1600 followers problem

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 x^{2n} + a_{2n-1} (-1)^{2n-1} x^{2n-1} + \ldots + 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 = 1600 a_0 + a_1 + a_2 + \ldots + a_{2n-1} = 1600 and it has non-negative integer roots r 1 , r 2 , , r 2 n r_1, r_2, \ldots , r_{2n} .

For distinct prime p 1 , p 2 , , p 2 n p_1,p_2, \ldots, p_{2n} , 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 . p_1 ^{r_1} \times p_2 ^{r_2} \times \cdots \times p_{2n} ^{r_{2n}}.


This problem is original. Follow me for more :)


The answer is 1601.

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3 solutions

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.

Yes , this is the expected solution. Good job!

Nihar Mahajan - 5 years, 8 months ago

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But it is self explanatory.

Aakash Khandelwal - 5 years, 8 months ago

Same solution, I did it mentally. This is the speciality of Nihar's questions, they are simple yet beautiful.

Kushagra Sahni - 5 years, 8 months ago

Great!!!!!!!!

Dev Sharma - 5 years, 7 months ago

You must mention that it is the sum of "the absolute values" of coefficients.

Nihar Mahajan - 5 years, 8 months ago
Aaghaz Mahajan
Mar 13, 2018

@Nihar Mahajan Nice question, but could we really generate such a polynomial?? Or the number with these prime factors??

Kun Pakawat
Oct 15, 2015

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.

Kushagra Sahni - 5 years, 7 months ago

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