View source for Mock AIME 3 Pre 2005 Problems/Problem 8

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

<math>\left|a_1^{2} - a_2^{2}\right| = 10</math>

<math>\left|a_2^{2} - a_3^{2}\right| = 20</math>

<math>\cdots</math>

<math>\left|a_7^{2} - a_8^{2}\right| = 70</math>

<math>\left|a_8^{2} - a_1^{2}\right| = 80</math>


Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>.

==Solution==
{{solution}}

Here are some thoughts on the problem:

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

<math>\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6</math>

which equals either <math>-1,1,15</math>.

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

Now if the expression equals <math>1</math> its symmetric to the case where it equals <math>-1</math> so lets just consider 

<math>\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6=1</math>

The signs in <math>\pm 5 \pm 6</math> cannot both be positive. If they are both negative we get <math>\pm 1 \pm 2 \pm 3 \pm 4=10</math> and there is obviously <math>1</math> choice here only. Otherwise <math>\pm 5 \pm 6=\pm 1</math> so

<math>\pm 1 \pm 2 \pm 3 \pm 4=0</math>

or

<math>\pm 1 \pm 2 \pm 3 \pm 4=2</math>

The latter case means <math>\pm 1 \pm 2 \pm 3 =6</math> (obviously 1 choice) or <math>\pm 1 \pm 2 \pm 3 =2</math> (1 choice). Thus <math>2</math> choices total for the latter case. In the former case

==See Also==
{{Mock AIME box|year=Pre 2005|n=3|num-b=7|num-a=9}}