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 can simplify it by expressing the bases as prime numbers and attempting to create powers of 10.

The given expression is:

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

First, let's write the bases, 625 and 128, in their prime factorized forms:

$625=5^4$

$128=2^7$

Substituting these prime factorizations into the original expression gives:

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

Using the exponent rule ${\left(a^m\right)}^n=a^{m\times n}$, we multiply the exponents:

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

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

To easily determine the digits of a very large number, it helps to represent part of the number as a power of 10. Since $10=5\times 2$, we need an equal number of factors of 5 and 2. We can separate $5^{260}$ into $5^8\times 5^{252}$ so that its exponent matches the exponent of 2:

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

Now, combine the terms sharing the exponent of 252 by using the rule $a^n\times b^n={(a\times b)}^n$:

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

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

Next, we calculate the exact value of the remaining coefficient, $5^8$:

$5^8={\left(5^4\right)}^2={625}^2=390625$

So the full number simplifies to:

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

This means the number is written as the digits 390625 followed by 252 trailing zeros. Because adding zero does not alter a sum, the sum of the digits of this massive number is exactly the same as the sum of the digits of just the 390625 part.

Sum of digits $=3+9+0+6+2+5=25$