2006 AIME II Problems/Problem 13

Problem

How many integers $N$ less than 1000 can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$ ?

Solution

Let the first odd integer be $2n+1$ with $n\geq 0$ .

The final odd integer is $2n+1 + 2(j-1) = 2(n+j) - 1$

The odd integers form an arithmetic sequence with sum $j\left(\frac{(2n+1) + (2(n+j)-1)}{2}\right) = j(2n+j)$

so $N = j(2n+j)$ . $j$ is a factor of $N$ .

Since $n\geq 0$ , it follows that $2n+j \geq j$ and $j\leq \sqrt{N}$

Since there are exactly $5$ values of $j$ that satisfy the equation, there must be either $9$ or $10$ factors to $N$ . This means $N=p_1^2p_2^2$ or $N=p_1p_2^4$ . Unfortunately, we cannot simply observe prime factorizations of $N$ because the factor $(2n+j)$ does not cover all integers for any given value of $j$

Instead we do some casework:

If $N$ is odd, then $j$ must also be odd. For every odd value of $j$ , $2n+j$ is also odd, making this case valid for all odd $j$ . Looking at the forms above and the bound of 1000, N must be

$(3^2\cdot5^2)$ , $(3^2\cdot7^2)$ , $(3^4\cdot5)$ , $(3^4\cdot7)$ or $(3^4\cdot 11)$

Those give $5$ possibilities for odd $N$

If $N$ is even, then $j$ must also be even. Substituting $j=2k$ , we get

$N = 4k(n+k)$

$\frac{N}{4} = k(n+k)$

Now we can just look at all the prime factorizations since $(n+k)$ cover the integers for any $k$ . Note that our upper bound is now 250

$\frac{N}{4} = (2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13),$ or $(3^4\cdot2)$

Those give $10$ possibilities for even $N$

The total number of integers $N$ is $5 + 10 = \boxed{015}$

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
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2006 AIME II Problems/Problem 13

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