$\pounds_1 + 2\pounds_2 + 3\pounds_3 + 4\pounds_4 + 5\pounds_5 + 6\pounds_6 + 7\pounds_7 + 8\pounds_8$

$\pounds_1 + 2\pounds_2 + 4\pounds_4 + 5\pounds_5 + 7\pounds_7 + 8\pounds_8 + 3(\pounds_3 + 2\pounds_6)$

Add and Subtract $\pounds_2$ , $\pounds_5$ and $\pounds_8$

$(\pounds_1 - \pounds_2) + (\pounds_4 - \pounds_5) + (\pounds_7 - \pounds_8) + 3(\pounds_2 + 2\pounds_5 + 3\pounds_8) + 3(\pounds_3 + 2\pounds_6)$

Number of ways to choose $\pounds_3$ and $\pounds_6$ is $2 \times 2 = 4$

Number of ways to choose $\pounds_1, \pounds_2, \pounds_4, \pounds_5, \pounds_7$ and $\pounds_8$ noticing that $\pounds_i \in$ ${-1, 1}$ , are:

$\textbf{Case I:}$

$(\pounds_1 - \pounds_2) = (\pounds_4 - \pounds_5) = (\pounds_7 - \pounds_8) = 0$

Since there are $2$ ways for each bracket to be $0$ , the number of ways for choosing $\pounds_1, \pounds_2, \pounds_4, \pounds_5, \pounds_7$ and $\pounds_8$ are $2 \times 2 \times 2 = 8$

$\textbf{Case II:}$

$(\pounds_1 - \pounds_2) = (\pounds_4 - \pounds_5) = (\pounds_7 - \pounds_8) = 2$

Since there is only $1$ way for each bracket to be $0$ , the number of ways for choosing $\pounds_1, \pounds_2, \pounds_4, \pounds_5, \pounds_7$ and $\pounds_8$ is $1 \times 1 \times 1 = 1$

$\textbf{Case III:}$

$(\pounds_1 - \pounds_2) = (\pounds_4 - \pounds_5) = (\pounds_7 - \pounds_8) = -2$

Since there is only $1$ way for each bracket to be $0$ , the number of ways for choosing $\pounds_1, \pounds_2, \pounds_4, \pounds_5, \pounds_7$ and $\pounds_8$ is $1 \times 1 \times 1 = 1$

$\textbf{Case IV:}$

Any of the brackets be equal to $2, -2, 0$

As there are $3!$ ways to arrange these values in the brackets, and only one way to choose for $2$ and $-2$ as well as 2 ways to choose for $0$

Hence, the number of ways for choosing $\pounds_1, \pounds_2, \pounds_4, \pounds_5, \pounds_7$ and $\pounds_8$ is $3! \times 1 \times 1 \times 2 = 12$

Therefore, the total number of 8 tuples $\pounds_1, \pounds_2, \ldots \pounds_8$ are $4 \times (8 + 1 + 1 + 12) = \boxed{88}$