Photoelectric Primes

Algebra Level 5

Shivamani thinks that he has found a cubic polynomial $f(x)$ such that $f\left( n \right) =n$ -th prime number. He found that

$\begin{cases} {f(1) = 2} \\ {f(2) = 3 } \\ {f(3) = 5} \\ { f(4) = 7} \end{cases}$

Help him find the value of $f(5)$ .

This problem is original.

The answer is 8.

2 solutions

Patrick Corn
Jul 8, 2015

Let $h(x) = f(x)-2x+1$ . Then $h(1) = 1, h(2) = h(3) = h(4) = 0$ . Since $h$ is cubic, it must equal $-\frac16(x-2)(x-3)(x-4)$ . So $f(x) = 2x-1-\frac16(x-2)(x-3)(x-4)$ , so $f(5) = \fbox{8}$ .

Moderator note:

Nice loophole! I wonder how would we solve for $f(6)$ if we were further given that $f(5 ) = 11$ .

I used a chart to find the third difference then went backwards.

Anthony Pham - 5 years, 11 months ago

Ya method of differences can be used here:)

shivamani patil - 5 years, 11 months ago

Then a cubic polynomial wouldn't have sufficed You would need a biquadratic polynomial (5 unknowns)

Karan Jain - 5 years, 11 months ago

Answer to challenge master

$f(6)=22$ in that case

Michael Mendrin - 5 years, 11 months ago

How did you find it?

shivamani patil - 5 years, 8 months ago

I used the method of differences!

Swapnil Das - 5 years, 11 months ago

Đơn Giản Là Đạt
Aug 18, 2015

I solved in a kinda-longway solution, and Mr Corn solution was fantastic, I don't know how he came up with that 2x - 1, but here's my solution: f(x) is a cubic polynomial so it can be written as : ax^3 + bx^2 + cx + d (a, b, c, d are real numbers ) with every varies of f(x) we receive an equation, we have for equations, four unknowns (a,b,c,d ). After solving the above you will have a = -1/6 , b = 3/2 , c= -7/3, d = 3. Sub x=5 with the cubic polynomial you've just found you receice f(5) = 8

i did it the same way

A Former Brilliant Member - 4 years, 7 months ago

Photoelectric Primes

This problem is original.

The answer is 8.

2 solutions

Moderator note:

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