Habitual

Using the Carmichael function, we find that $5^8\equiv1 \pmod{32}$ . Thus $5^{7994}=5^{8*999+2}\equiv5^2\pmod{2^5}$ . Multiplying through with $5^5$ we find that $5^{7999}\equiv5^7=\boxed{78125}\pmod{10^5}.$

Since $gcd(5^{7999} , 100000) \neq 1$ , we will use CRT.

Obviously, $5^{7999} \equiv 0 \text{ ( mod 3125) }$

We must find $5^{7999} \text{ (mod 32) }$

We know that $\phi(32) = 16 \implies 5^{16} \equiv 1 \text{ (mod 32)}$

$\implies 5^{7999} \equiv 5^{15} \text{ (mod 32)}$

Let $x = 5^{7999}$ .

Since $x \equiv 5^{15}$ this implies that $5x \equiv 5^{16} \equiv 1 \text{ (mod 32)}$

$5x \equiv 1 \text{ (mod 32)}$ - Multiply 13 on both sides.

$x \equiv 13 \text{ (mod 32)}$

Use CRT on

$x \equiv 13 \text{ (mod 32)}$ $x \equiv 0 \text{ (mod 3125)}$

To get that $\boxed{x \equiv 78125 \text{ ( mod 100000)} }$

5^8=...90625, now 90625^2=...890625 so,90625^(2*500)=.....890625, then 5^8000=...890625 now,5^7999=890625/5=78125

If you don't come up with any solution, I suggest researching the way it repeats the last digits. For example, we find out that $5^{6} = ...15625$ $5^{7} = ...78125$ $5^{8} = ...90625$ $5^{9} = ..53125$ $5^{10} = ...65625$ $5^{11} = ...28125$ $5^{12} = ...40625$ $5^{13} = ...03125$ $5^{14} = ...15625$ $5^{15} = ...78125$ ... We could observe that the repeating round has 8 elements, and $7999 = 999 \times 8 + 7$ . Hence, the last five digits of $5^{7999}$ are $\boxed{78125}.$

This solution is based on the lemma below. First we note that $5^7 = 78125$ , and $5^{8} = ...90625$ . Further, $5^{7}5^{8} = ....78125$ and $90625^{2} =...90625.$ Thus $5^{8k} = ...90625$ for any whole number k. Hence $5^{7999}= 5^{7}(5^{8})^{999}$ $=....78125\times...90625 = ...78125.$ In conclusion $5^k$ will end in 78125 when ever k mod 8 = 7.

Lemma. Let A and B be whole numbers with n digits each. Further, let P be a large whole number whose last n digits are A, and let Q be a large whole number whose last n digits are B. Then the last n digits of the product PQ will be the same as the last n digits of the product AB.

Proof: Let t be the number of digits of P and u be the number of digits of Q, naturally t>n and u>n. Then P = $10^{t}+A$ and Q = $10^{u}+B$ . Distributing, we get PQ = $10^{t+u}+10^{t}B+10^{u}A+AB$ , a number whose last n digits are precisely the last n digits of AB.