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If 10^50 - 74 is written as an integer in base 10 notation

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If 10^50 - 74 is written as an integer in base 10 notation [#permalink] New post 24 Aug 2007, 07:04
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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
[Reveal] Spoiler: OA
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Re: PS 10 notation [#permalink] New post 24 Aug 2007, 07:33
minnu wrote:
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...


the question is little confusing:

= (48x9) + 2 + 6
= 432+2+6
= 440
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 [#permalink] New post 24 Aug 2007, 08:06
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solving for easier numbers ---> 10^3 and 74

(10^4) - 74 =

(10^4) - 10*7.4 =

10*[(10^3) - 7.4] =

10*[10^3 - 7.4] = 992.6*10 = 9926 = 9*2+2+6 = 26

Note that 10^4 will yield two nines a six and a two.

so solving for 10^50 and 74 will give 48 nines a six and a two:

10*[10^49 - 7.4] = 9*48+2+6 = 440

the answer is (C)

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Re: PS 10 notation [#permalink] New post 24 Aug 2007, 12:29
minnu wrote:
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
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10^50 - 74 [#permalink] New post 18 May 2010, 22:08
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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
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Re: 10^50 - 74 [#permalink] New post 18 May 2010, 22:38
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C. 440

10^x - 74 --> last 2 digits are always 2, 6.

10^2 - 74 = 26

10^3 - 74 = 926

10^4 - 74 = 9926 and so on....

If x > 1,
the sum of the digits --> (x-2) * 9 + 2 + 6. hence, (50-2) * 9 + 8 --> 440.
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Re: 10^50 - 74 [#permalink] New post 18 May 2010, 23:41
nsp007 wrote:
C. 440

10^x - 74 --> last 2 digits are always 2, 6.

10^2 - 74 = 26

10^3 - 74 = 926

10^4 - 74 = 9926 and so on....

If x > 1,
the sum of the digits --> (x-2) * 9 + 2 + 6. hence, (50-2) * 9 + 8 --> 440.


great approach but why are you using x here ? as you used ..."If x > 1," ..
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Re: 10^50 - 74 [#permalink] New post 19 May 2010, 00:06
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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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Re: 10^50 - 74 [#permalink] New post 19 May 2010, 10:00
great approach but why are you using x here ? as you used ..."If x > 1," ..


I meant to say that only when x > 1, for 10^x - 74.. the last 2 digits are 2, 6.
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Re: 10^50 - 74 [#permalink] New post 09 Jun 2010, 00:03
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"?
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Re: 10^50 - 74 [#permalink] New post 09 Jun 2010, 05:38
bibha wrote:
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

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Re: If (10^50) 74 is written as an integer in base 10 notation, [#permalink] New post 25 Jan 2012, 08:09
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Re: If (10^50) 74 is written as an integer in base 10 notation, [#permalink] New post 25 Jan 2012, 09:41
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Baten80 wrote:
What does "in base 10 notation" mean?


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.

Answer: C.

Similar questions:
value-of-n-126388.html
10-126300.html

Hope it helps.
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Re: If 10^50 - 74 is written as an integer in base 10 notation [#permalink] New post 24 Aug 2013, 22:23
minnu wrote:
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


Very good question and the explanation was nicely written :-)
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Re: If 10^50 - 74 is written as an integer in base 10 notation [#permalink] New post 02 Sep 2013, 19:05
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10^1 = 10
10^2 - 74 = 026
10^3 - 74 = 926
10^4 - 74 = 9926

Basically for 10^n , its 9999....(n-2)26.

So for 10^50-74, it is 99999....4826

48times9 + 2+6 = 440.
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Re: If 10^50 - 74 is written as an integer in base 10 notation [#permalink] New post 22 Sep 2014, 10:30
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Re: 10^50 - 74 [#permalink] New post 31 Jan 2016, 11:44
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Re: 10^50 - 74   [#permalink] 31 Jan 2016, 11:44
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