Remainder of division when $p$ is not prime

Question

Calculate the remainder of the division of $1160^{602}$ by $35$.
What I thought of doing right away is to factor out $602$ until I get to a point where I can apply Fermat's theorem. However, $35$ is not a prime number. How to proceed in this case?

Answer 1

$$1160\equiv5\pmod{35}\implies1160^{602}\equiv5^{602}$$

Like How to find last two digits of $2^{2016}$,

As $(5^n,35)=5$ for $n\ge1,$

let us find $5^{602-1}\pmod{35/5}$

As $(5,7)=1$ and $601\equiv1\pmod6$ $$5^{602-1}\equiv5^1\pmod7$$

Now multiply $5$ throughout

$$5^{602-1}\cdot5\equiv5^1\cdot5\pmod{7\cdot5}$$

Answer 2

we know that $35=5.7$ and

$1160^{602} \equiv 0 \pmod{5}$

$1160 \equiv 5 \pmod{7}$

$a^{p-1} \equiv 1 \pmod{p} \Rightarrow$ $1160^6 \equiv 1 \pmod{7}$

$1160^6 \equiv 1 \pmod{7}$ $\Rightarrow$ $1160^{602}\equiv 1160^2\equiv 4 \pmod{7}$

$1160^{602}\equiv 4 \pmod{7}$ $\Rightarrow$ $1160^{602}=7k+4$ for some integer $k$ also we know that

$7k+4 \equiv 0 \pmod{5}$ $\Rightarrow 7k \equiv 1 \pmod{5}$

$7k \equiv 1 \pmod{5} \Rightarrow k \equiv 3 \pmod{5}$ and that means there exists an integer $m$ such that $k=5m+3$

$k=5m+3$ $\Rightarrow 7k+4=35m+25$ so

$1160^{602} \equiv 25 \pmod{35}$

Answer 3

$\ \ \ \,5^{\large 2+6n}\!\bmod 35 = \color{#0a0}5(5^{\large 1+\color{#c00}6n}\!\bmod 7) = 5(5^{\large 1}),\,$ by $\,5^{\color{#c00}{\large 6}}\equiv 1\pmod{\!7}\,$ by Fermat, and

using $\, ab\bmod ac = \color{#0a0}a(b\bmod c),\,$ the mod Distributive Law to factor out $\,a= \color{#0a0}{5}$ from mod.

Answer 4

Hint:

Use the inverse isomorphism of the Chinese remainder theorem and lil' Fermat:

$1160\equiv0\mod 5$ and $1160\equiv 5\mod 7$, so $1160^n\equiv 0\bmod 5$ for all $n>0$, and $$1160^{602}\equiv 5^{602\bmod 6}=5^2=4\bmod 7.$$

Now the inverse isomorphism $\mathbf Z/5\mathbf Z\times\mathbf Z/7\mathbf Z\longrightarrow \mathbf Z/35\mathbf Z$ is deduced from a Bézout's relation between $5$ and $7$: as we have $3\cdot 5-2\cdot 7=1$, it is given by $$(x\bmod 5,y\bmod 7)\longmapsto [y\cdot3\cdot 5-x\cdot 2\cdot 7]\bmod 35. $$

Answer 5

Let $N \equiv 1160^{602} \pmod{35} \equiv 5^{602} \pmod {35}$.

Clearly $N \equiv 0 \pmod 5$

By Euler's theorem, $5^6 \equiv 1 \pmod 7$.

So $N \equiv 5^{600} \cdot 5^2 \pmod 7 \equiv 1^{100} \cdot25 \pmod 7 \equiv 4 \pmod 7$.

Ww now seek a multiple of $5$ that is congruent to $1$ modulo $7$.

$$5 \equiv 5 \pmod 7, \qquad 10 \equiv 3 \pmod 7, \qquad \color{red}{ 15\equiv 1 \pmod 7}$$

Next we seek a multiple of $7$ that is congruent to $1$ modulo $5$.

$$7 \equiv 2 \pmod 5, \qquad 14 \equiv 4 \pmod 5, \qquad \color{red}{21 \equiv 1 \pmod 5}$$

It follows that $N \equiv 4 \cdot 15 + 0 \cdot 21 \equiv 60 \equiv 25\pmod{35}$