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

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== Solution ==
 
== Solution ==
Suppose that the <math>n</math>th term of the sequence <math>S_k</math> is 2005. 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 12.
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Suppose that the <math>n</math>th term of the sequence <math>S_k</math> is 2005. 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 12, and the answer is 012.
  
 
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>2^23^1167^1</math>.
 
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>2^23^1167^1</math>.

Revision as of 02:38, 8 March 2007

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 012.

Alternatively, notice that the formula for the number of divisors states that there are $(2 + 1)(1 + 1)(1 + 1) = 12$ divisors of $2^23^1167^1$.

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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