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==Solution 4 (Logic)== | ==Solution 4 (Logic)== |
Revision as of 17:57, 12 July 2020
Contents
Problem
In how many ways can be written as the sum of an increasing sequence of two or more consecutive positive integers?
Solution 1
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
Let and
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
.
Solution 4 (Logic)
The consecutive sums can be written as where is the number of terms in a sequence, and is the first term. Then, and . Evaluating the sum and rearranging yields .
The prime factorization of is . Then, , , and are also divisors. As odd divisors of , note that they all produce integer solutions to as . Only is not valid, as . Similarly, quickly notice that is a solution. Multiplying an odd divisor by always yields an integer solution (see below). As such, the even integer solutions are .
Note that the function is decreasing for increasing values of and that for . Thus, when is negative, is only slightly more than . Recall . Since the next highest solution, is twice , . Thus, the remaining solutions are when .
- This is proven by the fact that an odd number ( in this case) divided by an divisor always yields an odd number. Dividing this by produces a number in the form , where z is an integer. Since an even number multiplied by an odd number also yields an even number, similarly is in the form , producing an integer.
(Solution by BJHHar)
Solution 5
We're dealing with an increasing arithmetic progression of common difference 1. Let be the number of terms in a summation. Let be the first term in a summation. The sum of an arithmetic progression is the average of the first term and the last term multiplied by the number of terms. The problem tells us that the sum must be 345.
In order to satisfy the constraints of the problem, x and y must be positive integers. Maybe we can make this into a Diophantine thing! In fact, if we just factor out that ... voilà!
There are 16 possible factor pairs to try (for brevity, I will not enumerate them here). Notice that the expression in the right parenthesis is more than the expression in the parenthesis on the left. is at least 1. Thus, the expression in the right parenthesis will always be greater than the expression on the left. This eliminates 8 factor pairs. The problem also says the "increasing sequence" has to have "two or more" terms, so . This eliminates the factor pair . With brief testing, we find that the the other 7 factor pairs produce 7 viable ordered pairs. This means we have found ways to write 345 in the silly way outlined by the problem.
Solution 6
By the sum of an arithmetic sequence... this ultimately comes to .
Quick testing (would take you roughly a minute)
We see that the first 7 values of that work are
.
We see that each one of them works.
Hence, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.