Difference between revisions of "2006 AMC 12A Problems/Problem 8"
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<cmath>15 = \frac{n}{2} \cdot (2a + n - 1)</cmath> | <cmath>15 = \frac{n}{2} \cdot (2a + n - 1)</cmath> | ||
<cmath>2an + n^2 - n = 30</cmath> | <cmath>2an + n^2 - n = 30</cmath> | ||
+ | <cmath>n^2 + n(2a - 1) - 30 = 0</cmath> | ||
+ | |||
+ | Now we need to find the number of possible sets of values of a, n which satisfy this equation. Now <math>a</math> cannot be 15 as we need 2 terms. So a can only be less the 15. | ||
+ | |||
+ | Trying all the values of a from 1 to 14 we observe that <math>a = 4</math> is the only option which gives a real solution. | ||
+ | Putting <math>a = 4</math> the equation becomes | ||
+ | |||
+ | <cmath>n^2 + 7n - 30 = 0</cmath> | ||
+ | <cmath>n^2 + 10n - 3n - 30 = 0</cmath> | ||
+ | <cmath>n(n+10)-3(n+10) = 0</cmath> | ||
+ | <cmath>(n-3)(n+10) = 0</cmath> | ||
== See also == | == See also == |
Revision as of 01:44, 13 January 2024
- The following problem is from both the 2006 AMC 12A #8 and 2008 AMC 10A #9, so both problems redirect to this page.
Contents
Problem
How many sets of two or more consecutive positive integers have a sum of ?
Solution 1
Notice that if the consecutive positive integers have a sum of , then their average (which could be a fraction) must be a divisor of
. If the number of integers in the list is odd, then the average must be either
or
, and
is clearly not possible. The other two possibilities both work:
If the number of integers in the list is even, then the average will have a . The only possibility is
, from which we get:
Thus, the correct answer is
Question: (RealityWrites) Is it possible that the answer is , because
should technically count, right?
Answer: (IMGROOT2) It isn't possible because the question asks for positive integers, and this means that negative integers or zero aren't allowed.
Note to readers: make sure to always read the problem VERY carefully before attempting; it could mean the difference of making the cutoff.
Solution 2
Any set will form a arithmetic progression with the first term say . Since the numbers are consecutive the common difference
.
The sum of the AP has to be 15. So,
Now we need to find the number of possible sets of values of a, n which satisfy this equation. Now cannot be 15 as we need 2 terms. So a can only be less the 15.
Trying all the values of a from 1 to 14 we observe that is the only option which gives a real solution.
Putting
the equation becomes
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 |
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.