The sum of all the digits of the number $(10^{50}+10^{25}-123)$ , is

324

212

255

221

✅ Correct Option: 4

$10^{50} + 10^{25} - 123 = 10^{50} + (10^{25} - 123)$

$10^{25}$ is $1$ followed by $25$ zeros.

Subtracting $123$ from $10^{25}$ : the last $3$ zeros become $877$ , and the remaining $25 - 3 = 22$ zeros each become $9$ .

$10^{25} - 123 = \underbrace{99\ldots9}_{22} \ 877$

This is a $25$ -digit number.

$10^{50}$ is $1$ followed by $50$ zeros, which is a $51$ -digit number.

Adding $10^{50}$ places a $1$ in the $51$ st digit position, with $25$ zeros filling the gap before the $25$ -digit number.

$10^{50} + 10^{25} - 123 = 1\underbrace{00\ldots0}_{25}\underbrace{99\ldots9}_{22} \ 877$

Sum of digits $= 1 + (25 \times 0) + (22 \times 9) + 8 + 7 + 7$

$= 1 + 0 + 198 + 22$

$= 221$

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