Find a Shorter Solution for f(7) with Given Constraints | Elementary Question

[icystrike]

Homework Statement

I've posted it many months ago.. now i manage to get the answer but I am wondering if anyone has a shorter solution?

The question goes like this(without a single omission):

Let f(x)=$a_0+a_1x+a_2x^2+...+a_nx^n$ ,where $a_0,a_1,a_2,...,a_n$
are nonnegative integers. If f(1)=8 and f(35)=$6^6$ , find f(7).

Answer is 512.

Homework Equations

The Attempt at a Solution

Given $$a_{i}$$ is nonnegative , it can be 0 ,1 ,2 ,3 , ... , 7

Since $$35^{4}>6^{6}$$,
$$a_{4},a_{5},a_{6}... = 0$$
$$f(35)=6^{6} \equiv1 mod(35)$$ by fermat's little theorem.
$$a_0 =1$$
$$a_3=0 or 1 \because 2\times35^3 > 6^6$$
therefore ,
$$a_1 + a_2 = 7 (rej \because a_2 > 38)$$
or
$$a_1 + a_2 =6$$

Hence, we can have a two by two simultaneous eqn:
$$a_1 + a_2 = 6$$
$$35a_1 + 35^2 a_2 =3780$$
$$a_1=3 ,a_2=3$$

therefore , $$f(7)=1+3\times7+3\times7^{2}+7^{3}=512$$

 

[chaoseverlasting]

This is a fairly simple question. Another way to do this:

f(x) looks like the binomial expansion of $$f(x)=(1+x)^n$$
In this case, since it is given that f(1)=8, n=3.

Putting x=35 confirms this as we get $$f(35)=(1+35)^3$$.

Thus you can easily get f(7)=512.

 

[icystrike]

This is a fairly simple question. Another way to do this:

f(x) looks like the binomial expansion of $$f(x)=(1+x)^n$$
In this case, since it is given that f(1)=8, n=3.

Putting x=35 confirms this as we get $$f(35)=(1+35)^3$$.

Thus you can easily get f(7)=512.

But how you manage to find out that f(x) is actually the binomial expansion from the given information...

 

[chaoseverlasting]

Because of the increasing powers of x. It starts with x^0 and ends with x^n. It resembles a binomial expansion. Since f(1) was a power of 2, that furthur strengthened the assumption. Thats because the sum of binomial coefficients is 2^n, so it seemed to fit. From there on, I just checked if the given expansion fit the data and from there, the answer.