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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 8"
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==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
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Here are some thoughts on the problem:
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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:
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<math>\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6</math>
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which equals either <math>-1,1,15</math>.
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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.
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Now if the expression equals <math>1</math> its symmetric to the case where it equals <math>-1</math> so lets just consider
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<math>\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6=1</math>
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If there are <math>x</math> possibility in this case then the answer to the problem should be <math>2x+2</math>. Unfortunately I've got this far but I don't exactly know how to find <math>x</math> without bashing.
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--------------------------------------------------
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So, how do we find the number of choices of <math>\pm</math> signs such that
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<math>\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6=1</math>
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and can we generalize from <math>6</math> to <math>n</math>?
  
 
==See Also==
 
==See Also==
 
{{Mock AIME box|year=Pre 2005|n=3|num-b=7|num-a=9}}
 
{{Mock AIME box|year=Pre 2005|n=3|num-b=7|num-a=9}}

Revision as of 22:08, 5 March 2015

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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

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 $n$?

See Also

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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