Difference between revisions of "1985 AHSME Problems/Problem 27"
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==Problem== | ==Problem== | ||
− | Consider a sequence <math> x_1, x_2, x_3, \ | + | Consider a sequence <math>x_1,x_2,x_3,\dotsc</math> defined by: |
− | + | <cmath>\begin{align*}&x_1 = \sqrt[3]{3}, \\ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*}</cmath> | |
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and in general | and in general | ||
+ | <cmath>x_n = \left(x_{n-1}\right)^{\sqrt[3]{3}} \text{ for } n > 1.</cmath> | ||
− | + | What is the smallest value of <math>n</math> for which <math>x_n</math> is an integer? | |
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− | What is the smallest value of <math> n </math> for which <math> x_n </math> is an | ||
<math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }27 </math> | <math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }27 </math> | ||
==Solution== | ==Solution== | ||
− | + | Firstly, we will show by induction that <cmath>x_n = \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}.</cmath> For the base case, we indeed have <cmath>\begin{align*}x_1 &= \sqrt[3]{3} \\ &= \left(\sqrt[3]{3}\right)^1 \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^0\right)},\end{align*}</cmath> and for the inductive step, if our claim is true for <math>x_n</math>, then <cmath>\begin{align*}x_{n+1} &= \left(x_n\right)^{\sqrt[3]{3}} \\ &= \left(\left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}\right)^{\sqrt[3]{3}} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\cdot\sqrt[3]{3}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^n\right)},\end{align*}</cmath> which completes the proof. | |
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− | + | We now rewrite our formula for <math>x_n</math> as follows: | |
+ | <cmath>\begin{align*}x_n &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(3^{\frac{n-1}{3}}\right)} \\ &=3^{\left(\frac{1}{3} \cdot 3^{\frac{n-1}{3}}\right)} \\ &= 3^{\left(3^{\left(\frac{n-1}{3}-1\right)}\right)} \\ &= 3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)},\end{align*}</cmath> | ||
+ | and as <math>3</math> is not a perfect power, we deduce that <math>x_n</math> is an integer if and only if the exponent, <math>3^{\left(\frac{n-4}{3}\right)}</math>, is itself an integer. By precisely the same argument, this reduces to <math>\frac{n-4}{3}</math> being an integer, so the smallest possible (positive) value of <math>n</math> is <math>\boxed{\text{(C)} \ 4}</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 02:06, 20 March 2024
Problem
Consider a sequence defined by:
and in general
What is the smallest value of for which
is an integer?
Solution
Firstly, we will show by induction that For the base case, we indeed have
and for the inductive step, if our claim is true for
, then
which completes the proof.
We now rewrite our formula for as follows:
and as
is not a perfect power, we deduce that
is an integer if and only if the exponent,
, is itself an integer. By precisely the same argument, this reduces to
being an integer, so the smallest possible (positive) value of
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.