Difference between revisions of "2005 AIME I Problems/Problem 2"
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== Problem == | == Problem == | ||
− | For each positive integer <math> k | + | 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 == | ||
− | Suppose that the | + | 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>. |
+ | |||
+ | 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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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>. | ||
+ | |||
+ | 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 == | ||
− | + | {{AIME box|year=2005|n=I|num-b=1|num-a=3}} | |
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:07, 13 June 2020
Contents
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.
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.