Question

How many zeros are at the end of the binary expansion of ${\mathbf{100}}_{\mathbf{10}}$

Short answer

16 zero’s are at the end of the binary expansion.

Step-by-step solution

Step 1

Definitions :

$$\begin{gathered} \mathrm{is}\mathrm{the}\mathbf{b}\left\{\mathrm{base}b\mathrm{reprsentation}\mathrm{of}n\right\}\mathbf{}\mathbf{is}\mathbf{}{\mathbf{a}}_{\mathbf{k}}\mathbf{\dots }{\mathbf{a}}_{\mathbf{2}}{\mathbf{a}}_{\mathbf{1}}{\mathbf{a}}_{\mathbf{0}}\mathbf{}\mathrm{then} \\ \mathbf{n}={\mathbf{a}}_{\mathrm{k}}{\mathbf{b}}^{\mathrm{k}}+{\mathbf{a}}_{\mathrm{k}-1}{\mathbf{b}}^{\mathrm{k}-1}+\dots +{\mathbf{a}}_1\mathbf{b}+{\mathbf{a}}_0 \end{gathered}$$

Step 2

Solutions :

${\mathbf{100}}_{\mathbf{10}}\mathbf{!}$

By the definition of baseb representation of nwithb = 10! and n = 100

${\mathbf{100}}_{\mathbf{10}}\mathbf{!}\mathbf{=}\mathbf{1}\mathbf{\cdot }\mathbf{(}\mathbf{10}\mathbf{!}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{0}\mathbf{+}\mathbf{10}\mathbf{!}\mathbf{+}\mathbf{0}\mathbf{\cdot }\mathbf{1}\mathbf{=}\mathbf{(}\mathbf{10}\mathbf{!}{\mathbf{)}}^{\mathbf{2}}$

By the definition of factorials n! = n.(n - 1)....1

$\begin{array}{r}\mathbf{10}\mathbf{!}\mathbf{=}\mathbf{10}\mathbf{\cdot }\mathbf{9}\mathbf{\cdot }\mathbf{8}\mathbf{\cdot }\mathbf{7}\mathbf{\cdot }\mathbf{6}\mathbf{\cdot }\mathbf{5}\mathbf{\cdot }\mathbf{4}\mathbf{\cdot }\mathbf{3}\mathbf{\cdot }\mathbf{2}\mathbf{\cdot }\mathbf{1}\\ \mathbf{=}\mathbf{2}\mathbf{\cdot }\mathbf{5}\mathbf{\cdot }\mathbf{9}\mathbf{\cdot }{\mathbf{2}}^{\mathbf{3}}\mathbf{\cdot }\mathbf{7}\mathbf{\cdot }\mathbf{2}\mathbf{\cdot }\mathbf{3}\mathbf{\cdot }\mathbf{5}\mathbf{\cdot }{\mathbf{2}}^{\mathbf{3}}\mathbf{\cdot }\mathbf{3}\mathbf{\cdot }\mathbf{2}\mathbf{\cdot }\mathbf{1}\\ \mathbf{=}{\mathbf{2}}^{\mathbf{8}}\mathbf{\cdot }\mathbf{5}\mathbf{\cdot }\mathbf{9}\mathbf{\cdot }\mathbf{7}\mathbf{\cdot }\mathbf{3}\mathbf{\cdot }\mathbf{5}\mathbf{\cdot }\mathbf{3}\mathbf{\cdot }\mathbf{1}\end{array}$

We then note that 10! contains 8 two’s in it’s factorization, which implies that ${\left(10!\right)}^{\mathbf{2}}$ will contain 8.2 = 16 two’s in its factorization and thus the binary expansion of ${\mathbf{100}}_{\mathbf{10}}\mathbf{!}$will contain zeros at the end.

Note: the binary expansion of ${\mathbf{100}}_{\mathbf{10}}\mathbf{!}$ is

$\mathbf{10111111100111110101010000010000000000000000}$