The sum of digits of the number $\left(625\right){}^{65}\times \left(128\right){}^{36}$  is

The correct answer is 25.

To find the sum of the digits of the given expression, we first simplify the expression by expressing the bases as powers of prime numbers (specifically 2 and 5).

The given expression is:

$(625{)}^{65}\times (128{)}^{36}$

First, express the base numbers 625 and 128 as powers of primes:

$625=5^4$

$128=2^7$

Now, substitute these into the main expression:

$(5^4{)}^{65}\times (2^7{)}^{36}$

Apply the exponent rule $(a^m{)}^n=a^{m\times n}$ :

$5^{4\times 65}=5^{260}$

$2^{7\times 36}=2^{252}$

So the expression becomes:

$5^{260}\times 2^{252}$

Next, group powers of 2 and 5 together to form powers of 10 (since $2\times 5=10$ ):

$5^{260}\times 2^{252}=5^8\times 5^{252}\times 2^{252}$

$=5^8\times (5\times 2{)}^{252}$

$=5^8\times {10}^{252}$

Now, evaluate $5^8$ :

$5^8=390625$

Therefore, the number is:

$390625\times {10}^{252}$

Multiplying 390625 by ${10}^{252}simply\; appends\; 252\; zeros\; to\; the\; right\; of\; 390625.\; Since\; zeros\; do\; not\; contribute\; to\; the\; sum\; of\; digits,\; we\; only\; need\; to\; sum\; the\; digits\; of\; 390625:$

$\text{Sum of digits}=3+9+0+6+2+5=25$

Thus, the sum of the digits of the given number is 25.