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

Let \[ N = (625)^{65} \times (128)^{36}. \] Step 1: Express numbers in prime powers. \[ 625 = 5^4, \qquad 128 = 2^7. \] Substitute: \[ N = (5^4)^{65} \times (2^7)^{36}. \]
Step 2: Simplify exponents. \[ (5^4)^{65} = 5^{260}, \qquad (2^7)^{36} = 2^{252}. \] Thus, \[ N = 5^{260} \times 2^{252}. \]
Step 3: Pair powers of 2 and 5 to form tens. Match \(2^{252}\) with \(5^{252}\): \[ N = 5^{260} \cdot 2^{252} = 5^{8} \cdot (5^{252} \cdot 2^{252}) = 5^{8} \cdot 10^{252}. \]
Step 4: Compute \(5^8\). \[ 5^4 = 625,\qquad 5^8 = 625 \times 625 = 390625. \] So, \[ N = 390625 \times 10^{252}. \] This is the number \(390625\) followed by 252 zeros.
Step 5: Sum the digits. Zeros contribute nothing, so compute: \[ 3 + 9 + 0 + 6 + 2 + 5 = 25. \] Thus, the sum of digits of the number is \[ \boxed{25}. \]