This appeared in the Area Level of the 19th Philippine Mathematical Olympiad. Electronic calculators were not allowed during the competition of course. The closest I got to was to express it as:  $$ \dfrac{1}{\sqrt[6]{2}(\sqrt[6]{7813}-\sqrt[3]{6}\sqrt[6]{217})}. $$  The answer using an electronic calculator is 9375. Can anyone show how this can be approximated without the use of electronic calculators?
123422
Find the integer which is closest to the value of   $\frac{1}{\sqrt[6]{5^6+1}- \sqrt[6]{5^6-1}}$  I have tried putting them between n and n+1 and tried manipulating  n but I cant seem to find a strategy.
106305
It is straight forward to show that $\sqrt[n]{n!} &lt; n$ from:   $2 &gt; \sqrt{2}$ $\frac{n+1}{n} &gt; \frac{\sqrt[n+1]{n+1}}{\sqrt[n]{n}}$ since $\frac{n+1}{\sqrt[n+1]{n+1}} &gt; \frac{n}{\sqrt[n]{n}}$    Is this the best that can be done?  Is it possible to find a better upper bound for $\sqrt[n]{n!}$
71721
Do you a digit efficient way to approximate $\pi$? I mean representing many digits of $\pi$ using only a few numeric digits and some sort of equation. Maybe mathematical operations also count as penalty.  For example the well known $\frac{355}{113}$ is an approximation, but it gives only 7 correct digits by using 6 digits (113355) in the approximation itself. Can you make a better digit ratio?  EDIT: to clarify the "game" let's assume that each mathematical operation (+, sqrt, power, ...) also counts as one digit. Otherwise one could of course make artifical infinitely nested structures of operations only. And preferably let's stick to basic arithmetics and powers/roots only.  EDIT: true. logarithm of imaginary numbers provides an easy way. let's not use complex numbers since that's what I had in mind. something you can present to non-mathematicians :)
98951
There are so many available bases. Why is the strange number $e$ preferred over all else?  Of course one could integrate $\frac{1}x$ and see this. But is there more to the story?
7677
I  would need to be able to compute logarithms without using a calculator, just on paper. The result should be a fraction so it is the most accurate. For example I have seen this in math class calculated by one of my class mates without the help of a calculator.     $$\log_8128 = \frac 73$$  How do you do this?
9253
The question states:    I'm not sure what to do here. I used the geometric sum to do $2^{-n}$, but I can't think of a way to sum $\sin(n\theta)$ to get the required RHS.    Further working     Apologies for maybe being slow or missing something obvious here... I don't know how to get it into required form.
120670
While solving , I discovered that there is a relationship between the Fibonacci sequence and the golden ratio.  After I got the correct answer via brute force, I discovered this relationship.  One of the posters said this:     The nth Fibonacci number is $[\phi^n / \sqrt{5}\;]$,  where the brackets   denote "nearest integer".       So we need $\phi^n/\sqrt{5} &gt; 10^{999}$   $n \log(\phi) - \log{5}/2 &gt; 999 \log(10)$   $n \log(\phi) &gt; 999 \log(10) + \log(5)/2$   $n &gt; (999 \log(10) + \log(5) / 2) / \log(\phi)$       A handheld calculator shows the right hand side to be 4781.8593, so   4782 is the first integer n with the desired property.   I can't quite make sense of this.  How does the poster know that the $n$-th term is $\phi^n / \sqrt{5}$?  Can you also walk me through the operations?
92669
For these six numbers, $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$, how to arrange them in the increasing order?  This problem is taken from the today test:   I think we can consider $$f(x)=\dfrac{\ln{x}}{x}$$  or perhaps other methods, so I am looking forward to seeing other methods to solve this problem. Thank you.
67560
I am stuck on the 4 th one.I have done the rest.I have found out the value of a_n.But not getting how to proceed for the 4 th one.
13294
Say you have a number $x^{\sqrt 2}$.  Is there any way to represent this number so that there's no root (or irrational) as the exponent (so that it's easier to understand for me)? I just can't wrap my head around this.  I was thinking something like $$x^{2^{1/2}}$$ and extending on that idea, but I don't know how.  Also, what about for $x^{\sqrt 3}$? Or $x^{\sqrt n}$?
27849
I need some help with this problem:  $$439^{233} \mod 713$$  I can't calculate $439^{223}$ since it's a very big number, there must be a way to do this.  Thanks.
13240
Is there a direct way to determine how many digits a power of 2 will contain without actually performing the multiplication?  An estimation would help as well if there is no absolute solution.  EDIT: In both decimal and binary bases.
92025
I am trying to calculate the last two digits of $3^{400}$, but working $\bmod {100}$ is proving intractable. I did the last digit case pretty handily, and now I'm stuck at last two.
119275
Question from Model Question Paper for B.Math/B.Stat:  by   Q. Assume the following inequalities for positive k:  $$\frac{1}{2\sqrt{k+1}}&lt; \sqrt{k+1}-\sqrt k&lt;\frac{1}{2\sqrt k}$$  The Integer part of:  $$\sum_{k=2} ^{9999}\frac{1}{\sqrt k}$$ equals:  A. $198$  B. $197$  C. $196$  D. $195$  My approach:  $$S=\sum_{k=2} ^{9999}\frac{1}{\sqrt k}$$ $$S=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+ \cdots + \frac{1}{\sqrt{9999}}$$ $$S=2\left(\frac{1}{2\sqrt 2}+\frac{1}{2\sqrt 3}+ \cdots + \frac{1}{2\sqrt{9999}}\right)$$ $$S\approx 2\left((\sqrt 2- \sqrt 1) + (\sqrt3- \sqrt2)+\cdots +(\sqrt{9999}-\sqrt{9998})\right)$$ $$[S]\approx2(\sqrt{9999}-\sqrt1)$$ $$[S]\approx 2(98)$$ $$[S]=196$$  I know my method is not an efficient one. Also I don't know if the answer is correct. Any help will be appreciated.
96648
It is given $23!=2585201xy38884976640000$. Now it is required to find the value of $x$ and $y$. I know I could find it by using divisibility rules and solving simultaneous equations. Is there any other way to solve it (without computing it by a calculator)? This question is just out of curiosity.
105962
How can I check if $$12^{2013} + 7^{2013}$$ is divisible by $19$?  Also, how can I format my questions to allow for squares instead of doing the ^ symbol.
80740
Using 5000-digit precision in PARI/GP, I discovered that the fractional part of $(\sqrt{2}+\sqrt{3})^{2008}$ is extremely small, less than $10^{-999}$.  Is there a simple explanation for this fact ?  This looks like a Pisot number issue (similar questions have already been studied on MSE, see for example ), but it’s a more complicated situation.   Related :