Difference between revisions of "2006 AIME II Problems/Problem 13"

Revision as of 09:40, 14 July 2022

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 $2n+1$ , $n\geq 0$ . Then the final odd integer is $2n+1 + 2(j-1) = 2(n+j) - 1$ . The odd integers form an arithmetic sequence with sum $N = j\left(\frac{(2n+1) + (2(n+j)-1)}{2}\right) = j(2n+j)$ . Thus, $j$ 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 of $N$ . 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

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

Those give $5$ possibilities for odd $N$ .

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

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

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 $250$ :

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

Those give $10$ possibilities for even $N$ .

The total number of integers $N$ is $5 + 10 = \boxed{15}$ .

Solution 2

Let the largest odd number below the sequence be the $q$ th positive odd number, and the largest odd number in the sequence be the $p$ th positive odd number. Therefore, the sum is $p^2-q^2=(p+q)(p-q)$ by sum of consecutive odd numbers. Note that $p+q$ and $p-q$ have the same parity, and $q$ can equal $0$ . We then perform casework based on the parity of $p-q$ .

If $p-q$ is odd, then $p^2-q^2$ must always be odd. Therefore, to have 5 pairs of odd factors, we must have either $9$ (in which case the number is a perfect square) or $10$ factors. Considering the upper bound, the only way this can happen is $p_1^4\cdot{p_2}$ or $p_1^2\cdot{p_2^2}$ . N must then be one of $(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)$ So, there are $5$ solutions when $(p+q)(p-q)$ is odd.

If $p-q$ is even, then $(p+q)(p-q)$ must have at least two factors of $2$ , so we can rewrite the expression as $4(k)(k-q)$ where $k=\frac{p+q}{2}$ . We can disregard the $4$ by dividing by $4$ and restricting our upper bound to $250$ . Since $k$ and $q$ 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 $(2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)$ as the possibilities.

Therefore, there are $10+5=\boxed{015}$ possibilities for $p^2-q^2=N$ ~sigma

@@ Line 23: / Line 23: @@
 :Those give <math>10</math> possibilities for even <math>N</math>.
-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{15}</math>.
+== Solution 2==
+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
+<cmath>(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)</cmath>
+So, there are <math>5</math> solutions when <math>(p+q)(p-q)</math> is odd.
+If <math>p-q</math> is even, then <math>(p+q)(p-q)</math> must have at least two factors of <math>2</math>, so we can rewrite the expression as <math>4(k)(k-q)</math> where <math>k=\frac{p+q}{2}</math>. We can disregard the <math>4</math> by dividing by <math>4</math> and restricting our upper bound to <math>250</math>. Since <math>k</math> and <math>q</math> 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
+<cmath>(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), (3^4\cdot2)</cmath>
+as the possibilities.
+Therefore, there are <math>10+5=\boxed{015}</math> possibilities for <math>p^2-q^2=N</math> ~sigma
 == See also ==
 {{AIME box|year=2006|n=II|num-b=12|num-a=14}}

2006 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2006 AIME II Problems/Problem 13"

Revision as of 09:40, 14 July 2022

Contents

Problem

Solution

Solution 2

See also