Find remainder when $8^{13^{48}}$ divided by 1000

Question

Find the remainder when $8^{(13^{48})}$ divided by $1000$.

First I use $\phi$ function properties but stuck how can I do?

Remark : $\phi (1000)$ is relative prime that lower than $1000$ and is integer

Do you mean $(8^{13})^{48}$ or $8^{(13^{48})}$? The power operation is not associative... — Klaus, Jan 21 '19 at 15:46
Do you mean $\left( 8^{13}\right)^{48}$ or $8^{\left( 13^{48}\right)}$? — For the love of maths, Jan 21 '19 at 15:47
How dd you use the $\varphi$ function properties, exactly? If you don't tell us, we won't know where you went wrong. — TonyK, Jan 21 '19 at 15:56
It seems clear by the asker's edit that they mean $\displaystyle 8^{(13^{48})}$. — jordan_glen, Jan 21 '19 at 15:57
The $\varphi$ which you are referring to is the Euler totient function. You might want to add this as a tag and include it somewhere in the question. — R. Burton, Jan 21 '19 at 16:15
You could have a look at: How do I compute $a^b,\bmod c$ by hand?. — Martin Sleziak, Jan 21 '19 at 17:31

score 4 · Accepted Answer · answered Jan 21 '19 at 16:23

Here is a crude solution:

Finding $(8^{13})^{48}$ mod 1000 is relatively straightforward. So this answer will be geared toward answering what $8^{(13^{48})}$ mod 1000 is, the methods can be adapted to answer the previous question.

The Euler generalization of Fermat's theorem states: if $\gcd(a, m) = 1$, then we have that $$a^{\phi(m)} \equiv 1 \pmod{m},$$ where $\phi(m)$ is the Euler Totient function. As a consequence of this theorem, we have that the powers of a form a "cycle" of $\phi(m)$. So if we want to figure out what $a^k$ is congruent to modulo m, it suffices to figure out what $k$ is congruent to modulo $\phi(m)$, since it is easier to calculate this power.

We can't immediately use the Euler generalization here, since $\gcd(8, 1000) = 8 \neq 1$. However, we can reduce the original problem to an application of the Fermat Generalization as follows: Let $x$ be the number which is the remainder of $8^{(13^{48})}$ when divided by 1000. This number satisfies the following system of congruences: $$x \equiv 0 \pmod{8}$$$$x \equiv ? \pmod{125}.$$ Since $\gcd(125, 8) = 1$, we can use the Fermat Generalization now to find $?$.

Since $\phi(1000) = 400$, we find what $13^{48}$ is mod $400$. To do this, we find what $13^{16}$ and $13^{32}$ are mod $400$, this only takes 5 multiplications to do. They are $241$ and $81$ respectively, so we figure out that $$13^{48} = 13^{32}\times 13^{16} \equiv 241 \times 81 \equiv 321 \pmod{400}.$$ So we reduce the problem to finding $8^{321} \pmod{125}$. We have that $\phi(125) = 100$. So $$8^{321} \equiv 8^{21} \pmod{125}.$$ Finding the remainders of $8$, $8^4$, and $8^{16}$ modulo 125, which takes 4 multiplications, we find that $8^{21} \equiv 58 \pmod{125}$.

So now we have the following: $$x \equiv 0 \pmod{8}$$$$x\equiv 58 \pmod{125}.$$

There is only one number less than 1000 satisfying this property, and it is $808$. Hence the remainder is $808$ and we are done.

Find remainder when $8^{13^{48}}$ divided by 1000

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