Question

The remainder on dividing ${121}^n-{25}^n+{1900}^n-(-4{)}^n$ by $2000$ is :

• A. $1$
• B. $1000$
• C. $100$
• D. $0$

Answer 1

The correct option is D $0$

Let $N=$ ${121}^n$$-$${25}^n$$+$${1900}^n$$-$$(-4{)}^n$

We know that $(a-b)$ will always divide $(a{)}^n$$-$$(b{)}^n$

for all values of $n$.

$121-(-4)$ will divide $(121{)}^n$$-$$(-4{)}^n$

similarly, $1900-25$ will divide $(1900{)}^n$$-$$(25{)}^n$

Hence, $1875+125$ will divide $N$

$1875+125$$=$$2000$

Hence, remainder is $0$ and option D is correct.