# Tagged Questions

Use this tag when you want to determine the thinking that is needed to solve a certain type of problem, as opposed to looking for a specific answer to a question.

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### How many rationals of the form $\large \frac{2^n+1}{n^2}$ are integers?

This was Problem 3 (first day) of the 1990 IMO. A full solution can be found here. How many rationals of the form $\large \frac{2^n+1}{n^2},$ $(n \in \mathbb{N} )$ are integers? The possible ...
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### How could I improve this approximation?

In a computer application, I need to solve trillions of times an equation which can be reduced to $$f(x)=\sin(x)-a x=0$$ Newton methods (quadratic and higher orders) are used for the solution. ...
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### Permutation of 1…9 with no ascending or descending subsequence of length 4

Arrange the numbers $1,2,...,9$ in such an order that no four of them appear (adjacently or otherwise) in ascending or descending order. Show that there is no arrangement of the numbers $1,2,...,10$ ...
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### Not able to solve $({\frac{1}{2}})^p + ({\frac{1}{3}})^p + ({\frac{1}{7}})^p - 1 = 0.$

I'm not able to solve $$({\frac{1}{2}})^p + ({\frac{1}{3}})^p + ({\frac{1}{7}})^p - 1 = 0.$$ If you put values of $p$ (like $\frac{1}{2}$ or 2) back in the equation it doesn't satisfy! So please ...
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### Multiplication Table with a frame and picture of equal sum

Is there an $n \times n$ multiplication table such that if you form a border of width $k$ ("the frame") and sum its elements, the total will equal the sum of the remaining elements ("the picture")? ...
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### In △ ABC, D is the midpoint of AB, while E lies on BC satisfying BE = 2EC. If m∠ADC=m∠BAE, what is the measure of ∠BAC in degrees?

In △ABC, D is the midpoint of AB, while E lies on BC satisfying BE = 2EC. If m∠ADC=m∠BAE, what is the measure of ∠BAC in degrees? I know already that angle A and angle D are congruent because ...
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### Penguin Brainteaser : 321-avoiding permutations

There are $k$ penguins, $k\ge 3$. They are all different heights. How many ways are there to order the penguins in a line, left to right, so that we cannot find any three that are arranged tallest to ...
I don't have a clue on how to start this question. I have a feeling I will need to use the Clairaut's theorem: $f_xy=f_yx$ Can anyone advise?