Find the remainder when $7^{98} $ is divided by $5$.
What I am doing here is expanding ${(5+2)}^{98} $ using binomial theorem and writing it as $5k + 2$, where $k$ is a positive integer but the answer is $4$ and I'm getting $2.$
They are expanding ${(50 – 1)}^{49} $ using binomial theorem and then writing $50k – 1$ and getting $4$ as a reminder.
How those two methods are different?
Best Answer
The binomial theorem tells you $$(5+2)^{98} = \binom{98}{0}\cdot 5^{98}+\binom{98}{1}\cdot 5^{91}2^1+ \binom{98}{2}\cdot 5^{90}2^2+\cdots + \binom{98}{97}\cdot 5\cdot 2^{97}+ \binom{98}{98}\cdot 2^{98} = 5k+2^{98}.$$ So you need to find the remainder when $2^{98}$ is divided by $5$, not the remainder when $2$ is divided by $5$.
The method using $(50-1)^{49}$ is a little easier because $(-1)^{49}=-1$.