Question

The remainder, if $1+2+2^2+2^3+...+2^{2019}$ is divided by $5$, is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (A)

$0$

Since $1+2+2^2+2^3+..........+2^{2}019$ is a G.P

Sum of 2020 terms of this G.P is $1\frac{(2^{2020}-1)}{(2-1)}=2^{2020}-1$

Now,

$2^1=2$

$2^2=4$

$2^3=8$

$2^4=16$

$2^5=32$

$2^6=64$

$2^7=128$

$2^8=256$

We see that $2^{4n}$ always has $6$ in its once place and thus $2^{4n}-1[n=1,2,\mathrm{3.....}]$is always divisible by $5$.

And $\frac{2020}{4}=505$ is an integer thus $2^{4\times 504}-1$ is divisible by $5$ and hence remainder is $0$