Mock AIME 3 Pre 2005 Problems/Problem 8

Problem

Let $N$ denote the number of $8$ -tuples $(a_1, a_2, \dots, a_8)$ of real numbers such that $a_1 = 10$ and

$\left|a_1^{2} - a_2^{2}\right| = 10$

$\left|a_2^{2} - a_3^{2}\right| = 20$

$\cdots$

$\left|a_7^{2} - a_8^{2}\right| = 70$

$\left|a_8^{2} - a_1^{2}\right| = 80$

Determine the remainder obtained when $N$ is divided by $1000$ .

Solution

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Here are some thoughts on the problem:

We can call $a_1^2$ through $a_8^2$ by $b_1$ through $b_8$ and the only restriction is that the $b_i$ 's are positive. We can express $b_1=100$ , $b_2=100 \pm 10$ , ... $b_5=100 \pm 10 \pm 20 \pm 30 \pm 40$ and also $b_8=100 \pm 80$ . Note that $b_7$ is either $90,110,250$ . Note that regardless of how we choose these $\pm$ 's all the $b_i$ 's I've listed are positive so no restrictions are imposed here. There are restrictions imposed by $b_6$ being equal to $b_7 \pm 60$ . We can now write $b_7/(10) \pm 6=b_6/(10)=10 \pm 1 \pm 2 \pm 3 \pm 4 \pm 5$ so the only restrictions are imposed by $10 \pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6$ being equal to either $9,11,25$ . If we find all the $\pm$ in this expression then $b_1$ through $b_8$ are all determined. We can reformulate now as find the number of choices of $\pm$ signs in the expression below:

$\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6$

which equals either $-1,1,15$ .

If the expression equals $15$ then note that $\pm 1 \pm 2 \pm 3 \pm 4 \pm 5$ is at most 15 so we must have $\pm 1 \pm 2 \pm 3 \pm 4 \pm 5=9$ , which forces $\pm 1 \pm 2 \pm 3 \pm 4 =4$ which forces $\pm 1 \pm 2 \pm 3 =0$ for which there are two possibilities of signs.

Now if the expression equals $1$ its symmetric to the case where it equals $-1$ so lets just consider

$\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6=1$

If there are $x$ possibility in this case then the answer to the problem should be $2x+2$ . Unfortunately I've got this far but I don't exactly know how to find $x$ without bashing.

So, how do we find the number of choices of $\pm$ signs such that

$\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6=1$

and can we generalize from $6$ to $n$ ?

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Mock AIME 3 Pre 2005 Problems/Problem 8

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