MHB Finding $f(1)$ in a Polynomial of Integer Coefficients $\leq$ 4

AI Thread Summary
The discussion revolves around finding the value of $f(1)$ for a polynomial $f(x)$ with integer coefficients less than 4, given that $f(4) = 2009$. Participants confirm that the polynomial's coefficients must be integers in the range of 0 to 3. The calculations reveal that the correct result for $f(1)$ is 11. The method used by a participant named kaliprasad is acknowledged as correct and effective. The focus remains on the polynomial's constraints and the derived values.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Given $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_0, a_a,\cdots,a_n$ are all smaller than 4 and are integer, $a_n \in (0, 1, 2,\cdots)$.

Given that $f(4)=2009$, find $f(1)$.
 
Mathematics news on Phys.org
anemone said:
Given $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_0, a_a,\cdots,a_n$ are all smaller than 4 and are integer, $a_n \in (0, 1, 2,\cdots)$.

Given that $f(4)=2009$, find $f(1)$.

= 1 + 3 + 3 + 1 + 2 + 1 = 11

as f(x) = x^5 + 3x^4 + 3x^3 + x^2 +2x +1

as no coefficient is >4 and we are given f(4) subtract the highest power of 4 as many times as it can go

2009 = 1024 + 985
985 = 256 * 3 + 217
217 = 64 * 3 + 25
25 = 16 + 9
9 = 2 *4 + 1
 
kaliprasad said:
= 1 + 3 + 3 + 1 + 2 + 1 = 11

as f(x) = x^5 + 3x^4 + 3x^3 + x^2 +2x +1

as no coefficient is >4 and we are given f(4) subtract the highest power of 4 as many times as it can go

2009 = 1024 + 985
985 = 256 * 3 + 217
217 = 64 * 3 + 25
25 = 16 + 9
9 = 2 *4 + 1

Hey kaliprasad,

Thanks for participating and yes, the answer is correct and your method is great and nice!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Back
Top