1986 AJHSME Problems/Problem 7


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

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


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.


See Also

1986 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 AJHSME/AMC 8 Problems and Solutions

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