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Found in: Page 307

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

# Convert ${\left(}{7206}{\right)}_{8}{}{\mathrm{and}}{}{\left(}{\mathrm{A}}{0}{\mathrm{EB}}{\right)}_{16}$ to a binary representation

${\left(}{111010000110}{\right)}_{2}$

${\left(}{1010000011101011}{\right)}_{2}$

See the step by step solution

## Step 1

${\left(}{7206}{\right)}_{8}{=}{7}{\ast }{{8}}^{{3}}{+}{2}{\ast }{{8}}^{{2}}{+}{6}{\ast }{{8}}^{{0}}\phantom{\rule{0ex}{0ex}}{=}\left({2}^{2}+{2}^{1}+{2}^{0}\right){{2}}^{{9}}{+}{{2}}^{{1}}{\ast }{{2}}^{{6}}{+}\left({2}^{2}+{2}^{1}\right){{2}}^{{0}}\phantom{\rule{0ex}{0ex}}{=}{{2}}^{{11}}{+}{{2}}^{{10}}{+}{{2}}^{{9}}{+}{{2}}^{{7}}{+}{{2}}^{{2}}{+}{{2}}^{{1}}\phantom{\rule{0ex}{0ex}}{=}{\left(}{111010000110}{\right)}_{2}$

## Step 2

${\left(}{\mathrm{A}}{0}{\mathrm{EB}}{\right)}_{16}{=}{10}{\ast }{{16}}^{{3}}{+}{{14}}^{{\ast }}{{16}}^{{1}}{+}{{11}}^{{\ast }}{{16}}^{{0}}\phantom{\rule{0ex}{0ex}}{=}\left({2}^{3}+{2}^{1}\right){{2}}^{{1}}{2}{+}\left({2}^{3}+{2}^{2}+{2}^{1}\right){{2}}^{{4}}{+}\left({2}^{3}+{2}^{1}+{2}^{0}\right){{2}}^{{0}}\phantom{\rule{0ex}{0ex}}{=}{{2}}^{{15}}{+}{{2}}^{{13}}{+}{{2}}^{{6}}{+}{{2}}^{{5}}{+}{{2}}^{{3}}{+}{{2}}^{{1}}{+}{{2}}^{{0}}\phantom{\rule{0ex}{0ex}}{=}{\left(}{1010000011101011}{\right)}_{2}$