Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 8"
Hashtagmath (talk | contribs) m (→Solution) |
Hashtagmath (talk | contribs) m (→Solution) |
||
Line 16: | Line 16: | ||
==Solution== | ==Solution== | ||
− | |||
− | |||
Here are some thoughts on the problem: | Here are some thoughts on the problem: | ||
Line 41: | Line 39: | ||
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. | ||
+ | |||
+ | |||
+ | {{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 12:55, 11 January 2019
Problem
Let denote the number of -tuples of real numbers such that and
Determine the remainder obtained when is divided by .
Solution
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
The signs in cannot both be positive. If they are both negative we get and there is obviously choice here only. Otherwise so
or
The latter case means (obviously 1 choice) or (1 choice). Thus choices total for the latter case. In the former case the number of choices is twice the number of choices for which forces for which there is 1 choice. Thus 2 choices total for the former case. Thus the number of choices when the expression equals is . So the answer is , 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 |