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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+

Last edited by Bunuel on 29 Oct 2012, 00:59, edited 2 times in total.

Re: The Sum of the digits of(10^x)^y [#permalink]
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. _________________

I will rather do nothing than be busy doing nothing - Zen saying

Re: The Sum of the digits of(10^x)^y [#permalink]
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) _________________

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+

Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 18:44

1

This post received KUDOS

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

Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 19:42

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

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+

Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 20:47

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

Re: The Sum of the digits of(10^x)^y [#permalink]
28 Oct 2012, 20:49

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. _________________

Re: The sum of the digits of [(10^x)^y]-64=279. What is the [#permalink]
29 Oct 2012, 01:26

1

This post received KUDOS

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)

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