Dgdrxrt

I stumbled across this problem

Find $,10^large 5^102$ modulo $35$, i.e. the remainder left after it 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?

The "logic" is that we can use a mod distributive law to pull out a common factor $,c=5,,$ i.e.

$$ cabmod cn =, c(abmod n)quadqquad $$

This decreases the modulus from $,cn,$ to $,n, ,$ simplifying modular arithmetic. Also it may eliminate CRT = Chinese Remainder Theorem 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.

$$quad qquad caequiv c(abmod n) rm if color#d0fcnequiv 0 pmod! m$$

because we have: $, c(abmod n) equiv c(a! +! kn)equiv ca+k(color#d0fcn)equiv capmod!m$

e.g. in the OP: $ Ige 1,Rightarrow, 10^large I+N!equiv 10^large I(10^large N!bmod 7) rm by 10^I 7equiv 0,pmod35$

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

i.e. notice that $ color#c00rm mod, 7!:, 10^large 6equiv, 1,Rightarrow, color#c0010^large 6Jequiv 1. $ Thus if $ I ge 1 $ then as above

$phantomrm mod, 35!:, color#0a010^large I+6J!equiv 10^large I 10^large 6J!equiv 10^large I(color#c0010^large 6J!bmod 7)equiv color#0a010^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#0a010^large 1+6J!equiv color#0a010^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 $,cn = 14^2cdotcolor#c0025equiv 0pmod100$

$beginalign&color#c00rm mod 25!: 14equiv 8^large 2Rightarrow, 14^large 10equiv overbrace8^large 20equiv 1^rmlarge Euler phi,Rightarrow, color#0a014^large 10Nequivcolor#c00bf 1\[1em]
&rm mod 100!:, 14^large 2+10Nequiv 14^large 2, color#0a014^large 10N! equiv 14^large 2!! underbrace(color#c00bf 1 + 25k)_largecolor#0a014^Large 10N!bmodcolor#c0025!!! equiv 14^large 2 equiv, 96endalign$

This distributive law is actually equivalent to CRT as we sketch below, with $,m,n,$ coprime

$beginalign x&equiv a!!!pmod! m\ color#c00x&equivcolor#c00 b!!!pmod! nendalign$
$,Rightarrow, x!-!abmod mn, =, mleft[dfraccolor#c00x-ambmod nright] = mleft[dfraccolor#c00b-ambmod nright]$

which is exactly the same solution given by Easy CRT. But the operational form of this law often makes it more convenient to apply in computations versus the classical CRT formula.

First, note that $10^7equiv10^1pmod35$.

Therefore $n>6implies10^nequiv10^n-6pmod35$.

Let's calculate $5^102bmod6$ using Euler's theorem:

• $gcd(5,6)=1$


• Therefore $5^phi(6)equiv1pmod6$


• $phi(6)=phi(2cdot3)=(2-1)cdot(3-1)=2$


• Therefore $colorred5^2equivcolorred1pmod6$


• Therefore $5^102equiv5^2cdot51equiv(colorred5^2)^51equivcolorred1^51equiv1pmod6$

Therefore $10^5^102equiv10^5^102-6equiv10^5^102-12equiv10^5^102-18equivldotsequiv10^1equiv10pmod35$.

Carrying on from your calculation:
$$beginalign
10^3&equiv 6 bmod 7 \
&equiv -1 bmod 7 \
implies 10^6 = (10^3)^2&equiv 1 bmod 7
endalign$$
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 $5^102bmod 6$, and here again $5equiv -1 bmod 6 $ so $5^textevenequiv 1 bmod 6$. (Again there are other ways to the same conclusion, but spotting $-1$ is often useful).

Thus $10^large 5^102equiv 10^6k+1equiv 10^1equiv 3 bmod 7$.

Now the final step uses the Chinese remainder theorem for uniqueness of the solution (to congruence):
$$left .beginalign
x&equiv 0 bmod 5 \
x&equiv 3 bmod 7 \
endalign
right}implies xequiv 10 bmod 35 $$