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The sum of the digits of [(10^x)^y]64=279. What is the
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Updated on: 29 Oct 2012, 00:59
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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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Originally posted by rajathpanta on 28 Oct 2012, 09:18.
Last edited by Bunuel on 29 Oct 2012, 00:59, 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]64=79. What is the value
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29 Oct 2012, 00:59
rajathpanta wrote: The sum of the digits of [(10^x)^y]64=79. What is the value of xy
A. 28 B. 29 C. 30 D. 31 E. 32 The question should read: The sum of the digits of [(10^x)^y]64=279. What is the value of xyA. 28 B. 29 C. 30 D. 31 E. 32 Also, it should be mentioned that xy is a positive integers. First of all \((10^x)^y=10^{xy}\). \(10^{xy}\) has \(xy+1\) digits: 1 and \(xy\) zeros. For example: 10^2=100 > 3 digits: 1 and 2 zeros; \(10^{xy}64\) will have \(xy\) digits: \(xy2\) 9's and 36 in the and. For example: 10^449=10,00049=9,951 > 4 digits: 42=two 9's and 51 in the end; We are told that the sum of all the digits of \(10^{xy}64\) is 279 > \(9(xy2)+3+6=279\) > \(9(xy2)=270\) > \(xy=32\). Answer: E. Similar questions to practice: thesumofallthedigitsofthepositiveintegerqisequal126388.html1025560isdivisiblebyallofthefollowingexcept126300.htmlif105074iswrittenasanintegerinbase10notation51062.htmlHope it's clear.
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Re: The sum of the digits of [(10^x)^y]64=279. What is the
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29 Oct 2012, 01:26
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
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28 Oct 2012, 10:13
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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Re: The Sum of the digits of(10^x)^y
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28 Oct 2012, 10:22
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)



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Re: The Sum of the digits of(10^x)^y
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28 Oct 2012, 18:44
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) = (n2)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]64=79. What is the value
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24 Nov 2013, 18:36
Hi Bunuel, Could you please xplain the last bit oft he equations which takes us to a 279? Thanks quote="Bunuel"] rajathpanta wrote: The sum of the digits of [(10^x)^y]64=79. What is the value of xy
A. 28 B. 29 C. 30 D. 31 E. 32 The question should read: The sum of the digits of [(10^x)^y]64=279. What is the value of xyA. 28 B. 29 C. 30 D. 31 E. 32 Also, it should be mentioned that xy is a positive integers. First of all \((10^x)^y=10^{xy}\). \(10^{xy}\) has \(xy+1\) digits: 1 and \(xy\) zeros. For example: 10^2=100 > 3 digits: 1 and 2 zeros; \(10^{xy}64\) will have \(xy\) digits: \(xy2\) 9's and 36 in the and. For example: 10^449=10,00049=9,951 > 4 digits: 42=two 9's and 51 in the end; We are told that the sum of all the digits of \(10^{xy}64\) is 279 > \(9(xy2)+3+6=279\) > \(9(xy2)=270\) > \(xy=32\). Answer: E. Similar questions to practice: thesumofallthedigitsofthepositiveintegerqisequal126388.html1025560isdivisiblebyallofthefollowingexcept126300.htmlif105074iswrittenasanintegerinbase10notation51062.htmlHope it's clear.[/quote]



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Re: The sum of the digits of [(10^x)^y]64=79. What is the value
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25 Nov 2013, 01:52
Shibs wrote: Hi Bunuel, Could you please xplain the last bit oft he equations which takes us to a 279? Thanks quote="Bunuel"] rajathpanta wrote: The sum of the digits of [(10^x)^y]64=79. What is the value of xy
A. 28 B. 29 C. 30 D. 31 E. 32 The question should read: The sum of the digits of [(10^x)^y]64=279. What is the value of xyA. 28 B. 29 C. 30 D. 31 E. 32 Also, it should be mentioned that xy is a positive integers. First of all \((10^x)^y=10^{xy}\). \(10^{xy}\) has \(xy+1\) digits: 1 and \(xy\) zeros. For example: 10^2=100 > 3 digits: 1 and 2 zeros; \(10^{xy}64\) will have \(xy\) digits: \(xy2\) 9's and 36 in the and. For example: 10^449=10,00049=9,951 > 4 digits: 42=two 9's and 51 in the end; We are told that the sum of all the digits of \(10^{xy}64\) is 279 > \(9(xy2)+3+6=279\) > \(9(xy2)=270\) > \(xy=32\). Answer: E. Similar questions to practice: thesumofallthedigitsofthepositiveintegerqisequal126388.html1025560isdivisiblebyallofthefollowingexcept126300.htmlif105074iswrittenasanintegerinbase10notation51062.htmlHope it's clear. \(10^{xy}64\) will have \(xy\) digits: \(xy2\) 9's and 36 in the and. Threfore the sum of the digits is \(9(xy2)+3+6=279\). Hope it's clear.
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Re: The sum of the digits of [(10^x)^y]64=279. What is the
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08 Jan 2014, 03:19
Hi, this is my process (edited to be the most efficient possible): \(100064= 936\). Whatever XY is you finish with \(36 ==> 3+6=9\) Therefore, \(2799=270\) and \(270/9=30\) Now you add the last two digits (3 and 6) Answer is \(30+2=32\) Hope it helps
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Re: The sum of the digits of [(10^x)^y]64=279. What is the
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30 Jan 2016, 06:00
rajathpanta wrote: 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 Sum of the digits of (10^x)^y  64 = 279. We know that (X^a)^b = X^(ab) Thus, the given form becomes = 10^(xy)  64. Now start with xy=2, 10^2  64 = 36, sum of the digits = 6+3=9 (realize that the sum of digits is 9*(xy1)) xy=3 > 100064=936 = 18 (realize that the sum of digits is 9*(xy1)) xy=4 > 1000064=9936 = 27 (realize that the sum of digits is 9*(xy1)) xy=5 > 10000064 = 99936 (realize that the sum of digits is 9*(xy1))... etc so this is your pattern. The sum of the digits = 9*(xy1) Now sum of the digits given = 279. Thus, based on the pattern above, the sum of the digits must be > 9*(xy1) = 279 > xy = 32. E is thus the correct answer. Hope this helps.



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Re: The sum of the digits of [(10^x)^y]64=279. What is the
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24 Feb 2016, 17:32
gosh..the structure of the question is hideous... i solved it this way.. the sum is 279. the last 2 digits must be 3 and 6, or 9. so the rest will be a bunch of 9's. how many 9's? 30. now, 30 of nines + 36 > 32 digits +1 since we need to round up  33 in total, we thus must have 10^32.



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Re: The sum of the digits of [(10^x)^y]64=279. What is the
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27 Mar 2018, 10:43
Hi All, This question has some awkward wording to it, but here's its intent: the number 10^(xy)  64 has digits that add up to 279. So, we need to figure out what THAT number is. Here's what you need to "see" to solve this problem: IF xy = 3, then 1,000  64 = 936 and the sum of digits = 18 = (2)(9) If xy = 4, then 10,000  64 = 9,936 and the sum of digits = 27 = (3)(9) If xy = 5, then 100,000  64 = 99,936 and the sum of digits = 36 = (4)(9) Notice the pattern? The sum of digits is increasing by 9 every time. So, how many times does 9 divide into 279? 31 times As a reminder of the pattern: xy = 3 > (2)(9) xy = 4 > (3)(9) xy = 5 > (4)(9) So, (31)(9) > xy = 32 Final Answer: GMAT assassins aren't born, they're made, Rich
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