Difference between revisions of "2006 AMC 12A Problems/Problem 8"

Revision as of 23:29, 17 December 2024

The following problem is from both the 2006 AMC 12A #8 and 2008 AMC 10A #9, so both problems redirect to this page.

Problem

How many sets of two or more consecutive positive integers have a sum of $15$ ?

$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

Solution 1

Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$ , and $1$ is clearly not possible. The other two possibilities both work:

$1 + 2 + 3 + 4 + 5 = 15$
$4 + 5 + 6 = 15$

If the number of integers in the list is even, then the average will have a $\frac{1}{2}$ . The only possibility is $\frac{15}{2}$ , from which we get:

$15 = 7 + 8$

Thus, the correct answer is $\boxed{\textbf{(C) }3}.$

Question: (RealityWrites) Is it possible that the answer is $4$ , because $0+1+2+3+4+5$ should technically count, right?

Answer: (IMGROOT2) It isn't possible because the question asks for positive integers, and this means that negative integers or zero aren't allowed.

Note to readers: make sure to always read the problem VERY carefully before attempting; it could mean the difference of making the cutoff.

Solution 2

Any set will form a arithmetic progression with the first term say $a$ . Since the numbers are consecutive the common difference $d = 1$ .

The sum of the AP has to be 15. So,

$S_n = \frac{n}{2} \cdot (2a + (n-1)d)$ $S_n = \frac{n}{2} \cdot (2a + (n-1)1)$ $15 = \frac{n}{2} \cdot (2a + n - 1)$ $2an + n^2 - n = 30$ $n^2 + n(2a - 1) - 30 = 0$

Now we need to find the number of possible sets of values of a, n which satisfy this equation. Now $a$ cannot be 15 as we need 2 terms. So a can only be less the 15.

Trying all the values of a from 1 to 14 we observe that $a = 4$ , $a = 7$ and $a = 1$ provide the only real solutions to the above equation.The three possibilites of a and n are.

$(a,n) = (4,3),(7, 2),(1, 6)$

The above values are obtained by solving the following equations obtained by substituting the above mentioned values of a into $n^2 + n(2a - 1) - 30 = 0$ ,

$n^2 + 7n - 30 = 0$ $n^2 + 13n - 30 = 0$ $n^2 - n - 30 = 0$

Since there are 3 possibilities the answer is $\boxed{\textbf{(C) }3}.$

Solution 3

!!! I might have a solution but am not sure if this is a correct way to solve it. !!!

We want some consecutive numbers that add up. We can use casework to solve this. We know that any set of 6 consecutive, positive integers cannot be greater than 5. Therefore the max amount of elements is 5.

Case 1: There are 5 numbers

-We know that 15/5 = 3, so if we put 3 as our middle number and expand evenly, we get {1,2,3,4,5}

Case 2: There are 4 numbers

-By a bit of bashing we find that there are no solutions

Case 3: There are 3 numbers

-We know that 15/3 = 5, so if we put 5 as our middle number and expand evenly, we get {4,5,6}

Case 4: There are 2 numbers

-We'll say x is the lower number. Thus x + (x + 1) = 15. Solving for x, x = 7. Thus getting {7,8}

Adding up all the cases we get 1 + 0 + 1 + 1 = 3

@@ Line 49: / Line 49: @@
 Since there are 3 possibilities the answer is <math>\boxed{\textbf{(C) }3}.</math>
 ==Solution 3==
-!!!I might have a solution but am not sure if this is a correct way to solve it. !!!
+!!! I might have a solution but am not sure if this is a correct way to solve it. !!!
 We want some consecutive numbers that add up. We can use casework to solve this. We know that any set of 6 consecutive, positive integers cannot be greater than 5. Therefore the max amount of elements is 5.

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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