Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 8"
Viperstrike (talk | contribs) (→Solution) |
Hashtagmath (talk | contribs) m (→Solution) |
||
(3 intermediate revisions by 2 users not shown) | |||
Line 16: | Line 16: | ||
==Solution== | ==Solution== | ||
− | |||
− | |||
Here are some thoughts on the problem: | Here are some thoughts on the problem: | ||
Line 32: | Line 30: | ||
<math>\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6=1</math> | <math>\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6=1</math> | ||
− | If | + | 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 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 |