Difference between revisions of "1986 AJHSME Problems/Problem 7"

Revision as of 20:11, 3 July 2013

Problem

How many whole numbers are between $\sqrt{8}$ and $\sqrt{80}$ ?

$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution

No... of course you're not supposed to know what the square root of 8 is, or the square root of 80. There aren't any formulas, either. Approximation seems like the best strategy.

Clearly it must be true that for any positive integers $a$ , $b$ , and $c$ with $a>b>c$ , $\sqrt{a}>\sqrt{b}>\sqrt{c}$

If we let $a=9$ , $b=8$ , and $c=4$ , then we get $\sqrt{9}>\sqrt{8}>\sqrt{4}$ $3>\sqrt{8}>2$

Therefore, the smallest whole number between $\sqrt{8}$ and $\sqrt{80}$ is $3$ .

Similarly, if we let $a=81$ , $b=80$ , and $c=64$ , we get $\sqrt{81}>\sqrt{80}>\sqrt{64}$ $9>\sqrt{80}>8$

So $8$ is the largest whole number between $\sqrt{8}$ and $\sqrt{80}$ .

So we know that we just have to find the number of integers from 3 to 8 inclusive. If we subtract 2 from every number in this set (which doesn't change the number of integers in the set at all), we find that now all we need to do is find the number of integers there are from 1 to 6, which is obviously 6.

$\boxed{\text{B}}$

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 [[Category:Introductory Algebra Problems]]
 [[Category:Introductory Combinatorics Problems]]
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Difference between revisions of "1986 AJHSME Problems/Problem 7"

Revision as of 20:11, 3 July 2013

Problem

Solution

See Also