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Let $f(x)=a_0+a_1x+a_2x^2....+a_nx^n$, where $a_i$ are non-negative integers for $i=0, 1, 2,...., n$. If $f(1)=21$ and $f(25)=78357$, find the value of $\frac{f(10)+3}{100}$

My work--

Let $f(x)-21=g_1(x)\Rightarrow g_1(1)=0$. This implies that $1$ is a root of $g_1(x)\Rightarrow g_1(x)=(x-1)p_1(x)\Rightarrow f(x)-21=(x-1)p_1(x)\Rightarrow f(x)=(x-1)p_1(x)+21$. Similarly $f(x)=(x-25)p_2(x)+78357\Rightarrow (x-1)p_1(x)+21=(x-25)p_2(x)+78357\Rightarrow (1-1)p_1(1)+21=(1-25)p_2(1)+78357\Rightarrow p_2(1)=3264$

And in the same way $p_1(25)=3264$.

I am unable to proceed beyond this..I hope that we can't say $p_1(x)=p_2(x)$. Any other method is appreciated.

An Alien
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    Hint: Use $a_i$ are nonnegative integers to bound the degree $n$ etc. In fact, converting 78357 to base 25 gives ... – user10354138 Oct 17 '21 at 01:22

1 Answers1

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Following the hint in the comment, noting that $25^3 < 78357 < 25^4$ shows that $n \le 3$, i.e. $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$.

From here, just do long division to find $a_3, a_2, a_1, a_0$. For instance, $78357 = 25^3 \cdot 5 + 232$, and the remainder $232$ is $232 = 25^2 \cdot 0 + 25 \cdot 9 + 7$.

angryavian
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