Difference between revisions of "2005 AIME I Problems/Problem 2"
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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. | + | 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 , let denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is . For example, is the sequence For how many values of does contain the term 2005?
Solution
Suppose that the th term of the sequence is 2005. 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 12, and the answer is 012.
Alternatively, notice that the formula for the number of divisors states that there are divisors of .
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 |