x 2 n + a 2 n − 1 ( − 1 ) 2 n − 1 x 2 n − 1 + … + a 2 ( − 1 ) 2 x 2 + a 1 ( − 1 ) 1 x + a 0
Consider the polynomial above such that a 0 + a 1 + a 2 + … + a 2 n − 1 = 1 6 0 0 and it has non-negative integer roots r 1 , r 2 , … , r 2 n .
For distinct prime p 1 , p 2 , … , p 2 n , compute the number of factors of the integer which has the form
p 1 r 1 × p 2 r 2 × ⋯ × p 2 n r 2 n .
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Yes , this is the expected solution. Good job!
Same solution, I did it mentally. This is the speciality of Nihar's questions, they are simple yet beautiful.
Great!!!!!!!!
You must mention that it is the sum of "the absolute values" of coefficients.
@Nihar Mahajan Nice question, but could we really generate such a polynomial?? Or the number with these prime factors??
The problem should have said 'positive' factors, otherwise the answer for this problem should be 3202 which was my original answer.
Oh yes you're right.
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Total no. Of factors of no. Are (1+r1)(1+r2).......(1+r2n)
P(x)=(x-r1)(x-r2)......(x-r2n)
P(-1)=(1+r1)(1+r2)....(1+r2n) as there are even terms all negatives become positive.
But P(-1)=1 + sum of coefficients if polynomial . Hence answer = 1+1600= 1601.