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{15}</math>. | The total number of integers <math>N</math> is <math>5 + 10 = \boxed{15}</math>. | ||
== Solution 2== | == Solution 2== | ||
− | Let the largest odd number below the sequence be the <math> | + | Let the largest odd number below the sequence be the <math>q</math>th positive odd number, and the largest odd number in the sequence be the <math>p</math>th positive odd number. Therefore, the sum is <math>p^2-q^2=(p+q)(p-q)</math> by sum of consecutive odd numbers. Note that <math>p+q</math> and <math>p-q</math> have the same parity, and <math>q</math> can equal <math>0</math>. We then perform casework based on the parity of <math>p-q</math>. |
If <math>p-q</math> is odd, then <math>p^2-q^2</math> must always be odd. Therefore, to have 5 pairs of odd factors, we must have either <math>9</math> (in which case the number is a perfect square) or <math>10</math> factors. Considering the upper bound, the only way this can happen is <math>p_1^4\cdot{p_2}</math> or <math>p_1^2\cdot{p_2^2}</math>. N must then be one of | If <math>p-q</math> is odd, then <math>p^2-q^2</math> must always be odd. Therefore, to have 5 pairs of odd factors, we must have either <math>9</math> (in which case the number is a perfect square) or <math>10</math> factors. Considering the upper bound, the only way this can happen is <math>p_1^4\cdot{p_2}</math> or <math>p_1^2\cdot{p_2^2}</math>. N must then be one of |
Latest revision as of 10:40, 14 July 2022
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
Let the largest odd number below the sequence be the th positive odd number, and the largest odd number in the sequence be the
th positive odd number. Therefore, the sum is
by sum of consecutive odd numbers. Note that
and
have the same parity, and
can equal
. We then perform casework based on the parity of
.
If is odd, then
must always be odd. Therefore, to have 5 pairs of odd factors, we must have either
(in which case the number is a perfect square) or
factors. Considering the upper bound, the only way this can happen is
or
. N must then be one of
So, there are
solutions when
is odd.
If is even, then
must have at least two factors of
, so we can rewrite the expression as
where
. We can disregard the
by dividing by
and restricting our upper bound to
. Since
and
don't have to be the same parity, we can include all cases less than 250 that have 9 or 10 factors. We then have
as the possibilities.
Therefore, there are possibilities for
~sigma
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.