1986 AJHSME Problems/Problem 7

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.

We know that $\sqrt 9$ is 3, and is the smallest square number above $\sqrt 8$ that will be an integer.

We also know that $\sqrt 4$ is 2, and is the largest square number lower than $\sqrt 8$ that will be an integer.

From this, we know that $\sqrt 8$ should be somewhere between 2 and 3. We can do the same thing for $\sqrt 80$ . $\sqrt 81$ is 9, and is the smallest square number above $\sqrt 80$ that is an integer, and that $\sqrt 64$ is 8, and is the largest square number under $\sqrt 8$ that is an integer.

So we know that we just have to find the number of integers from 3 to 8. 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.

6 is B

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1986 AJHSME Problems/Problem 7

Solution