Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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=79. What is the value [#permalink]
29 Oct 2012, 00:59

3

This post received KUDOS

Expert's post

5

This post was BOOKMARKED

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 xy

A. 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: \(xy-2\) 9's and 36 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of \(10^{xy}-64\) is 279 --> \(9(xy-2)+3+6=279\) --> \(9(xy-2)=270\) --> \(xy=32\).

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)

Re: The sum of the digits of [(10^x)^y]-64=79. What is the value [#permalink]
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 xy

A. 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: \(xy-2\) 9's and 36 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of \(10^{xy}-64\) is 279 --> \(9(xy-2)+3+6=279\) --> \(9(xy-2)=270\) --> \(xy=32\).

Re: The sum of the digits of [(10^x)^y]-64=79. What is the value [#permalink]
25 Nov 2013, 01:52

1

This post received KUDOS

Expert's post

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 xy

A. 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: \(xy-2\) 9's and 36 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of \(10^{xy}-64\) is 279 --> \(9(xy-2)+3+6=279\) --> \(9(xy-2)=270\) --> \(xy=32\).

Re: The sum of the digits of [(10^x)^y]-64=279. What is the [#permalink]
21 Jan 2015, 22:41

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...