CAT 2025

Slot 3

QA

Question & Solution

$$(10^{50} + 10^{25} - 123)$$ can be written as $$10^{50} + (10^{25}-123)$$

$$1000-123$$ gives $$877$$. 
$$10000-123$$ gives $$9877$$.
$$100000-923$$ gives $$99877$$.
$$1000000-923$$ gives $$999877$$.
And so on.

Considering $$10^n-123$$, which has a total of $$n$$ digits; in general, of the $$n$$ total digits; $$2$$ are the digit $$7$$, $$1$$ is the digit $$8$$, and $$(n-3)$$ are the digit $$9$$.

Also, when added to $$10^{50}$$, the cumulative increase in the sum of the digit of the result would be by $$1$$, this would come from the leftmost digit of the result. All other newly added digits will be zero.

Thus, the sum of the digits of the result will be given by;

$$[7*2] + [1*8] + [(25-3)*9] + [25*0] + [1*1] = 14+8+198+1 = 221$$

The correct answer is option B.