Difference between revisions of "2011 AIME I Problems/Problem 7"
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== See also == | == See also == | ||
| − | {{AIME box|year=2011|n=I| | + | {{AIME box|year=2011|n=I|num-b=6|num-a=8}} |
* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
Revision as of 06:41, 29 March 2011
Problem 7
Find the number of positive integers 
for which there exist nonnegative integers 
, 
, 
, 
such that
![\[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\]](http://latex.artofproblemsolving.com/3/7/c/37c3a4d8143a7bc88f260f55b43ac3f3fbb7678e.png)
Solution
NOTE: This solution is incomplete. Please help make it better.
This formula only works if is exactly 1 more than a factor of 2010. Since 2010 factors as 
, there are 
such factors.
First I show that 
must divide 
. Consider the desired equation 
. The left side is 
, whereas the right side is 
. Thus, we have 
, so must divide 2010.
I will consider the example of 
to give a sense of why will work so long as divides 2010. We can write 
. Exchanging three 
terms for a 
term leaves the value on the right the same and decreases the number of terms by 2. Thus, we can write 
using 
terms, or 
, where the pattern is that the number of possible terms is 
. Since 
, is a value of for which we can obtain the desired sum. If we run out of terms, we can start exchanging three terms for a 
term. In general, this exchange will take terms of 
and replace them with one 
term, thus reducing the number of terms by 
.
See also
| 2011 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
