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Difference between revisions of "1997 CEMC Pascal Problems/Problem 10"

(Created page with "==Problem== The sum of nine consecutive positive integers is <math>99</math>. The largest of these integers is <math> \text{ (A) }\ 9 \qquad\text{ (B) }\ 11 \qquad\text{ (C)...")
 
 
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<math>n = 7</math>
 
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The problem asks us for the highest number in the list, which is <math>n + 8 = 7 + 8 = 15</math>.
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The problem asks us for the highest number in the list, which is <math>n + 8 = 7 + 8 = \boxed {\textbf {(E) } 15}</math>.
  
 
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<math>9n = 135</math>
 
<math>9n = 135</math>
  
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<math>n = 15</math>
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<math>n = \boxed {\textbf {(E) } 15}</math>
  
 
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<math>n = 11</math>
 
<math>n = 11</math>
  
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The largest term in the list is <math>n + 4 = 11 + 4 = 15</math>.
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The largest term in the list is <math>n + 4 = 11 + 4 = \boxed {\textbf {(E) } 15}</math>.
  
 
~anabel.disher
 
~anabel.disher

Latest revision as of 14:06, 22 June 2025

Problem

The sum of nine consecutive positive integers is $99$. The largest of these integers is

$\text{ (A) }\ 9 \qquad\text{ (B) }\ 11 \qquad\text{ (C) }\ 19 \qquad\text{ (D) }\ 7 \qquad\text{ (E) }\ 15$

Solution 1

Let $n$ be the starting number. The full list of numbers is then:

$n, n + 1, n + 2, n + 3, n + 4, n + 5, n + 6, n + 7, n + 8$

We can now set up an equation involving the sum:

$n + n + 1 + n + 2 + n + 3 + n + 4 + n + 5 + n + 6 + n + 7 + n + 8 = 99$

Using the fact that $0 + 1 + 2 + 3 + ... + x = \frac{x(x + 1)}{2}$ for some non-negative integer $x$, and the fact that $n$ appears $9$ times, we have:

$9n + \frac{8(8 + 1)}{2} = 99$

$9n + \frac{8 \times 9}{2} = 99$

$9n + \frac{72}{2} = 99$

$9n + 36 = 99$

We can either subtract $36$ from both sides or factor out a $9$: $9n = 99 - 36 = 63$

$n = 7$

or

$9(n + 4) = 99$

$n + 4 = 11$

$n = 7$

The problem asks us for the highest number in the list, which is $n + 8 = 7 + 8 = \boxed {\textbf {(E) } 15}$.

~anabel.disher

Solution 2

Let $n$ be the highest number in the list of consecutive integers. The full list is then:

$n - 8, n - 7, n - 6, n - 5, n - 4, n - 3, n - 2, n - 1, n$

Using a similar process to solution 1, we get:

$9n - 36 = 99$

$9n = 135$

$n = \boxed {\textbf {(E) } 15}$

~anabel.disher

Solution 3

Let $n$ be the middle term in the list. The full list is then:

$n - 4, n - 3, n - 2, n - 1, n, n + 1, n + 2, n + 3, n + 4$

We can then write an equation for the sum:

$n - 4 + n - 3 + n - 2 + n - 1 + n + n + 1 + n + 2 + n + 3 + n + 4 = 99$

We can see that all of the constant terms cancel out, and $n$ appears $9$ times, so we have:

$9n = 99$

$n = 11$

The largest term in the list is $n + 4 = 11 + 4 = \boxed {\textbf {(E) } 15}$.

~anabel.disher

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