Difference between revisions of "2006 AIME II Problems/Problem 13"
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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
[hide]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.