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The sum of the digits of [(10^x)^y]-64=279. What is the [#permalink]
 28 Oct 2012, 10:18
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The sum of the digits of [(10^x)^y]-64=279. What is the value of xy ? A. 28 B. 29 C. 30 D. 31 E. 32
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Last edited by Bunuel on 29 Oct 2012, 01:59, edited 2 times in total.
Renamed the topic and edited the question.
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Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 11:13
I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here.
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Senior Manager
Status: Prevent and prepare. Not repent and repair!!
Joined: 13 Feb 2010
Posts: 263
Location: India
Concentration: Technology, General Management
GPA: 3.75
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Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 11:22
Pansi wrote: I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here. Well, It is a multiple of 9. How will you arrive at xy with that approach? Try finding patterns. (thats the clue)
_________________
I've failed over and over and over again in my life and that is why I succeed--Michael Jordan
Kudos drives a person to better himself every single time. So Pls give it generously
Wont give up till i hit a 700+
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Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 19:44
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Pansi wrote: I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here. Well, simple reason is that the question is incorrect. rajathpanta wrote: Well, It is a multiple of 9. How will you arrive at xy with that approach?
Try finding patterns. (thats the clue) Question is: 10^xy -64 = N, where sum of digits of N=79 The pattern is like this: 100 -64 = 36 1000 -64 = 936 10000 -64 =9936 or, 1 followed by (n times 0) = (n-2)times 9 followed by 36 Therefore sumof digits on right side is always a multiple of 9 [9s and 6+3 =9] However in question stem RHS is 79, which is not divisible by 9. And therefore you can not arrive at any of the answer choices given. Rajathpanta- on a lighter note - if this too is from Aristotle, I'd suggest please change the source of questions. :D Hope it helps!
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Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 20:40
still confuse with question any more explanation
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Senior Manager
Status: Prevent and prepare. Not repent and repair!!
Joined: 13 Feb 2010
Posts: 263
Location: India
Concentration: Technology, General Management
GPA: 3.75
WE: Sales (Telecommunications)
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Kudos [?]:
10
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Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 20:42
Vips0000 wrote: Pansi wrote: I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here. Well, simple reason is that the question is incorrect. rajathpanta wrote: Well, It is a multiple of 9. How will you arrive at xy with that approach?
Try finding patterns. (thats the clue) Question is: 10^xy -64 = N, where sum of digits of N=79 The pattern is like this: 100 -64 = 36 1000 -64 = 936 10000 -64 =9936 or, 1 followed by (n times 0) = (n-2)times 9 followed by 36 Therefore sumof digits on right side is always a multiple of 9 [9s and 6+3 =9] However in question stem RHS is 79, which is not divisible by 9. And therefore you can not arrive at any of the answer choices given. Rajathpanta- on a lighter note - if this too is from Aristotle, I'd suggest please change the source of questions. :D Hope it helps! Hi Vips00, This is from the veritas prep questions set! Thanks.
_________________
I've failed over and over and over again in my life and that is why I succeed--Michael Jordan
Kudos drives a person to better himself every single time. So Pls give it generously
Wont give up till i hit a 700+
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Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 21:47
rajathpanta wrote: Vips0000 wrote: Pansi wrote: I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here. Well, simple reason is that the question is incorrect. rajathpanta wrote: Well, It is a multiple of 9. How will you arrive at xy with that approach?
Try finding patterns. (thats the clue) Question is: 10^xy -64 = N, where sum of digits of N=79 The pattern is like this: 100 -64 = 36 1000 -64 = 936 10000 -64 =9936 or, 1 followed by (n times 0) = (n-2)times 9 followed by 36 Therefore sumof digits on right side is always a multiple of 9 [9s and 6+3 =9] However in question stem RHS is 79, which is not divisible by 9. And therefore you can not arrive at any of the answer choices given. Rajathpanta- on a lighter note - if this too is from Aristotle, I'd suggest please change the source of questions. :D Hope it helps! Hi Vips00, This is from the veritas prep questions set! Thanks. Hmmm, but you get that the question is incorrect and why?
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Director
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Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 21:49
Aristocrat wrote: still confuse with question
any more explanation What is confusing? I already explained it in detail. If there is any particular thing you could not understand let me know, would try to explain further.
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Re: The sum of the digits of [(10^x)^y]-64=79. What is the value [#permalink]
29 Oct 2012, 01:59
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Re: The sum of the digits of [(10^x)^y]-64=279. What is the [#permalink]
29 Oct 2012, 02:26
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As Question is (10^x)^y - 64 . Let say (10^x)^y as Number1 Say Number1 - 64 = Number2 ==> 100 - 64 = 36 [ Number1: No. of zeroes = 2 , Number2: No. of 9's = zero ] and Sum of digits of Number 2 : 9*0 + (3+6) = 1*9 = 9 1000 - 64 = 936 [ Number1: No. of zeroes = 3 , Number2: No. of 9's = 1] and Sum of digits of Number 2 : 9*1 + (3+6) = 9 + 9 = 2*9 = 18 10000 - 64 = 9936 [ Number1: No. of zeroes = 4 , Number2: No. of 9's = 2] and Sum of digits of Number 2 : 9*2 + (3+6) = 18 + 9 = 3*9= 27 100000 - 64 = 99936 [ Number1: No. of zeroes = 5 , Number2: No. of 9's = 3] and Sum of digits of Number 2 : 9*3 + (3+6) = 27 + 9 =4*9= 36 so lets go from right to left for the sum of digits of number2 i.e given as 279 so 279 = 31*9 = 9*30 + (3+6) => Number2: Number of 9's = 30 ==> Number1: Number of zeros = 32 So the Number1 i.e. (10^x)^y = 10000.....(32 zeroes) Now, as we now, 10^1 = 10 (1 zero) 10^2 = 100 (2 zeroes) 10^3 = 1000 (3 zeroes) same way, 10000.....(32 zeroes) = 10^32(10^x)^y = 10^(xy) = 10^32==> xy = 32
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Re: The sum of the digits of [(10^x)^y]-64=279. What is the
[#permalink]
29 Oct 2012, 02:26
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