dimitri92 wrote:

If 10^50 - 74 is written as an integer in a base 10 notation.What is the sum of the digits in that integer?

a. 424

b. 433

c. 440

d. 449

e. 467

C. 440

another approach is:

We know that 10^50 is ending 00, so 10^50-74=9....9926

total number of digits in 10^50-74 is 50, or 48 digits of 9 and two digits 2 and 6.

answer choice is 48*9+8=440

plugging numbers:

let represent the sum of the digits in that integer as Y, with the reminder 8, we can represent it in form Y=X*9+8, where X number of digits in 10^50-74 and 8=2+6.

Start with C and than move to B or D.

B. 433=X*9+1, X=48

C. 440=X*9+8, X=48 -

correct as we have the reminder 8 and 48 number of digits (50-2), 2 digits are 26. D. 449=X*9+8, X=49

Personally, I like NSP007's approach. My approaches are easy to comprehend.

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