Only last 2 ;)

$(90-1)^{47}$

$= 90^{47} + {47 \choose 1} 90^{46} (-1)^{1} + \ldots + {47\choose 46}90^{1}(-1)^{46} + (-1)^{47}$

The terms before the last two terms are multiples of 100, hence, they do not contribute to the last two digits.

$= \ldots + 47(90) - 1$

$= \ldots + 4230 - 1$

$= \ldots + 4229$

Hence, the last two digits are $\boxed{29}$

The problem is to find $x$ in $89^{47} \equiv x \pmod {100}$ . Since $89$ and $100$ are coprimes, we can use Euler's totient theorem and $89^{\phi (100)} \equiv 89^{40} \equiv 1 \pmod {100}$ . Therefore, we have:

$\begin{aligned} 89^{47} & \equiv 89^{40} 89^7 \pmod {100} \\ & \equiv 89^7 \pmod {100} \\ & \equiv (100-11)^7 \pmod {100} \\ & \equiv -11^7 \pmod {100} \\ & \equiv -(10+1)^7 \pmod {100} \\ & \equiv -({\color{#3D99F6} 70+1}) \pmod {100} & \small \color{#3D99F6} \text{Last two terms of the binomial expansion.} \\ & \equiv \boxed{29} \pmod {100} & \small \color{#3D99F6} \text{Earlier terms are divisible by 100.} \end{aligned}$

My solution :

We know that $89^2$ ends with 21 because it is similar to $11^2$ which ends with 21.

We can split it as 89* $21^{23}$ = 89 * 61 Since it is based on Powers of numbers ending with 1 .

Hence we can find last 2 digits as 89 * 61 = $\boxed{29}$ which can be found quickly with Vertically and crosswise method .

One important thing is to stop with last 2 digits which means left vertical is not        required to avoid unnecessary calculation.