Dhaka Higher Secondary 2010/5

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Higher Secondary 2010/5

Unread post by BdMO » Tue Jan 18, 2011 2:05 pm

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

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Tahmid Hasan
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Re: Dhaka Higher Secondary 2010/5

Unread post by Tahmid Hasan » Tue Feb 01, 2011 11:29 pm

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)
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Hasib
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Re: Dhaka Higher Secondary 2010/5

Unread post by Hasib » Tue Feb 01, 2011 11:34 pm

$11x^36-21x^10+26x^2=48$
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.

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