Last day of 2015 problem

Algebra Level 5

Sriram says to Hari, " I am thinking of polynomial whose roots are all positive integers. The polynomial has the form $P(x)=2x^3-2ax^2 +(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ? "

After some calculations, Hari says, " There is more than one such polynomial."

Sriram says, " You're right. Here is the value of $a$ ." He writes a positive integer and asks, " Can you find the value of $c$ ? "

Hari says, " There are still two possible values of c ."

Find the sum of two possible values of c.

The answer is 440.

1 solution

Shaun Leong
Dec 31, 2015

Let the three natural roots be $p,q$ and $r$

By Vieta's Formulas ,

$a=p+q+r$ -------------------------- (1)

$a^2-81=2(pq+qr+pr)$ -------------------------- (2)

Substitute (1) into (2):

$p^2+q^2+r^2+2(pq+qr+pr)-81=2(pq+qr+pr)$

$\Rightarrow p^2+q^2+r^2=81$

WLOG let $p \geq q \geq r$ .

$p \geq 6$ since $25+25+25 < 81$ .

We proceed by trying $p=6,7,8$ .

Case 1: p=8,q=4,r=1,a=13

Case 2: p=7,q=4,r=4,a=15

Case 3: p=6,q=6,r=3,a=15

Since there are 2 possible values of c, Sriram gave $a=15$ .

By Vieta's Formulas, $c=2pqr$ .

$\Rightarrow c_1+c_2=2*7*4*4+2*6*6*3=224+216=\boxed{440}$

Exactly , I messed up with the cases and inputed the wrong answer ...Btw Great solution.(+1) @Shaun Leong

A Former Brilliant Member - 5 years, 5 months ago

MLG, very good solution and very concise.

A Former Brilliant Member - 5 years, 5 months ago

nice problem but i have seen the same problem 3 times on brilliant and it is aime ii problem 6...

Aareyan Manzoor - 5 years, 5 months ago

Last day of 2015 problem

The answer is 440.

1 solution

0 pending reports