I just learnt something new!

So, here's a solution based on the Binomial Theorem.

We have that : ${ 79 }^{ 50 }={ (80-1) }^{ 50 }$ .

And based on the Binomial Theorem, the above expression is equal to:

${ (80-1) }^{ 50 }=\displaystyle \sum _{ k=0 }^{ 50 }{ \binom{50}{k}{ 80 }^{ k }{ (-1) }^{ 50-k } }$ .

Notice that the expansion consists $51$ terms, where the first $50$ are easily proved to be divisible by $1000$ , and the last term is equal to $1$ .

Hence, $\displaystyle \sum _{ k=0 }^{ 50 }{ \binom{50}{k}{ 80 }^{ k }{ (-1) }^{ 50-k } } \equiv 1 \equiv {79}^{50} ( \text {mod } 1000)$ .

Thus, the sum of the last three digits of ${79}^{50}$ is $\boxed {1}$ .

According to Euler's theorem, x^{\phi n} is congruent to 1(mod N) where x and N are co-prime. Finding the last three digits is same as finding the remainder when the number is divided by 1000.

\phi 1000 = N(1-1/a)(1-1/b)(1-1/c)... where N = a^{p} X b^{q} X c^{r} and so on where a, b, c,... are distinct primes.

So, \phi 1000 = 1000(1-1/2)(1-1/5) = 400.

Now, by Euler's theorem, 79^{400} is congruent to 1(mod 1000)........equation 1

And 79^{8} is congruent to 1(mod 1000)...equation 2

Dividing equation 1 with equation 2, we get

79^{50} is congruent to 1(mod 1000.)

So, the last three digits of 79^{50} are 001.

And there sum is \boxed{1}