If \(10^{50}-74\) is written as an integer in base \(10\) notation, what is the sum of the digits in that integer?

A. 424

B. 433

C. 440

D. 449

E. 467

Answer: C

Source: Official Guide

## If \(10^{50}-74\) is written as an integer in base \(10\) notation, what is the sum of the digits in that integer?

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One approaches to look for a pattern...

10^3 - 74 = 1,000 - 74 = 926 (1 nine)

10^4 - 74 = 10,000 - 74 = 9926 (2 nines)

10^5 - 74 = 100,000 - 74 = 99926 (3 nines)

10^6 - 74 = 1,000,000 - 74 = 999926 (4 nines)

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In general, we can see that 10^n - 74 will feature n-2 9's followed by 26

So, 10^50 - 74 will feature 48 9's followed by 26

This means the sum of its digits = 48(9) + 2 + 6 = 432 + 2 + 6 =

**440**

Answer: C