# remainder of $a^2+3a+4$ divided by 7

If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7

(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$

if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$

if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$

thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.

is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)

• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23

$a = 6 \quad(\mathrm{mod} 7)$

$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$

$3a = 18 = 4\quad (\mathrm{mod} 7)$

$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$

If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.

Hence, $a^2+3a+4 = (7n+b)^2+3(7n+b)+4$ $= 49n^2 + 14nb + b^2 + 21n + 3b + 4$ $= 7(7n^2+2nb+3n) + (b^2+3b+4)$.

So, the remainder when $a^2+3a+4$ is divided by $7$ will be the same as the remainder when $b^2+3b+4$ is divided by $7$.

For the specific case when $b = 6$, we get that $a^2+3a+4 = 7(7n^2+12n+3n)+58$ $= 7(7n^2+12n+3n+8)+2$.
So the remainder when $a^2+3a+4$ is divided by $7$ is $2$.

The remainder of $a^2+3a+4$ divided by $7$ is sum of the remainder of each terms, modulo $7$.

So $a^2\equiv 1 \pmod{7}$ since $a=7k+6$ then $a^2=7l+1$; $\quad$ $3a\equiv 4 \pmod{7}$ since $3a=21k+18=21k+14+4$ and clearly $4\equiv 4 \pmod{7}$.

Finally $1+4+4 \equiv 2 \pmod{7}$ then the remainder is $2$.

• To obtain $a \equiv b \pmod{n}$, type a \equiv b \pmod{n} in math mode. Jul 28 '15 at 13:43

$a^2 + 3a + 4 \equiv a^2 - 4a + 4 \equiv (a-2)^2 \pmod 7$

If $a\equiv b \pmod 7$, then $a^2 + 3a + 4 \equiv (b-2)^2 \pmod 7$