Darian Hubbard

2022-07-15

If $b\equiv 3\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}12\right)$, what is $8b\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}12\right)$?
We can write $b=12q+3$. Then with some algebra $8b=12\left(8q+2\right)$.
This is causing me problems because there is not number plus $12\left(8q+2\right)$ to tell us the remainder. Is it just zero?

cindysnifflesuz

Expert

Step 1
We have $b\equiv 3\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}12$
Step 2
Then $8b\equiv 24\equiv 0\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}12$

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