2005 AIME I Problems/Problem 2
Problem
For each positive integer , let denote the increasing arithmetic sequence of integers whose first term is and whose common difference is . For example, is the sequence For how many values of does contain the term ?
Solution
Suppose that the th term of the sequence is . Then so . The ordered pairs of positive integers that satisfy the last equation are ,, , , , , ,, , , and , and each of these gives a possible value of . Thus the requested number of values is , and the answer is .
Alternatively, notice that the formula for the number of divisors states that there are divisors of .
Solution 2
Any term in the sequence can be written as 1+kx. If this is to equal 2005, then the remainder when 2005 is divided by k is 1.
Now all we have to do is find the numbers of factors of 2004. There are divisors of .
Note that although the remainder when 2005 divided by 1 is not 1, it still works- would be the sequence of all positive integers, in which 2005 must appear.
Video Solution by OmegaLearn
https://youtu.be/qL0OOYZiaqA?t=83
~ pi_is_3.14
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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