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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

Thanks guys, I am not sure what this question asks...

C.
I am not sure if the GMAT test "base" concept, but basically, base 10 is just regular number.
Find out some numbers:
(10^3)-74 = 926
(10^4)-74 = 9926
(10^5)-74 = 99926
Look at the pattern, the sum of the digits = 9*(50-2) + 2 + 6 = 440

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.
_________________

Hey Pkit, Please explain your approach. How do you know "total number of digits in 10^50-74 is 50, or 48 digits of 9 and two digits 2 and 6"?

Approach is easy Just look at the following example, what is the sum of digits of 10^3-5 ? you know that 10^3=1000 (thus 10^50 is a figure that begins with 1 anf has 50 zeros) and 1000-5=995, so I have two "9" and one "5". the sum is 9+9+5=23

Based 10 notation, or decimal notation, is just a way of writing a number using 10 digits: 1, 2, 3, 4, 5, 6, 7, 8, and 0 (usual way), in contrast, for example, to binary numeral system (base-2 number system) notation.

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

\(10^{50}\) has 51 digits: 1 followed by 50 zeros; \(10^{50}-74\) has 50 digits: 48 9's and 26 in the end;

So, the sum of the digits of \(10^{50}-74\) equals to 48*9+2+6=440.

Re: If 10^50 - 74 is written as an integer in base 10 notation [#permalink]

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22 Sep 2014, 11:30

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