(C) Let $n = 10^{10}$. The expression becomes $n(n+1)(n+2)$.This is the product of $3$ consecutive integers.The product of any $k$ consecutive integer

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(C) Let $n = 10^{10}$. The expression becomes $n(n+1)(n+2)$.This is the product of $3$ consecutive integers.The product of any $k$ consecutive integers is always divisible by $k!$.Therefore,$n(n+1)(n+2)$ is divisible by $3! = 3 \times 2 \times 1 = 6$.Since the expression is perfectly divisible by $6$,the remainder is $0$.

Class 11 Mathematics Binomial Theorem Divisibility problems

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The remainder when $10^{10} \cdot (10^{10} + 1) \cdot (10^{10} + 2)$ is divided by $6$ is

KCET 2013 All KCET

A
$2$
B
$4$
C
$0$
D
$6$

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Binomial Theorem

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