2008 iTest Problems/Problem 68
Problem
Let be the term of the sequence
where the first term is the smallest positive integer that is more than a multiple of , the next two terms are the next two smallest positive integers that are each two more than a multiple of , the next three terms are the next three smallest positive integers that are each three more than a multiple of , the next four terms are the next four smallest positive integers that are each four more than a multiple of , and so on:
Determine .
Solution
First, observe that the difference between consecutive terms within a grouping will always equal Second, since all terms in a group with terms are congruent to modulo and all terms in a group with terms are congruent to modulo the difference between the first term of the group with terms and the last term of the group with terms is This means that the difference between the last terms of a grouping have the same second difference, so the series of numbers can be modeled by a quadratic function.
Let be the number of terms in a group, and let be the last term in a group with terms. We can write a system of equations to find a quadratic function.
Solving the system yields making the function
Note that the last term of the group with terms is term in the sequence. The largest such that is and Since the term of the sequence is This means the term is and with some basic algebra (or skip counting), the term is
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 67 |
Followed by: Problem 69 | |
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