Difference between revisions of "1989 AJHSME Problems/Problem 18"
5849206328x (talk | contribs) (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...) |
Duoduoling0 (talk | contribs) (→Solution 2) |
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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 (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 <math>2 = \boxed{\text{B}}.</math> | ||
+ | |||
+ | ~DuoDuoling0 | ||
==See Also== | ==See Also== |
Latest revision as of 13:25, 28 December 2021
Problem
Many calculators have a reciprocal key that replaces the current number displayed with its reciprocal. For example, if the display is
and the
key is depressed, then the display becomes
. If
is currently displayed, what is the fewest number of times you must depress the
key so the display again reads
?
Solution
Let . We have
Thus, we need to iterate the key pressing twice to get the display back to the original
.
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
~DuoDuoling0
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 |