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2000 CEMC Pascal Problems/Problem 10

Problem

The number of integers between $-\sqrt{8}$ and $\sqrt{32}$ is

$\text{ (A) }\ 5 \qquad\text{ (B) }\ 6 \qquad\text{ (C) }\ 7 \qquad\text{ (D) }\ 8 \qquad\text{ (E) }\ 19$

Solution

We know that $\sqrt{25} = 5$, $\sqrt{36} = 6$, $-\sqrt{9} = -3$, and $-\sqrt{4} = -2$.

This means that $-3 < -\sqrt{8} < -2$ and $5 < \sqrt{32} < 6$, so the range of integers between the two numbers goes from $-2$ to $5$ (inclusive).

This gives $5 - (-2) + 1 = \boxed {\textbf {(D) } 8}$ integers.

~anabel.disher

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