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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]

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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]

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New post 24 Aug 2007, 08:33
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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]

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New post 24 Aug 2007, 09:06
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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]

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New post 24 Aug 2007, 13:29
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]

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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]

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New post 18 May 2010, 23:38
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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]

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New post 19 May 2010, 00:41
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]

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New post 19 May 2010, 01:06
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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]

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New post 19 May 2010, 11:00
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]

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New post 09 Jun 2010, 01:03
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]

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New post 09 Jun 2010, 06:38
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]

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New post 25 Jan 2012, 09:09

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

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New post 25 Jan 2012, 10:41
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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]

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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]

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New post 23 Mar 2017, 07:51
great question!

so, we have 10^50-74
In base 10 notation means in usual system
as 10^50 is too big number, let's take sth simple, e.g. 10^3=1000
1000-74=926
this means that 10^3 gets us one 9, hence 50-3=47+1(because including the numbers)=48
48*9+2+6=440
Answer is C

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Re: If 10^50 - 74 is written as an integer in base 10 notation   [#permalink] 23 Mar 2017, 07:51
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If 10^50 - 74 is written as an integer in base 10 notation

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