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

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== Problem ==
 
== Problem ==
For each positive integer <math> k, </math> let <math> S_k </math> denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is <math> k. </math> For example, <math> S_3 </math> is the sequence <math> 1,4,7,10,\ldots. </math> For how many values of <math> k </math> does <math> S_k </math> contain the term 2005?
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For each [[positive integer]] <math>k</math>, let <math> S_k </math> denote the [[increasing sequence | increasing]] [[arithmetic sequence]] of [[integer]]s whose first term is <math>1</math> and whose common difference is <math> k</math>For example, <math> S_3 </math> is the [[sequence]] <math> 1,4,7,10,\ldots. </math> For how many values of <math> k </math> does <math> S_k </math> contain the term <math>2005</math>?
  
 
== Solution ==
 
== Solution ==
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Suppose that the <math>n</math>th term of the sequence <math>S_k</math> is <math>2005</math>. Then <math>1+(n-1)k=2005</math> so <math>k(n-1)=2004=2^2\cdot 3\cdot 167</math>.  The [[ordered pair]]s <math>(k,n-1)</math> of positive integers that satisfy the last equation are <math>(1,2004)</math>,<math>(2,1002)</math>, <math>(3,668)</math>, <math>(4,501)</math>, <math>(6,334)</math>, <math>(12,167)</math>, <math>(167,12)</math>,<math>(334,6)</math>, <math>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math> and <math>(2004,1)</math>, and each of these gives a possible value of <math>k</math>. Thus the requested number of values is <math>12</math>, and the answer is <math>\boxed{12}</math>.
  
{{solution}}
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Alternatively, notice that the formula for the number of [[divisor]]s states that there are <math>(2 + 1)(1 + 1)(1 + 1) = 12</math> divisors of <math>2^2\cdot 3^1\cdot 167^1</math>.
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== Solution 2 ==
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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.
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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>.
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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.
  
 
== See also ==
 
== See also ==
* [[2005 AIME I Problems]]
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{{AIME box|year=2005|n=I|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 22:07, 13 June 2020

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 $\boxed{12}$.

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 $(2 + 1)(1 + 1)(1 + 1) = 12$ 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.

See also

2005 AIME I (ProblemsAnswer KeyResources)
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

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