# last 2 digits of a sequence

$x+\frac{1}{x} = 3$, what are the last 2 digits of $x^{2^{2013}}+\frac{1}{x^{2^{2013}}}$?

Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step

• The next step is exactly the same as the first step. – Arthur Aug 2 '18 at 9:33
• The last two digits alternate between $07$ and $47$. – Mohammad Zuhair Khan Aug 2 '18 at 9:35
• – lab bhattacharjee Aug 2 '18 at 9:36

Definitely you see the pattern : $x^{2m} + \frac 1{x^{2m}} = \left(x^m + \frac 1{x^m}\right)^2 - 2$ for all $m \geq 1$.

So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.

Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.

However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22\color{blue}{07}$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.

Therefore, the last two digits of the number $x^{2^n} + \frac 1{x^{2^n}}$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$

So now, it should be obvious which term appears at $n=2013$.

• Well the algebra certainly beats my trial and error explanation – Mohammad Zuhair Khan Aug 2 '18 at 9:41

The trick is $$(a+1/a)^2= a^2 + 1/{a^2} +2$$

That is $$a^2 + 1/{a^2}=(a+1/a)^2-2$$

Starting with 3 and working mod $(100)$ the algorithm goes like,

$$3\to 7 \to 47 \to 07 \to 47 \to 07 \to .....$$

Now you can finish the solution.