2003 CEMC Pascal Problems/Problem 16

Problem

If $n$ is any integer, $n + 3$ , $n - 9$ , $n - 4$ , $n + 6$ , and $n - 1$ are also integers. If $n + 3$ , $n - 9$ , $n - 4$ , $n + 6$ , and $n - 1$ are arranged from smallest to largest, the integer in the middle is

$\text{ (A) }\ n + 3 \qquad\text{ (B) }\ n - 9 \qquad\text{ (C) }\ n - 4 \qquad\text{ (D) }\ n + 6 \qquad\text{ (E) }\ n - 1$

Solution

We can see $n - 9 < n - 4 < n - 1 < n + 3 < n + 6$ from the fact that $-9 < -4 < -1 < 3 < 6$ .

The integer that is in the middle is $\boxed {\textbf {(E) } n - 1}$ .

~anabel.disher

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2003 CEMC Pascal Problems/Problem 16

Problem

Solution