Five digits in our right hand

Let our answer be $a$ .

Observe that, as $10^n\equiv 1\; (mod\; 9)$ ,

$7777^{7777}\equiv x\equiv y\equiv z\equiv a\; (mod\; 9)$ .

Thus, the easiest part of the solution is, $7777\equiv (7+7+7+7)\equiv 1\; (mod\; 9)$

$\implies 7777^ {7777}\equiv 1\; (mod\; 9)$ .

The answer could be $1,10,19,28,37,...$ To get the correct answer, we have to work out an inequality to frame a limit.

In this part, see that $7777^{7777}<(10^5)^{7777}=10^{38885}$ . Which implies, $7777^{7777}$ does not have more than $38886$ digits.

Thus, $x$ cannot have more than $9\times 38886=349974$ . So, the highest possible $y$ occurs when $x=299999$ .

And so, $y\le 47$ . Similarly, the highest possible $z$ occurs when $y=39$ . So, $z\le 12$ .

Consequently, $k<10$ .

This proves that our answer is $\boxed 1$ .

A quick Python solution:

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num = 7777**7777
def digit_sum(x):
    total = 0
    for digit in str(x):
        total+=int(digit)
    return total
x = digit_sum(num)
y = digit_sum(x)
z = digit_sum(y)
print digit_sum(z)