Difference between revisions of "2021 OIM Problems/Problem 3"
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Let <math>a_1, a_2, a_3, \cdots</math> be a sequence of positive integers and let <math>b_1, b_2, b_3, \cdots</math> be the sequence of real numbers given by | Let <math>a_1, a_2, a_3, \cdots</math> be a sequence of positive integers and let <math>b_1, b_2, b_3, \cdots</math> be the sequence of real numbers given by | ||
| − | < | + | <cmath>b_n=\frac{a_1a_2,\cdots a_n}{a_1+a_2+\cdots + a_n}, \text{ for } n \ge 1.</cmath> |
Show that if among every one million consecutive terms of the sequence <math>b_1, b_2, b_3, \cdots</math> there is at least one integer, then there is some <math>k</math> such that <math>b_k > 2021^{2021}</math>. | Show that if among every one million consecutive terms of the sequence <math>b_1, b_2, b_3, \cdots</math> there is at least one integer, then there is some <math>k</math> such that <math>b_k > 2021^{2021}</math>. | ||
Latest revision as of 02:57, 14 December 2023
Problem
Let 
be a sequence of positive integers and let 
be the sequence of real numbers given by
![\[b_n=\frac{a_1a_2,\cdots a_n}{a_1+a_2+\cdots + a_n}, \text{ for } n \ge 1.\]](http://latex.artofproblemsolving.com/c/e/8/ce84c9832d5d0d9aacdb6c835f0bcd63bc83d774.png)
Show that if among every one million consecutive terms of the sequence there is at least one integer, then there is some 
such that 
.
Solution
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