Question

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

Hint:

When multiplying large powers of 2 and 5, group them into powers of 10. This isolates a manageable non-zero block of digits followed by many zeros, simplifying digit-sum problems.

Answer 1

Correct Answer: 25

Step 1: Since \(625=5^4\) and \(128=2^7\), the product is \(5^{260}\times2^{252}\). As there are more \(5\)'s than \(2\)'s, exactly \(252\) pairs of \(2\) and \(5\) combine into \(10^{252}\), leaving \(5^{260-252}=5^8\) uncombined.
Step 2: Compute \(5^8\) by repeated squaring: \(5^2=25\), \(5^4=25^2=625\), \(5^8=625^2=390625\). So the number is \(390625\times10^{252}\).
Step 3: The digit sum is just the digit sum of \(390625\), since the trailing zeros add nothing: \(3+9+0+6+2+5=25\). As a quick check, \(390625\) reduces digit-by-digit to \(2+5=7\), and \(625^{65}\times128^{36} \equiv 7 \pmod 9\) as well, confirming the result.
\[ \boxed{25} \]