# When can you simplify the modulus? ($10^{5^{102}} \text{ mod } 35$)

I stumbled across this problem

Find the remainder when $10^{5^{102}}$ is divided by 35

Beginning, we try to find a simplification for $10$ to get:
$$10 \equiv 1 \text{ mod } 7\\ 10^2 \equiv 2 \text{ mod } 7 \\ 10^3 \equiv 6 \text{ mod } 7$$

..As these problems are meant to be done without a calculator, calculating this further is cumbersome. The solution, however, states that since $35 = 5 \cdot 7$, then we only need to find $10^{5^{102}} \text{ mod } 7$. I can see (not immediately) the logic behind this. Basically, since $10^k$ is always divisible by $5$ for any sensical $k$, then:
$$10^k – r = 5(7)k$$
But then it’s not immediately obvious how/why the fact that $5$ divides $10^k$ helps in this case.

My question is, is in general, if we have some mod system with $a^k \equiv r \text{ mod } m$ where $m$ can be decomposed into a product of numbers $a \times b \times c \ \times …$, we only need to find the mod of those numbers where $a, b, c…..$ doesn’t divides $a$? (And if this is the case why?) If this is not the case, then why/how is the solution justified in this specific instance?

#### Solutions Collecting From Web of "When can you simplify the modulus? ($10^{5^{102}} \text{ mod } 35$)"

The “logic” is that we can us a mod distributive law to pull out a common factor $\,c=5,\,$ i.e.

$$ca\bmod cn =\, c(a\bmod n)\quad\qquad$$

This decreases the modulus from $\,cn\,$ to $\,n, \,$ simplifying modular arithmetic. Also it may eliminate CRT = Chinese Remainder calculations, eliminating needless inverse computations, which are much more difficult than above for large numbers (or polynomials, e.g. see this answer).

This distributive law is often more convenient in congruence form, e.g.

$$\quad \qquad ca\equiv c(a\bmod n)\ \ \ {\rm if}\ \ \ \color{#d0f}{cn\equiv 0}\ \pmod{\! m}$$

because we have: $\,\ c(a\bmod n) \equiv c(a\! +\! kn)\equiv ca+k(\color{#d0f}{cn})\equiv ca\pmod{\!m}$

e.g. in the OP: $\ \ I\ge 1\,\Rightarrow\, 10^{\large I+N}\!\equiv 10^{\large I}(10^{\large N}\!\bmod 7)\ \ \ {\rm by}\ \ \ 10^I 7\equiv 0\,\pmod{35}$

Let’s use that. First note that exponents on $10$ can be reduced mod $\,6\,$ by little Fermat,

i.e. notice that $\ \color{#c00}{{\rm mod}\,\ 7}\!:\,\ 10^{\large 6}\equiv\, 1\,\Rightarrow\, \color{#c00}{10^{\large 6J}\equiv 1}.\$ Thus if $\ I \ge 1\$ then as above

$\phantom{{\rm mod}\,\ 35\!:\,\ }\color{#0a0}{10^{\large I+6J}}\!\equiv 10^{\large I} 10^{\large 6J}\!\equiv 10^{\large I}(\color{#c00}{10^{\large 6J}\!\bmod 7})\equiv \color{#0a0}{10^{\large I}}\,\pmod{\!35}$

Our power $\ 5^{\large 102} = 1\!+\!6J\$ by $\ {\rm mod}\,\ 6\!:\,\ 5^{\large 102}\!\equiv (-1)^{\large 102}\!\equiv 1$

Therefore $\ 10^{\large 5^{\large 102}}\!\! = \color{#0a0}{10^{\large 1+6J}}\!\equiv \color{#0a0}{10^{\large 1}} \pmod{\!35}\$

Remark $\$ For many more worked examples see the complete list of linked questions. Often this distributive law isn’t invoked by name. Rather its trivial proof is repeated inline, e.g. from a recent answer, using $\,cn = 14^2\cdot\color{#c00}{25}\equiv 0\pmod{100}$

\begin{align}&\color{#c00}{{\rm mod}\ \ 25}\!:\ \ \ 14\equiv 8^{\large 2}\Rightarrow\, 14^{\large 10}\equiv \overbrace{8^{\large 20}\equiv 1}^{\rm\large Euler\ \phi}\,\Rightarrow\, \color{#0a0}{14^{\large 10N}}\equiv\color{#c00}{\bf 1}\\[1em] &{\rm mod}\ 100\!:\,\ 14^{\large 2+10N}\equiv 14^{\large 2}\, \color{#0a0}{14^{\large 10N}}\! \equiv 14^{\large 2}\!\! \underbrace{(\color{#c00}{{\bf 1} + 25k})}_{\large\color{#0a0}{14^{\Large 10N}}\!\bmod{\color{#c00}{25}}}\!\!\! \equiv 14^{\large 2} \equiv\, 96\end{align}

First, note that $10^{7}\equiv10^{1}\pmod{35}$.

Therefore $n>6\implies10^{n}\equiv10^{n-6}\pmod{35}$.

Let’s calculate $5^{102}\bmod6$ using Euler’s theorem:

• $\gcd(5,6)=1$
• Therefore $5^{\phi(6)}\equiv1\pmod{6}$
• $\phi(6)=\phi(2\cdot3)=(2-1)\cdot(3-1)=2$
• Therefore $\color\red{5^{2}}\equiv\color\red{1}\pmod{6}$
• Therefore $5^{102}\equiv5^{2\cdot51}\equiv(\color\red{5^{2}})^{51}\equiv\color\red{1}^{51}\equiv1\pmod{6}$

Therefore $10^{5^{102}}\equiv10^{5^{102}-6}\equiv10^{5^{102}-12}\equiv10^{5^{102}-18}\equiv\ldots\equiv10^{1}\equiv10\pmod{35}$.

\begin{align} 10^3&\equiv 6 \bmod 7 \\ &\equiv -1 \bmod 7 \\ \implies 10^6 = (10^3)^2&\equiv 1 \bmod 7 \end{align}
We could reach the same conclusion more quickly by observing that $7$ is prime so by Fermat’s Little Theorem, $10^{(7-1)}\equiv 1 \bmod 7$.
So we need to know the value of $5^{102}\bmod 6$, and here again $5\equiv -1 \bmod 6$ so $5^{\text{even}}\equiv 1 \bmod 6$. (Again there are other ways to the same conclusion, but spotting $-1$ is often useful).
Thus $10^{\large 5^{102}}\equiv 10^{6k+1}\equiv 10^1\equiv 3 \bmod 7$.
\left .\begin{align} x&\equiv 0 \bmod 5 \\ x&\equiv 3 \bmod 7 \\ \end{align} \right\}\implies x\equiv 10 \bmod 35