First slide

Binomial theorem for positive integral Index

Question

Number of zeros at then end of $99^{1001} + 1$ is

Moderate

A

2
B

4
C

$1002$
D

$1004$

Solution

Let $n = 1001,$ we can write
$\begin{array}{l} 99^{1001} + 1 = (100 - 1)^{n} + 1 \\ =^{n} C_{0} 100^{n} -^{n} C_{1} 100^{n - 1} +^{n} C_{2} 100^{n - 2} - \dots \\ = 100 m +^{n} C_{n - 1} (100) - 1 + 1 \end{array}$
where $m =^{n} C_{0} 100^{n - 1} -^{n} C_{1} 100^{n - 2} +^{n} C_{2} 100^{n - 3} - \dots -^{n} C_{n - 2} 100 +^{n} C_{n - 1}$
As $^{n} C_{n - 1} = 1001,$ the units digit of $m$ is different from $0$ . Thus, the number of zeros at the end of $99^{1001} + 1$ is two.

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