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

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The correct option is 221.

To find the sum of all the digits of the number:

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

Let us break down the calculation step-by-step:

Step 1: Simplify the term ${10}^{25}-123$

The term ${10}^{25}$ represents 1 followed by 25 zeros (a 26-digit number). When we subtract 123 from ${10}^{25}$, we get a 25-digit number:

${10}^{25}-123=999\dots 999877$

Analyzing the digits of this 25-digit result:

• The last 3 digits are 8, 7, 7 (since 1000 − 123 = 877).

• The remaining 22 digits (from the 4th digit to the 25th digit from the right) are all 9s.

Step 2: Add ${10}^{50}$ to the result

The term ${10}^{50}$ adds a single 1 at the 51st position from the right. The positions between the 26th digit and the 50th digit are all 0s.

Thus, the full number consists of:

• One 1 at the highest place value (${10}^{50}$ position)

• 25 zeros between the 26th and 50th places

• 22 nines (9s)

• One 8

• Two 7s

Step 3: Calculate the sum of all digits

Sum of digits = (1) + (25 × 0) + (22 × 9) + 8 + 7 + 7

Sum of digits = 1 + 0 + 198 + 22

Sum of digits = 221

Therefore, the sum of all the digits of the given number is 221.