Find the remainder of $7^{6n+2}+5^{18n+15}$ left dividing by $9$

Question

$$ 7^{6n+2}+5^{18n+15} $$ find the remainder left dividing by $9$. I can solve this if you say the relevant topic to consider.

Does this answer your question? How do I compute $a^b,\bmod c$ by hand? — Jyrki Lahtonen, Sep 11 '20 at 05:32
Maybe reasonable starting point could be: What is $7^2\bmod 9$ and $7^3\bmod9$? Can you get from that also $7^6 \bmod 9$? And you could try similar things with powers of $5$. — Martin Sleziak, Sep 11 '20 at 05:34
Also if you know Euler's theorem, the fact that $\varphi(9)=6$ could help you. — Martin Sleziak, Sep 11 '20 at 05:38
See mod order reduciton for general methods for computation of large modular powers. $\ \ $ — Bill Dubuque, Dec 29 '24 at 06:24

KBi7700 · Accepted Answer · 2020-09-14T04:45:19.633

1

First of all with the use of Euler's theorem we have that : $5^{6}=5^{\phi(9)}\equiv 1 \mod 9$ ( $(5,9)=1$ ) .

$7^{6n+2}=7^{6n}\cdot 49\equiv 4\cdot [(-2)^{6}]^{n}\equiv 4\cdot 64^{n}\equiv 4\mod 9$
$5^{18n+15}\equiv (5^{6})^{3n}\cdot (5^{6})^{2}\cdot 5^{3}\equiv -1 \mod 9$

So finally we got that $7^{6n+2}+5^{18n+15}\equiv 3\mod 9$

edited Sep 14 '20 at 04:45

answered Sep 11 '20 at 06:09

KBi7700

507

Did you mean Euler's theorem? – J. W. Tanner Sep 13 '20 at 01:46
Sure, I''ll fix it – KBi7700 Sep 14 '20 at 04:45

score 1 · Answer 2 · answered Sep 11 '20 at 06:35

$\DeclareMathOperator{\bbN}{\mathbb{N}}$$\newcommand{\digsum}{\text{digitsum}} \DeclareMathOperator{\bbO}{\mathbb{O}} \DeclareMathOperator{\bbE}{\mathbb{E}}$$\blacksquare$ Notation: Let us denote:

$\bbO := $ the set of all odd natural numbers
$\bbE := $ the set of all even natural numbers

At first, note that $$ 5^{18n + 15 }= 125^{6n + 5} \equiv (-1)^{6n + 5} \bmod 9 \equiv -1 \bmod 9 \quad [\text{as } 6n + 5 \in \bbO ~\forall~n \in \mathbb{N}] $$

Lemma: For any $a \in \bbN$ and $n \in 3\mathbb{Z}$, The following is satisfied. $$ a^{n} \equiv -1/0/1 \bmod 9 $$ Proof: If $n \in 3\mathbb{Z}$, then $n = 3k$ for some $k \in \mathbb{Z}$. Thus, we have that $$ a^{3k} \equiv \begin{cases} 0 \bmod 9 & \text{if } (a \bmod 9) \text{ is of the form } 3m\\ 1 \bmod 9 & \text{if } (a \bmod 9) \text{ is of the form } 3m + 1 \\ (-1) \bmod 9 & \text{if } (a \bmod 9) \text{ is of the form } 3m + 2 \end{cases} $$ Hence, proved!

Again, by Lemma we have that $$ 7^{6n + 2} = 49 \cdot 7^{6n} \equiv 4 \bmod 9 $$ Thus we finally have that $$ 5^{18n + 15} + 7^{6n + 2} \equiv 3 \bmod 9 $$

Find the remainder of $7^{6n+2}+5^{18n+15}$ left dividing by $9$

2 Answers2