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Difference between revisions of "2006 AMC 12A Problems/Problem 8"
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #8]] and [[2006 AMC 10A Problems/Problem 9|2008 AMC 10A #9]]}}
 
== Problem ==
 
== Problem ==
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How many [[set]]s of two or more consecutive positive integers have a sum of <math>15</math>?
  
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How many sets of two or more consecutive positive integers have a sum of <math>15</math>?
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<math> \textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) }  5</math>
  
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<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ }  5</math>
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== Solution 1==
 +
Notice that if the consecutive positive integers have a sum of <math>15</math>, then their average (which could be a fraction) must be a divisor of <math>15</math>. If the number of integers in the list is odd, then the average must be either <math>1, 3, </math> or <math>5</math>, and <math>1</math> is clearly not possible. The other two possibilities both work:
  
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== Solution ==
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*<math>1 + 2 + 3 + 4 + 5 = 15</math>
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*<math>4 + 5 + 6 = 15</math>
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If the number of integers in the list is even, then the average will have a <math>\frac{1}{2}</math>. The only possibility is <math>\frac{15}{2}</math>, from which we get:
 +
 
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*<math>15 = 7 + 8</math>
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Thus, the correct answer is <math>\boxed{\textbf{(C) }3}.</math>
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 +
 
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Question: (RealityWrites) Is it possible that the answer is <math>4</math>, because <math>0+1+2+3+4+5</math> should technically count, right?
 +
 
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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 <math>a</math>. Since the numbers are consecutive the common difference <math>d = 1</math>.
 +
 
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The sum of the AP has to be 15. So,
 +
 
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<cmath>S_n = \frac{n}{2} \cdot (2a + (n-1)d)</cmath>
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<cmath>S_n = \frac{n}{2} \cdot (2a + (n-1)1)</cmath>
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<cmath>15 = \frac{n}{2} \cdot (2a + n - 1)</cmath>
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<cmath>2an + n^2 - n = 30</cmath>
 +
<cmath>n^2 + n(2a - 1) - 30 = 0</cmath>
 +
 
 +
Now we need to find the number of possible sets of values of a, n which satisfy this equation. Now <math>a</math> 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 <math>a = 4</math>, <math>a = 7</math> and <math>a = 1</math> provide the only real solutions to the above equation.The three possibilites of a and n are.
 +
 
 +
<cmath>(a,n) = (4,3),(7, 2),(1, 6)</cmath>
 +
 
 +
The above values are obtained by solving the following equations obtained by substituting the above mentioned values of a into <math>n^2 + n(2a - 1) - 30 = 0</math>,
 +
 
 +
<cmath>n^2 + 7n - 30 = 0</cmath>
 +
<cmath>n^2 + 13n - 30 = 0</cmath>
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<cmath>n^2 - n - 30 = 0</cmath>
 +
 
 +
Since there are 3 possibilities the answer is <math>\boxed{\textbf{(C) }3}.</math>
  
 
== See also ==
 
== See also ==
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* [[2006 AMC 12A Problems]]
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{{AMC12 box|year=2006|ab=A|num-b=7|num-a=9}}
 +
{{AMC10 box|year=2006|ab=A|num-b=8|num-a=10}}
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{{MAA Notice}}
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 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 01:00, 13 January 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 $\boxed{\textbf{(C) }3}.$

See also

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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