remainder when $48^{45}$ is divided by 35?

Question

What is the remainder when $48^{45}$ is divided by 35?

I tried using Euler's theorem, which led to Re$\frac{{13}^{21}}{35}$

I don't know how to move ahead.

Answer 1

We split the congruence equation to mod $5$ and mod $7$ as below:
$48^{45}\equiv (-1)^{45}\equiv -1\equiv 6 \pmod{7}$,
$48^{45}\equiv 3^{45}\equiv 3\times3^{2\times22}\equiv3\times9^{22}\equiv3\times(-1)^{22}\equiv3\pmod{5}$,
Which then we can combine as below:
Assume the remainder is $r$. Since $r\equiv6\pmod{7}$, there exist positive integer $t$ s.t.
$r=7t+6$
Then, we plug into the other equation to obtain:
$7t+6\equiv3\pmod{5}$
$2t+1\equiv3\pmod{5}$
$2t\equiv2\pmod5$
$t\equiv1\pmod5$
Since the remainder should not exceed $35$, we choose $t=1$ which then leads to $r=13$. Hence $48^{45}$ leaves a remainder of $13$ when divided by $35$.

Answer 2

Beginning like the answer of Matti P., observe that $48^2\equiv 13^2\equiv 29 \bmod 35$

Note that $29\equiv -6 \bmod 35$, so $48^4 \equiv (-6)^2\equiv 1 \bmod 35$.

From this we conclude that $48^{44}\equiv 1 \bmod 35$ and $48^{45}\equiv 48\equiv 13 \bmod 35$

Answer 3

Let's try computing the remainder for some smaller powers.

• $48 \equiv 13 \mod 45$
• $48^2 \equiv 13^2 = 169 \equiv 29 \mod 35$
• $48^5 \equiv 13^5 = (13^2)(13^2)(13^1) = (29)(29)(13) = 10~933 \equiv 13 \mod 35$
• $48^{45} \equiv 13^{45} = (13^5)^9 \equiv (13)^9 = \ldots$