Anyone can help me in solving this?

We wish to find $5^{5555} \pmod{10000}$

Let $5^{5555} \equiv A \pmod{5^4}$ and $5^{5555} \equiv B \pmod{2^4}$ It is obvious $A=0$ and note that $5^8 \equiv 1\pmod{2^4}$ Thus, $B\equiv 5^3\equiv 13\pmod{2^4}$ and $B=13$

Thus, we want to find a number $N$ less than $10000$ that is:

$0\pmod{625}$

$13\pmod{16}$

Testing all the multiples we eventually find $\boxed{8125}$

Surely, I will give you a solution.

Analyze the problem using modulo arithmetic. Finding remainder of divison by 10000 is same as finding last 4 digits. Thus we can write 5^5555 as $\left( 5^{6}\right) ^{925} * 5^5$ . This expression will give us That 5^5555 is congruent to -1 modulo 5^6 as we just need one more 5 to get an expression divisible by 5^6. Thus answer will be 5^7 mod 1000 as 6-(-1)=7. Thus 8125.

You can also do this using totient. Hope this helped. Please upvote if you liked it

$5^4=625\equiv 1\pmod{16}$ so $5^{5551}=5^{4\times 1387+3}\equiv 5^3 \equiv 13\pmod{16}$ . Multiply through with $5^4$ to find $5^{5555}\equiv 13\times 5^4=\boxed{6125}\pmod{10000}$ .

As 10000 is the divisor the remainder will be a 4-digit number.

So we must know the last four digits of $5^{5555}$ .

$5^{4}$ is 0625 and repeats itself every 4 times.

Divide 5555 by 4 remainder will be 3 so we can write the above question as $(5^{4})^{1388}$ * $5^{3}$ .

In the first part the last 4 digits will be 0625 as explained and the second part will be 125. On multiplying them the last 4-digits will be 8125.

Basically $5^{5}$ last 4 digits is 3125, $5^{6}$ last 4 digits is 5625, and $5^{7}$ last 4 digits is 8125.

So every time the remainder is 3 (4 mod (the power)) the last four digits will be 8125.

Because 4 mod 7 is 3 and $5^{7}$ = 78125 = $5^{4}$ * $5^{3}$