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Difference between revisions of "1989 AJHSME Problems/Problem 18"

(New page: ==Problem== Many calculators have a reciprocal key <math>\boxed{\frac{1}{x}}</math> that replaces the current number displayed with its reciprocal. For example, if the display is <ma...)
 
(→‎Solution)
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<cmath>f(f(x))=\frac{1}{\frac{1}{x}}=x</cmath>
 
<cmath>f(f(x))=\frac{1}{\frac{1}{x}}=x</cmath>
 
Thus, we need to iterate the key pressing twice to get the display back to the original <math>\rightarrow \boxed{\text{B}}</math>.
 
Thus, we need to iterate the key pressing twice to get the display back to the original <math>\rightarrow \boxed{\text{B}}</math>.
 +
 +
==Solution 2==
 +
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 <math>2 = \boxed{\text{B}}.</math>
  
 
==See Also==
 
==See Also==

Revision as of 13:24, 28 December 2021

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

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 $2 = \boxed{\text{B}}.$

See Also

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