Divisors and polynomials

Probability Level 3

Let $N$ be a positive integer which has exactly $n$ positive divisors, $d_{1}, d_{2}, d_{3}, ..........d_{n}$ . Consider two polynomials

$f(x) = x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ..............+ a_{2}x^{2} + a_{1}x + a_{0}$

$g(x) = x^{n} + b_{n-1}x^{n-1} + b_{n-2}x^{n-2} + ..............+ b_{2}x^{2} + b_{1}x + b_{0}$

Roots of $f(x)$ are $d_{1}, d_{2}, d_{3}, ..........d_{n}$ and roots of $g(x)$ are $1/d_{1}, 1/d_{2}, 1/d_{3}, ..........1/d_{n}$ .

Find the value of $\sum_{i=0}^{n-1}$ $\frac{a_{i}}{b_{i}}$ .

\frac{N(N^{n} - 1)}{N - 1}

\frac{N^{n-1} - 1}{N - 1}

\frac{N^{n} - 1}{N - 1}

\frac{N(N^{n-1} - 1)}{N - 1}

\frac{N^{n+1} - 1}{N - 1}

1 solution

Sahil Bansal
Jul 6, 2018

Let $d_{1}<d_{2}<d_{3}< ..........<d_{n}$

We can easily see that $d_{1} = \frac{N}{d_{n}}$ , $d_{2} = \frac{N}{d_{n-1}}$ .........concluding that $d_{k} = \frac{N}{d_{n+1-k}}$

From this, we can write the roots of $g(x)$ as $d_{1}/N, d_{2}/N, d_{3}/N..........d_{n}/N$ .

Hence from this we derive that $f(Nx)$ has the roots same as g(x).

Hence $g(x) = \frac{f(Nx)}{N^{n}}$ [make the coefficients of $x^{n}$ as $1$ ]

From this we get $b_{i} = \frac{a_{i}}{N^{n-i}}$

Hence required sum is $\sum_{i=0}^{n-1} N^{n-i}$ = $\frac{N(N^{n} - 1)}{N - 1}$