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Please send me the answer for this sum I am not able to solve the problem

Please send me the answer for this sum
I am not able to solve the problem

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Grade:7

1 Answers

Aditya Gupta
2081 Points
4 years ago
ques asks to find n such that n^2 + 2 divides n^6 + 206. or (n^6 + 206)= (n^2 + 2)*k where k is positive integer
now, we note that 
n^6 + 206= (n^2 + 2)(n^4 – 2n^2 + 4) + 198
or (n^6 + 206)/(n^2 + 2)= (n^4 – 2n^2 + 4) + 198/(n^2 + 2)
or k – (n^4 – 2n^2 + 4) = 198/(n^2 + 2)
let k – (n^4 – 2n^2 + 4) = m (say)
so m= 198/(n^2 + 2)
so, n^2 + 2 must divide 198.
but factors of 198 are 1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, 198.
so n^2 + 2 must be 1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99 or 198.
or n^2 can be – 1, 0, 1, 4, 7, 9, 16, 20, 31, 64, 97 or 196.
so positive integral n can be 1, 2, 3, 4, 8, 14.
kindly approve :)
 

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