2011 AIME I Problems/Problem 7
Problem 7
Find the number of positive integers for which there exist nonnegative integers , , , such that
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.
Now let us prove that any m where m-1 divides 2010 works. Suppose we have such an m. Consider the equation m^k = m^k, for sufficiently large k. Replace m^k on the right hand side by m^(k-1)+m^(k-1)+...m^(k-1), with k terms. We have increased the number of terms by k-1. By repeating this process, we can generate an expression on the rhs with 1+ A(k-1) terms, where A is any positive integer. Since k-1 divides 2010, we can choose A to make the value of 1+A(k-1) equal to 2011.
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 |