1989 AJHSME Problems/Problem 18

Problem

Many calculators have a reciprocal key $\boxed{\frac{1}{x}}$ that replaces the current number displayed with its reciprocal. For example, if the display is $\boxed{00004}$ and the $\boxed{\frac{1}{x}}$ key is depressed, then the display becomes $\boxed{000.25}$ . If $\boxed{00032}$ is currently displayed, what is the fewest number of times you must depress the $\boxed{\frac{1}{x}}$ key so the display again reads $\boxed{00032}$ ?

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution

Let $f(x)=\frac{1}{x}$ . We have $f(f(x))=\frac{1}{\frac{1}{x}}=x$ Thus, we need to iterate the key pressing twice to get the display back to the original $\rightarrow \boxed{\text{B}}$ .

Solution 2 (simpler, no use of functions)

In terms of mathematics, a number will always be changed back to its original number if you flip, or "reciprocal" it, twice. Therefore, no matter what number it will display, the answer will always be $2 = \boxed{\text{B}}.$

~DuoDuoling0

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1989 AJHSME Problems/Problem 18

Contents

Problem

Solution

Solution 2 (simpler, no use of functions)

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