Finding Remainder Using Binomial Theorem

Question

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?

Answer 1

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 $(-1)^{49}=-1$.

Answer 2

When you expand ${(5+2)}^{98}$, you will get $5k+2^{98}$, for some $k\in\Bbb N $.

If you know Fermat's little theorem, you have $2^4\equiv 1\pmod 5$. Hence $2^{98}\equiv 2^{2}\equiv 4\pmod 5$. So the remainder is $4$.

Answer 3

With your method, you have to check the divison by 5 of $2^{98} = 4\cdot (2^4)^{24} = 4\cdot (3\cdot 5 +1)^{24}$ and, using the Binomial theorem again, you end up with a rest after division of $4 \cdot 1^{24} = 4$ which is also the "other" result.