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For a given real number $r$ let $[r]$ denote the largest integer less than or equal to $r$ .Let $a>1$ be a real number which is not an integer,and let$k$ be the smallest possible positive integer such that $[a^k]>[a]^k$.then which of the following statement is always true----

A) $k \leq 2([a]+1)^2$

B) $k \leq ([a]+1)^4$

C) $k \leq 2^{([a]+1)}$

D) $k \leq \frac{1}{(a-[a])}+1$

I tried hard but couldn’t find out a purely algebraic way of solving the problem. Please help me in this regard. Thanks.

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Say $a = [a]+x, 0<x<1$. Then, $[a]^{k-1}=[a^{k-1}] = [([a]+x)^{k-1}] = [[a]^{k-1}+[a]^{k-2}x\binom{k-1}{1}+...+ x^{k-1}]$. In particular, $x(k-1)<x[a]^{k-2}(k-1)<1.$ Therefore, $k<1+\dfrac{1}{x} = 1+\dfrac{1}{(a - [a])}.$

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  • $\begingroup$ thanks for your answer.I think the equality sign in second line must be replaced by < sign.however I could not understand how to arrived at $x(k-1)<x[a]^{k-2}(k-1)<1$.please can you explain a little bit more. Thanks. $\endgroup$ – Navin Oct 29 '16 at 18:56
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    $\begingroup$ it's because $[a]$ >1. Also, the entire binomial expansion sum must be less than 1, so clearly just one of the term (namely the second term) is less than 1. $\endgroup$ – dezdichado Oct 30 '16 at 0:24

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