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 500,000 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:
Re: N = 1000x + 100y + 10z, remainder of N/9 [#permalink]
19 Oct 2012, 06:12
1
This post received KUDOS
Expert's post
Anichka wrote:
If N = 1000x + 100y + 10z, where x, y, and z are different positive integers less than 4, the remainder when N is divided by 9 is
(A) 2 (B) 4 (C) 6 (D) 8 (E) 9
Since we cannot have more than one correct answers in PS questions, then pick some numbers for x, y, and z and find the reminder when 1000x + 100y + 10z is divided by 9.
Say x=1, y=2, and z=3, then 1000x + 100y + 10z = 1,230 --> 1,230 divide by 9 yields the remainder of 6 (1,224 is divisible by 9 since the sum of its digit is a multiple of 9, thus 1,230, which is 6 more than a multiple of 9, yields the remainder of 6 when divided by 9).
Re: N = 1000x + 100y + 10z, remainder of N/9 [#permalink]
22 Oct 2012, 06:20
1
This post received KUDOS
Expert's post
crackatgmat wrote:
Hi Bunuel - But what happens when x = 2, y =3 and z = 1. Then the reminder is something different. Can we expect such questions in GMAT. Thanks
Since we cannot have more than one correct answer in PS questions, then for ANY values of x, y, and z which satisfy the given condition (different positive integers less than 4) we must get the same answer, otherwise the question would be flawed.
If you check the remainder when x = 2, y =3 and z = 1, then you'll see that it'll still be 6.
Re: N = 1000x + 100y + 10z, remainder of N/9 [#permalink]
22 Oct 2012, 07:51
3
This post received KUDOS
Anichka wrote:
If N = 1000x + 100y + 10z, where x, y, and z are different positive integers less than 4, the remainder when N is divided by 9 is
(A) 2 (B) 4 (C) 6 (D) 8 (E) 9
Since x, y, and z are distinct positive integers less than 4, they can be 1, 2, or 3, doesn't matter which one is what. Then N = 10(100x + 10y + z) = xyz0 is 10 times a three digit number formed by any of the permutations of the digits 1, 2, and 3. For example 1230, 3210, 3120,...in fact, is quite a short list, only 3! = 6 numbers. The divisibility rule by 9 says that the sum of the digits of the number and the number itself, leave the same remainder when divided by 9. For example, when the remainder is 0, the number and the sum of its digit, are both divisible by 9.
Since the sum of the digits of N is 1 + 2 + 3 = 6, the remainder when N is divided by 9, is 6.
Answer C. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: N = 1000x + 100y + 10z, remainder of N/9 [#permalink]
24 Oct 2012, 18:41
3
This post received KUDOS
1
This post was BOOKMARKED
mindmind wrote:
Is there any other method to solve the same.... instead of plugging in the values and dividing them by 9..
Since we have to divide N by 9, let's see N as = (999+1)x + (99+1)y + (9+1)z => N = 999x+99y+9z +x + y +z => N = 9(111x+11y+z) + (x + y + z)
We know that first part is always divisible by 9, since it is a multiple of 9. Remaining part is x+y+z; the remainder has to come from this part. But from the question, since these numbers are 1,2 and 3 (doesn't matter which is x or y or z), their sum is 6. So, remainder is 6.
Re: N = 1000x + 100y + 10z, remainder of N/9 [#permalink]
06 Jun 2013, 07:41
2
This post received KUDOS
For any number, it is divisible by 9 if all of the individual numbers in the number are divisible by 9 when added together. For example: 18 (9x2): 1+8 = 9/9 = 1, 36 (9x4): 3+6 = 9/9 = 1 396 (9x44): 3+6+9 = 18/9 = 2 657 (9x73): 6+5+7 = 18/9 = 2 999 (9x111): 9+9+9 = 27/9 = 3 etc.
The remainder when the individual numbers in a number are divided by 9 is the remainder when the number is divided by 9 as well For example: 19 (9x2 r1): 1+9 = 10/9 = 1 r1 38 (9x4 r2): 3+8 =11/9 = 1 r2 397 (9x44 r1): 3+9+7 = 19/9 = 2 r1 659 (9x73 r2): 6+5+9 = 20/9 = 2 r2 1002 (9x111 r3): 1+0+0+2 = 3/9 = 0 r3 etc.
That being said, in the problem if x, y, and z are all different positive integers less than 4, they can be 1,2, and 3 in any order: 1230, 1320, 2130, 2310, 3120, and 3210 if we apply the rule above: 1230: 1+2+3+0 = 6/9 = 0 r6 1320: 1+3+2+0 = 6/9 = 0 r6 2130: 2+1+3+0 = 6/9 = 0 r6 etc. etc. etc.
If you know this rule, then all you would need to know is x+y+z = 1+2+3 = 6
Re: N = 1000x + 100y + 10z, remainder of N/9 [#permalink]
26 Jun 2014, 18:45
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. _________________
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...
Ninety-five percent of the Full-Time Class of 2015 received an offer by three months post-graduation, as reported today by Kellogg’s Career Management Center(CMC). Kellogg also saw an increase...