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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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

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VJesus12 wrote:
Thu Sep 16, 2021 10:57 am
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
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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.
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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
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