The sum of digits of the number (625)^65 x (128)^36, is

Answer: 25. Solution: Expressing each base as a power of a prime number: 625 = 5^4 and 128 = 2^7 (5^4)^65 x (2^7)^36 = 5^260 x 2^252 Pairing up 2s and 5s to form powers of 10 (since 2 x 5 = 10): = (2 x 5)^252 x 5^(260-252) = 10^252 x 5^8 5^2 = 25 5^4 = 25^2 = 625 5^8 = 625^2 = 390625 10^252 x 390625 = 3906250000_(252 zeros) Multiplying by 10^252 simply adds 252 zeros at the end, so the significant digits remain 390625. Sum of digits = 3 + 9 + 0 + 6 + 2 + 5 + 0 + 0 + *s + 0 = 3 + 9 + 0 + 6 + 2 + 5 = 25

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

Entered answer:

✅ Correct Answer: 25

Expressing each base as a power of a prime number:

$625 = 5^4 \quad$ and $\quad 128 = 2^7$

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

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

Pairing up 2s and 5s to form powers of 10 (since $2 \times 5 = 10$ ):

$= (2 \times 5)^{252} \times 5^{260-252}$

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

$5^2 = 25$

$5^4 = 25^2 = 625$

$5^8 = 625^2 = 390625$

$10^{252} \times 390625 = 390625\underbrace{000\ldots0}_{252 \text{ zeros}}$

Multiplying by $10^{252}$ simply adds 252 zeros at the end, so the significant digits remain $390625$ .

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

$= 3 + 9 + 0 + 6 + 2 + 5$

$= 25$

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