Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AIME I Problems/Problem 2"

(→‎Solution)
(→‎Solution 2)
 
(6 intermediate revisions by 3 users not shown)
Line 10: Line 10:
 
Any term in the sequence <math>S_k</math> can be written as 1+kx.  If this is to equal 2005, then the remainder when 2005 is divided by k is 1.
 
Any term in the sequence <math>S_k</math> 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 <math>(2 + 1)(1 + 1)(1 + 1) = 12</math> divisors of <math>2^2\cdot 3^1\cdot 167^1</math>.
+
Now all we have to do is find the numbers of factors of 2004.  There are <math>(2 + 1)(1 + 1)(1 + 1) = \boxed{012}</math> divisors of <math>2^2\cdot 3^1\cdot 167^1</math>.
  
 
Note that although the remainder when 2005 divided by 1 is not 1, it still works- <math>S_1</math> would be the sequence of all positive integers, in which 2005 must appear.
 
Note that although the remainder when 2005 divided by 1 is not 1, it still works- <math>S_1</math> 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 ==
 
== See also ==

Latest revision as of 00:31, 5 December 2022

Problem

For each positive integer $k$, let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$. For example, $S_3$ is the sequence $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$?

Solution

Suppose that the $n$th term of the sequence $S_k$ is $2005$. Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\cdot 3\cdot 167$. The ordered pairs $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$,$(2,1002)$, $(3,668)$, $(4,501)$, $(6,334)$, $(12,167)$, $(167,12)$,$(334,6)$, $(501,4)$, $(668,3)$, $(1002,2)$ and $(2004,1)$, and each of these gives a possible value of $k$. Thus the requested number of values is $12$, and the answer is $\boxed{012}$.

Alternatively, notice that the formula for the number of divisors states that there are $(2 + 1)(1 + 1)(1 + 1) = 12$ divisors of $2^2\cdot 3^1\cdot 167^1$.

Solution 2

Any term in the sequence $S_k$ 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 $(2 + 1)(1 + 1)(1 + 1) = \boxed{012}$ divisors of $2^2\cdot 3^1\cdot 167^1$.

Note that although the remainder when 2005 divided by 1 is not 1, it still works- $S_1$ 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
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_2&oldid=183584"