Difference between revisions of "1985 AHSME Problems/Problem 27"
m (Fixed some typos) |
|||
(4 intermediate revisions by 3 users not shown) | |||
Line 4: | Line 4: | ||
<math> x_1=\sqrt[3]{3} </math> | <math> x_1=\sqrt[3]{3} </math> | ||
− | <math> x_2=\sqrt[3]{3}^\sqrt[3]{3} </math> | + | <math> x_2=(\sqrt[3]{3})^{\sqrt[3]{3}} </math> |
and in general | and in general | ||
− | <math> x_n=(x_{n-1})^\sqrt[3]{3} </math> for <math> n>1 </math>. | + | <math> x_n=(x_{n-1})^{\sqrt[3]{3}} </math> for <math> n>1 </math>. |
What is the smallest value of <math> n </math> for which <math> x_n </math> is an [[integer]]? | What is the smallest value of <math> n </math> for which <math> x_n </math> is an [[integer]]? | ||
Line 21: | Line 21: | ||
Now, we have <math> x_n=(x_{n-1})^{\sqrt[3]{3}}=\left(\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-2}\right)}\right)^{\sqrt[3]{3}}=\sqrt[3]{3}^{\left(\sqrt[3]{3}\cdot\sqrt[3]{3}^{n-2}\right)}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)} </math>. Thus the induction is complete. | Now, we have <math> x_n=(x_{n-1})^{\sqrt[3]{3}}=\left(\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-2}\right)}\right)^{\sqrt[3]{3}}=\sqrt[3]{3}^{\left(\sqrt[3]{3}\cdot\sqrt[3]{3}^{n-2}\right)}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)} </math>. Thus the induction is complete. | ||
− | We now get rid of the | + | We now get rid of the cube roots by introducing fractions into the exponents. |
<math> x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}=\sqrt[3]{3}^{\left(3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(\frac{1}{3}\cdot3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)} </math>. | <math> x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}=\sqrt[3]{3}^{\left(3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(\frac{1}{3}\cdot3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)} </math>. | ||
− | Notice that since <math> 3 </math> isn't a perfect power, <math> x_n </math> is integral if and only if the exponent, <math> 3^{\left(\frac{n-4}{3}\right)} </math>, is integral. By the same logic, this is | + | Notice that since <math> 3 </math> isn't a perfect power, <math> x_n </math> is integral if and only if the exponent, <math> 3^{\left(\frac{n-4}{3}\right)} </math>, is integral. By the same logic, this is integral if and only if <math> \frac{n-4}{3} </math> is integral. We can now clearly see that the smallest positive value of <math> n </math> for which this is integral is <math> 4, \boxed{\text{C}} </math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=26|num-a=28}} | {{AHSME box|year=1985|num-b=26|num-a=28}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:14, 3 April 2018
Problem
Consider a sequence defined by
and in general
for .
What is the smallest value of for which is an integer?
Solution
First, we will use induction to prove that
We see that . This is our base case.
Now, we have . Thus the induction is complete.
We now get rid of the cube roots by introducing fractions into the exponents.
.
Notice that since isn't a perfect power, is integral if and only if the exponent, , is integral. By the same logic, this is integral if and only if is integral. We can now clearly see that the smallest positive value of for which this is integral is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.