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How to find the remainder of $10^{5^{102}}$ mod $35$, in this question I cannot find "$a$" such that $10^a$ congruent to $1$ and use the basic trick. This makes me pretty confused, any help is really appreciated!

Hint: Find the quantity $bmod 5$ (shouldn't be hard given that the base is $10$) and then find it $bmod 7$ (using some $a$ so that $10^aequiv 1bmod 7$ like you mentioned). From here, splice them together with the Chinese remainder theorem.

Set $x= 10^{5^{102}}$. Clearly
$$
x=0 mod 5.
$$
You have to solve now:
$$
x=10^{5^{102}} mod 7,
$$
that is:
$$
x=3^{5^{102}} mod 7.
$$
Since $7$ is prime and $7$ is not a divisor of $3$ , Fermat's little theorem says
$$
3^6=1 mod 7,
$$
therefore:
$$
3^{5^{102}}=3^y mod 7 quad text{ if }quad 5^{102}=y mod 6.
$$
Now,
$$
y=5^{102} mod 6,\
y=(-1)^{102} mod 6,\
y=1 mod 6.
$$
This means:
$$
x=3^1=3 mod 7.\
$$
Now, using:
$$
x=0 mod 5,\
x=3 mod 7,\
$$
you should be able to prove:
$$
x = 10 mod 35.
$$
(Chinese remainder theorem tells you that this solution is unique mod 35)

You can't find it because $10$ and $35$ are not relatively prime and there is none.

As $gcd(10, 35)= 7$ every $10^k$ will be equivalent to a multiple of $5$. we can attempt to find $10^k equiv 10pmod {35}$....

But the Chinese Remeander Theorem is much easier.

CRT says: If $10^{5^{102}} equiv j pmod {5}$ and $10^{5^{102}} equiv m pmod 7$, there is unique equivalence class $n pmod{5*7}$ so that $nequiv j equiv 10^{5^{102}} pmod 5$ and $nequiv m equiv 10^{5^{102}}pmod 5$. And from $m,j$ will be able to find that $n$.

$10equiv 0 pmod 5$ so $10^{5^{102}} equiv 0 pmod 5$.

Now by Fermat Little Theorem $10^6 equiv 1 pmod 7$. And so "the trick" is if $a = 6b + r$ or in other words is $a equiv r pmod 6$ then $10^a = (10^6)^b10^r equiv 10^r pmod 7$ so

$10^{5^{102}} = 10^{(6-1)^{102}} equiv 10^{(-1)^{102}} =10^1equiv 3 pmod 7$.

So we nee to solve $n equiv 0 pmod 5$ and $nequiv 3 pmod 7$. By CRT there is one and also by CRT there is only one and as $10^{5^{102}}$ is a solution, $n equiv 10^{5^{102}} pmod 35$.

And as $10^k equiv 0 equiv 10 pmod 5$ and $10^{5^{102}}equiv 10 pmod 7$ we know.... $n equiv 10 mod 35$ is that solution. $10 equiv 0 pmod 5$ and $10equiv 3 pmod 7$.

So $10^{5^{102}}equiv 0 pmod 5$ and $10^{4^{102}}equiv 3pmod 7iff 10^{5^{102}} equiv 10 pmod 35$.

$$10^{large 25^{LARGE n}}!!bmod 35, =, overbrace{5left[ 10^{largecolor{#c00}{ 25}^{LARGE n}}!!/5bmod 7right], =, 5left[10^{large color{#c00}{bf 1}^{LARGE n}}!!/5bmod 7right]}^{ Large 10^{LARGE color{#0a0} 6}equiv, 1!pmod{!7}; color{#c00}{25 equiv 1}pmod{!color{#0a0}6}} =, 5[2], =, 10qquad$$