Difference between revisions of "2016 AMC 12B Problems/Problem 16"
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:59, 22 February 2016
Contents
Problem
In how many ways can be written as the sum of an increasing sequence of two or more consecutive positive integers?
Solution
We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.
For the first case, we can cleverly choose the convenient form of our sequence to be
because then our sum will just be . We now have
and
will have a solution when
is an integer, namely when
is a divisor of 345. We check that
work, and no more, because
does not satisfy the requirements of two or more consecutive integers, and when
equals the next biggest factor,
, there must be negative integers in the sequence. Our solutions are
.
For the even cases, we choose our sequence to be of the form:
so the sum is
. In this case, we find our solutions to be
.
We have found all 7 solutions and our answer is .
Solution 2
The sum from to
where
and
are integers and
is
If we factor into all of its factor groups
we will have several ordered pairs
where
.
The number of possible values for is half the number of factors of
which is
However, we have one extraneous case of because here,
and we have the sum of one consecutive number which is not allowed by the question.
Thus the answer is
.
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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