Difference between revisions of "2006 AIME II Problems/Problem 13"
Line 24: | Line 24: | ||
The total number of integers <math>N</math> is <math>5 + 10 = \boxed{015}</math>. | The total number of integers <math>N</math> is <math>5 + 10 = \boxed{015}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | It is a well known fact that the sum of the first <math>n</math> odd positive integers is equal to <math>n^2</math>. Let <math>a^2</math> be the sum of the odds from 1 to the integer included in the sum, and let <math>b^2</math> be the sum of odds from 1 to the largest odd less than the interval of length j. (For example, if we start at 5 and have <math> j = 3</math>, <math>a^2 = 1 + 3 + 5 + 7 + 9</math> and <math>b^2 = 1 + 3</math>). Then <math>a^2 - b^2 = 1000</math>, or <math>(a+b)(a-b) = 1000 = 2^3\cdot 5^3</math>. There are <math>4 \cdot 4 = 16</math> factors of 1000, but we want all <math>N < 1000</math>, thus the answer is 15. | ||
== See also == | == See also == |
Revision as of 01:53, 29 February 2016
Contents
Problem
How many integers less than can be written as the sum of consecutive positive odd integers from exactly 5 values of ?
Solution
Let the first odd integer be , . Then the final odd integer is . The odd integers form an arithmetic sequence with sum . Thus, is a factor of .
Since , it follows that and .
Since there are exactly values of that satisfy the equation, there must be either or factors of . This means or . Unfortunately, we cannot simply observe prime factorizations of because the factor does not cover all integers for any given value of .
Instead we do some casework:
- If is odd, then must also be odd. For every odd value of , is also odd, making this case valid for all odd . Looking at the forms above and the bound of , must be
- Those give possibilities for odd .
- If is even, then must also be even. Substituting , we get
- Now we can just look at all the prime factorizations since cover the integers for any . Note that our upper bound is now :
- Those give possibilities for even .
The total number of integers is .
Solution 2
It is a well known fact that the sum of the first odd positive integers is equal to . Let be the sum of the odds from 1 to the integer included in the sum, and let be the sum of odds from 1 to the largest odd less than the interval of length j. (For example, if we start at 5 and have , and ). Then , or . There are factors of 1000, but we want all , thus the answer is 15.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.