Last 2 digits of $\pi$ ?

We need to find $3^{14^{159}} \bmod 100$ . Since $3$ and $100$ are coprime integers or $\gcd(3,100) = 1$ , we can apply Euler's theorem and hence Carmichael's lambda function $\lambda(\cdot)$ . We note that $\lambda(100) = 20$ . Then we have:

$3^{14^{159}} \equiv 3^{14^{159} \bmod \lambda(100)} \equiv 3^{14^{159} \bmod 20} \text{ (mod 100)}$

We have to solve $14^{159} \bmod 20$ . Since $\gcd(14,20) \ne 1$ , we cannot use Euler's theorem. We have to use Chinese remainder theorem and consider the divisors $4$ and $5$ separately as follows:

• Divisor $4$ : $\ \ 14^{159} \equiv 0 \text{ (mod 4)}$
• Divisor $5$ : $\ \ 14^{159} \equiv (15-1)^{159} \equiv (-1)^{159} \equiv - 1 \equiv 4 \text{ (mod 5)}$

Therefore $14^{159} \equiv 5n + 4 \equiv 0 \text{ (mod 4)}$ $\implies n \equiv 0 \implies 14^{159} \equiv 4 \text{ (mod 20)}$ and

$3^{14^{159}} \equiv 3^4 \equiv \boxed{81} \text{ (mod 100)}$

After a while each number's powers will repeat the last digits.

• The last two digits of $3^{20k}$ , k-integer, always 01
• The last two digits of $14^{10k}$ , k-integer and bigger than 0, always 76

Therefore: $\begin{aligned} 3\displaystyle^{14\displaystyle^{159}}\pmod {100}&=1\times3\displaystyle^{\left (14\displaystyle^{159}\pmod{20}\right )}\pmod {100}\\ &=3\displaystyle^{\left (76\times14\displaystyle^{9}\pmod{10}\right )}\pmod {100}\\ &=3\displaystyle^{\left (76\times84\pmod{10}\right )}\pmod {100}\\ &=3\displaystyle^{84}\pmod{100}\\ &=3\displaystyle^{4}\pmod{100}\\ &=81 \end{aligned}$

Taking the last two digits of the result means to get the remainder of dividing it by 100. I experimented and found the loop section of the remainder of the power of three divided by 100:

Starting from the 20th power of 3, the remainder after dividing by 100 begins to loop from the beginning of the table, that is:

$3^n\equiv 3^{(n\mod 20)}\pmod{100}$

Then we only need to find out what is the remainder of ${14}^{159}$ divided by 20. For this reason, I did the following experiment and found the rule of the remainder of the power of 14 divided by 20:

$\begin{cases}14^0\equiv 1\pmod{20}\\14^1\equiv 14\pmod{20}\\14^{2a}\equiv 16\pmod{20}(a>0)\\14^{2a+1}\equiv 4\pmod{20}(a>0)\end{cases}$

So we can know that the remainder of ${14}^{159}$ divided by 20 is 4.

By comparing the above table again, we can know that the answer is $\boxed{81}$