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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 8"

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==Solution==
 
==Solution==
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{{alternate solutions}}
 
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Here are some thoughts on the problem:
 
Here are some thoughts on the problem:
  
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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 the number of choices is twice the number of choices for <math>\pm 1 \pm 2 \pm 3=4</math> which forces <math>\pm 1 \pm 2=1</math> for which there is 1 choice. Thus 2 choices total for the former case. Thus the number of choices when the expression equals <math>1</math> is <math>1+2+2=5</math>. So the answer is <math>2*5+1=11</math>, so actually the conditions of the problem were quite restrictive.
 
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 the number of choices is twice the number of choices for <math>\pm 1 \pm 2 \pm 3=4</math> which forces <math>\pm 1 \pm 2=1</math> for which there is 1 choice. Thus 2 choices total for the former case. Thus the number of choices when the expression equals <math>1</math> is <math>1+2+2=5</math>. So the answer is <math>2*5+1=11</math>, so actually the conditions of the problem were quite restrictive.
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{{alternate solutions}}
  
 
==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}}

Latest revision as of 11:55, 11 January 2019

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

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

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

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

$\pm 1 \pm 2 \pm 3 \pm 4=0$

or

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

The latter case means $\pm 1 \pm 2 \pm 3 =6$ (obviously 1 choice) or $\pm 1 \pm 2 \pm 3 =2$ (1 choice). Thus $2$ choices total for the latter case. In the former case the number of choices is twice the number of choices for $\pm 1 \pm 2 \pm 3=4$ which forces $\pm 1 \pm 2=1$ for which there is 1 choice. Thus 2 choices total for the former case. Thus the number of choices when the expression equals $1$ is $1+2+2=5$. So the answer is $2*5+1=11$, so actually the conditions of the problem were quite restrictive.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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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