Inspired by Mehul Chaturvedi

Algebra Level 5

$P(x)=\dfrac{ax^7}{b}+\dfrac{cx^6}{d}+\dfrac{ex^5}{f}+\dfrac{gx^4}{h}+\dfrac{ix^3}{j}+\dfrac{kx^2}{l}+\dfrac{mx}{n}-22$

Let a polynomial $P(x)$ be defined such that $P(n)$ is the number of open-chain structural isomers of $n$ carbon membered alkanes for $n\in [1,8]$ .

Given that $a$ , $c$ , $e$ , $g$ , $i$ , $k$ and $m$ are positive integers such that $\gcd(a,b)=$ $\gcd(c,d)=$ $\gcd(e,f)=$ $\gcd(g,h)=$ $\gcd(i,j)=$ $\gcd(k,l)=$ $\gcd(m,n)= 1$ .

Calculate $a+b+c+d+e+f+g+h+i+j+k+l+m+n-22$ .

The answer is 11266.

2 solutions

Pranjal Jain
Feb 26, 2015

The points are $(1,1),(2,1),(3,1),(4,2),(5,3),(6,5),(7,9),(8,18)$ .

The polynomial is $\dfrac{x^7}{504}+\dfrac{23x^6}{-360}+\dfrac{61x^5}{72}+\dfrac{427x^4}{-72}+\dfrac{421x^3}{18}+\dfrac{9181x^2}{-180}+\dfrac{1171x}{21}-22$

$1+504+23-360+61+72+427-72+421-18+9181-180+1171+21-22=11266$

You could've asked the value of a+b+c+d+e+f+g+h+i+j+k+l+m+n !! Why -22 ?

Satyajit Mohanty - 5 years, 11 months ago

Cuz I did some mistake while evaluating answer and later felt that its easier to change question than to change answer.

Pranjal Jain - 5 years, 11 months ago

How did you get to the polynomial?! I've been trying doing Lagrangian interpolation, but it fails for many-one functions. Did you simultaneously solve 7 equations to find the constants?

P.S.: I did manage to calculate the leasing coefficient via the method of differences, but not after that.

A Former Brilliant Member - 5 years, 4 months ago

John Ross
Apr 4, 2018

The question has been raised of how to get the polynomial form of the function passing through the given points. The following method works when we have points of the form (1,a), (2,b), (3,c), .... Make a difference table of your points. $\begin{array} {lrrrr} \\ 1 & 0 & 0 & 1 & -2 & 4 & -6 &10 \\ 1 & 0 & 1 & -1 & 2 & -2 & 4 \\ 1 & 1 & 0 & 1 & 0 & 2 \\ 2 & 1 & 1 & 1 & 2 \\ 3 & 2 & 2 & 3 \\ 5 & 4 & 5 \\ 9 & 9 \\ 18 \\ \\ \end{array}$ We are interested in the top row reading from right to left. We can write the desired polynomial as the sum of choose functions The coefficients to our choose functions will be the corresponding number on the top row minus the last coefficient that we found. Our polynomial will be equal to $10\binom{x}{7}-16\binom{x}{6}+20\binom{x}{5}-22\binom{x}{4}+23\binom{x}{3}-23\binom{x}{2}+23\binom{x}{1}-22\binom{x}{0}$ . (first coefficient = 10, second coefficient = -6-10, third coefficient = 4-(-16), fourth coefficient = -2-20, ...) This simplifies into the desired polynomial

Inspired by Mehul Chaturvedi

The answer is 11266.

2 solutions

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