Never Forget Ramanujan

Algebra Level 4

Let $f(x)=3^{g(x)}$ , where $g(x)$ is a quadratic polynomial , such that ,

$g(1)=6$ , $g(3)=14$ and $g(5)=30$ .

Then find last two digits of $f(1729)$

Note : $f(x)$ and $g(x)$ are two different functions.

The answer is 29.

2 solutions

Rishabh Jain
Feb 29, 2016

Seeing values of g at given points we can quickly identify $\large g(x)=x^2+5$ .
$\large \therefore f(x)=3^{x^2+5}$ $\large \begin{aligned}\therefore f(1729)&=3^{1729^2+5}\\&=3^{2989446}\\&=9^{1494723}\\&=(-1+10)^{1494723} \\&=\color{#D61F06}{-1+14947230+}\cdots\end{aligned}$ $~~~~~~~(\color{#20A900}{\textbf{Using Binomial expansion}})$

We only have to look at these two terms for predicting last two digits since the terms that follow have powers of 10 as two or larger and hence will contain $at~least$ two zeroes and hence no role in last two digits.
$\therefore \textbf{Last two digits }=\large 30-1=\huge\boxed{29}$

Nice application of Binomial Expansion!
Number theory approach would be :

$3^{2989446} \mod 100$ then using Euler's Theorem , $3^{\phi(100)} \equiv 1 \mod 100$ . We know that $\phi (100)=40$ and then continuing the pattern we get , $3^6 \equiv 29 \mod 100$ .

Hence 29 is the answer.

A Former Brilliant Member - 5 years, 3 months ago

Priyanshu Mishra
Feb 29, 2016

We have, by observation,

$g(x) = x^2 + 5$ . Hence, $f(x) = 3^{x^2 + 5}$ .

Thus it suffices to find the residue of $\large\ f(1729) \equiv 3^{1729^2 + 5} \pmod{100}$ .

We will be done if we calculate $\large\ 3^{1729^2} \pmod{100}$ which is $3$ .

Thus, $\large\ f(1729) \equiv 3^{1729^2 + 5} \equiv {3^6} \equiv {29} \pmod{100}$ .

Never Forget Ramanujan

The answer is 29.

2 solutions

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