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If 10^50  74 is written as an integer in base 10 notation [#permalink]
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24 Aug 2007, 08:04
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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
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Re: PS 10 notation [#permalink]
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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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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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24 Aug 2007, 13:29
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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...
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*(502) + 2 + 6 = 440



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10^50  74 [#permalink]
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18 May 2010, 23:08
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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]
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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 > (x2) * 9 + 2 + 6. hence, (502) * 9 + 8 > 440.
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Re: 10^50  74 [#permalink]
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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 > (x2) * 9 + 2 + 6. hence, (502) * 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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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 10^5074=9....9926 total number of digits in 10^5074 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 10^5074 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 (502), 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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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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09 Jun 2010, 01:03
Hey Pkit, Please explain your approach. How do you know "total number of digits in 10^5074 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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09 Jun 2010, 06:38
bibha wrote: Hey Pkit, Please explain your approach. How do you know "total number of digits in 10^5074 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 10^35 ? you know that 10^3=1000 (thus 10^50 is a figure that begins with 1 anf has 50 zeros) and 10005=995, so I have two "9" and one "5". the sum is 9+9+5=23 kudos!
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Re: If (10^50) 74 is written as an integer in base 10 notation, [#permalink]
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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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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 (base2 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: valueofn126388.html10126300.htmlHope 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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02 Sep 2013, 20:05
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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....(n2)26.
So for 10^5074, 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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23 Mar 2017, 07:51
great question!
so, we have 10^5074 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 100074=926 this means that 10^3 gets us one 9, hence 503=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]
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