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 21:08, 5 March 2015
Problem
Let denote the number of -tuples of real numbers such that and
Determine the remainder obtained when is divided by .
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 through by through and the only restriction is that the 's are positive. We can express , , ... and also . Note that is either . Note that regardless of how we choose these 's all the 's I've listed are positive so no restrictions are imposed here. There are restrictions imposed by being equal to . We can now write so the only restrictions are imposed by being equal to either . If we find all the in this expression then through are all determined. We can reformulate now as find the number of choices of signs in the expression below:
which equals either .
If the expression equals then note that is at most 15 so we must have , which forces which forces for which there are two possibilities of signs.
Now if the expression equals its symmetric to the case where it equals so lets just consider
If there are possibility in this case then the answer to the problem should be . Unfortunately I've got this far but I don't exactly know how to find without bashing.
So, how do we find the number of choices of signs such that
and can we generalize from to ?
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 |