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Suppose $a\equiv 7 \pmod 9 $, $b\equiv 1\pmod 6$. What is remainder when $a^2+2b$ is divided by 3 ?----- So, since $a=9m+7$, $b=6n+1$, $m,n\in \mathbb{Z}$, $a^2+2b=(9m+7)^2+2(6n+1)$,=$81m^2+12n+177$=$3(27m^2+4n+59)$....I am not sure what I am getting here.. Can I say the remainder is $0$?

Wes
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2 Answers2

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Yes, the remainder is $0$, since clearly $3 \mid 3(27m^2 + 4n + 59)$, and given $m, n \in \mathbb Z$, the quotient $27m^2 + 4n + 59$ is thereby integer

amWhy
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Another solution using the properties of modular arithmetic.

$a \equiv 7 \pmod 9 \Rightarrow a^2 \equiv 49 \equiv 4 \pmod 9$

$b \equiv 1 \pmod 6 \Rightarrow 2b \equiv 2 \pmod 6$

So $a^2 = 9n + 4$ and $2b = 6m + 2$, thus $a^2 + 2b = 9n + 6m + 6 = 3(3n + 2m + 2)$ which is definitely divisible by 3.

ejcqw
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  • I am not sure how you got $a^2 \equiv 49 \equiv 4\pmod 9$. – Wes Nov 13 '13 at 20:55
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    One of the properties of modular arithmetic is that if $a \equiv b \pmod m \Rightarrow a^n \equiv b^n \pmod m$ – ejcqw Nov 13 '13 at 20:58