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:
Arithmetic Progression formula (i.e. AP) is not a part of [#permalink]
29 Jun 2007, 01:57
3
This post received KUDOS
3
This post was BOOKMARKED
Arithmetic Progression formula (i.e. AP) is not a part of the OG prep material. I have never seen it on GmatPrep but maybe that’s because I never get questions that are above the 600 level (adaptive test, remember ).
I attached some questions that can be solved by using the AP formula but can be solved just as well by using common sense or other methods of solving.
If anyone encountered AP questions on the real GMAT or on GmatPrep, please let me know.
X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?
I like this topic. In my mind, the fact that you can do all of these without the AP formula is important. I always just use what I know about consecutive numbers to solve. But I guess it doesn't really matter - the first part of each solution is the same - still need to know about remainders, about number properties, and a few other things. Once you get the list of numbers written out, it's up to you how you want to solve it.
I like this topic. In my mind, the fact that you can do all of these without the AP formula is important. I always just use what I know about consecutive numbers to solve. But I guess it doesn't really matter - the first part of each solution is the same - still need to know about remainders, about number properties, and a few other things. Once you get the list of numbers written out, it's up to you how you want to solve it.
I agree ! Using fancy formulas is worthless unless you understand the logic behind them. Some of the above questions can be solved without AP and only by applying common sense. Question number three, can be solved by using simple trial and error:
I like this topic. In my mind, the fact that you can do all of these without the AP formula is important. I always just use what I know about consecutive numbers to solve. But I guess it doesn't really matter - the first part of each solution is the same - still need to know about remainders, about number properties, and a few other things. Once you get the list of numbers written out, it's up to you how you want to solve it.
I agree ! Using fancy formulas is worthless unless you understand the logic behind them. Some of the questions above can be solved without AP and only by apllying common sense. Question number three, can be solved by using simple trial and error:
8,29,50, ... n
21*22-8 = 454 21*23-8 = 475 21*24-8 = 496
n = 24
Once i figured out the list (8, 29, 50...) i realized that we want all multiples of 21 with remainder 8 below 500. So I divided 500 by 21, and got 23 remainder 17. So 23*21 is 483, and I added 8 to get to 491, which is the last one.
So then it was just a matter of counting number of numbers between 8 and 491 when the difference is 21. My way:
X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?
A. 18
B. 20
C. 22
D. 24
E. 25
x = 7k+1 =3m +2 for some k,m nonnegative integers
7k = 3m + 1
k = (3m + 1)/7
Since k must be an integer,
m = 7r + 2 where r is a nonnegative integer
so x = 21mr + 8
0 < 21mr + 8 < 500
-8 < 21mr < 492
mr is between 0 and 23 inclusive
24 possibilities
X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?
A. 18 B. 20 C. 22 D. 24 E. 25
x = 7k+1 =3m +2 for some k,m nonnegative integers
7k = 3m + 1
k = (3m + 1)/7 Since k must be an integer, m = 7r + 2 where r is a nonnegative integer
so x = 21mr + 8
0 < 21mr + 8 < 500 -8 < 21mr < 492 mr is between 0 and 23 inclusive 24 possibilities
can someone please explain this to me? _________________
Success is my only option, failure is not -- Eminem
X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?
A. 18 B. 20 C. 22 D. 24 E. 25
x = 7k+1 =3m +2 for some k,m nonnegative integers
7k = 3m + 1
k = (3m + 1)/7 Since k must be an integer, m = 7r + 2 where r is a nonnegative integer
so x = 21mr + 8
0 < 21mr + 8 < 500 -8 < 21mr < 492 mr is between 0 and 23 inclusive 24 possibilities
I got lost after that first part. Can somebody explain the part in red?
we know that x is an integer ... then 7K=3M+1 has to b an integer
which mean K is an integer..and
in order for k to be an integer (3m+1)/7 has to be an integer correct??
OK...
the only way 3m+1 can be an integer is m*3 is a multiple of 7 and 1 somehow becomes a multiple of 7
m=(7r+2)*3 ; suppose r=1
then 3(7 +2) +1= will always result in a multiple of 7!
hope this helps!!
Brilliant piece of work by Kevin..
beckee529 wrote:
kevincan wrote:
X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?
A. 18 B. 20 C. 22 D. 24 E. 25
x = 7k+1 =3m +2 for some k,m nonnegative integers
7k = 3m + 1
k = (3m + 1)/7 Since k must be an integer, m = 7r + 2 where r is a nonnegative integer
so x = 21mr + 8
0 < 21mr + 8 < 500 -8 < 21mr < 492 mr is between 0 and 23 inclusive 24 possibilities
I got lost after that first part. Can somebody explain the part in red?
Re: Using the Arithmetic Progression formula - 101 [#permalink]
14 Aug 2007, 09:34
KillerSquirrel wrote:
Arithmetic Progression formula (i.e. AP) is not a part of the OG prep material. I have never seen it on GmatPrep but maybe that’s because I never get questions that are above the 600 level (adaptive test, remember ).
I attached some questions that can be solved by using the AP formula but can be solved just as well by using common sense or other methods of solving.
If anyone encountered AP questions on the real GMAT or on GmatPrep, please let me know.
X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?
For question#1, do you consider 1 as an option?
I imagine you should start from 4 since by definition, 1/3 is really no remainder. Shouldn't the answer be 16?
Re: Using the Arithmetic Progression formula - 101 [#permalink]
15 Aug 2007, 00:41
bkk145 wrote:
For question#1, do you consider 1 as an option? I imagine you should start from 4 since by definition, 1/3 is really no remainder. Shouldn't the answer be 16?
No - to me 1/3 is zero (0) with remainder 1, since "3 goes into 1 zero (0) times" leaving a remainder of 1, and 3*0+1 = 1
Re: Using the Arithmetic Progression formula - 101 [#permalink]
30 Jan 2008, 01:16
KillerSquirrel wrote:
bkk145 wrote:
For question#1, do you consider 1 as an option? I imagine you should start from 4 since by definition, 1/3 is really no remainder. Shouldn't the answer be 16?
No - to me 1/3 is zero (0) with remainder 1, since "3 goes into 1 zero (0) times" leaving a remainder of 1, and 3*0+1 = 1
22/39 part/whole
22 is the remainder 39 parts comprise the total _________________
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson
Re: Arithmetic Progression formula (i.e. AP) is not a part of [#permalink]
16 May 2014, 05:20
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. _________________
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
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’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...