RMO 2015! #10
Find the remainder on dividing x 1 9 5 9 − 1 by ( x 2 + 1 ) ( x 2 + x + 1 ) .
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3 solutions
Sir, how did you get the remainder as 1 and x = i . Rather, can you please explain me your solution in short.
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Dividing ( x 1 9 5 9 − 1 ) / ( x 3 − 1 ) by x 2 + 1 we can write ( x 1 9 5 9 − 1 ) / ( x 3 − 1 ) = p ( x ) ( x 2 + 1 ) + a x + b , where a x + b is the remainder for some integers a and b . Plugging in x = i gives 1 = a i + b so that a = 0 , b = 1 .
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Sir, how did you choose ( x 1 9 5 9 − 1 ) / ( x 3 − 1 ) when it was given ( x 1 9 5 9 − 1 ) only.
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@Vishal Yadav – I did this to simplify the work. We know that x 2 + x + 1 divides x 3 − 1 which in turn divides x 1 9 5 9 − 1
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@Otto Bretscher – Thanks Sir. But how did you test that x 2 + x + 1 divides x 3 − 1 because if we use factor theorem then the value of x turns out to be i m a g i n a r y and substitution takes a long time. Please give me a quicker way if you have.
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@Vishal Yadav – 1 + x + . . . + x n − 1 always divides x n − 1 . The roots of the former polynomial are the n th roots of unity other than 1.
what do you mean by p(x)?
I know what u want to tell but you should writhe the whole process of Division Algorithm.Then plugg value of x=i which is nothing but root-1.
Can i simply divide it by highest power of x i.e X^4 ....i see that 1959%4 ==3 there by there has to be an x^3 option and i see only one
Como o divisor é um polinômio de 4° grau, então o resto é terceiro grau.
Q(x).(x^2 + 1).(x^2 + x + 1) = (x^1959 - 1) + R(x)
Q(x) é do grau 1955
Logo, R(x) = ax^3 + bx^2 + cx + d
Então a única alternativa é x^3 - 1
Dividing the polynomial ( x 1 9 5 9 − 1 ) / ( x 3 − 1 ) by x 2 + 1 gives ( x 1 9 5 9 − 1 ) / ( x 3 − 1 ) = p ( x ) ( x 2 + 1 ) + 1 ; plug in x = i to find the remainder. Thus x 1 9 5 9 − 1 = p ( x ) ( x 2 + 1 ) ( x − 1 ) ( x 2 + x + 1 ) + x 3 − 1 .