## Dhaka Higher Secondary 2010/5

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
BdMO
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Joined: Tue Jan 18, 2011 1:31 pm

### Dhaka Higher Secondary 2010/5

Find primes greater than $5$ satisfying the equation: $11x^{36} - 21x^{10} + 26 x^{2} = 48$

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Dhaka Higher Secondary 2010/5

let's take in common
$x^2(11x^{34}-21x^8+26)=48$
so,$x^2 \mid 48$
any value of $x$ as a prime here greater than $5$ is none.
ans.no soln.
one can also do it using trial and error(only putting 1 value will contradict everything)
বড় ভালবাসি তোমায়,মা

Hasib
Posts: 238
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Location: খুলনা, বাংলাদেশ
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### Re: Dhaka Higher Secondary 2010/5

$11x^36-21x^10+26x^2=48$
or, $x^2(11x^34-21x^8+26)=48$
so, $x^2|48$
hence, $48=2^4 \times 3$
so, we can only have a square
number(4) which divides 48.(N.B yet dont know if it is a solution)
But, $x>5$.
So, there is no solution! A man is not finished when he's defeated, he's finished when he quits.