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:
14,15,16 =45 + already we have n=2 and n=6 So insufficient
Together
we have atleast 2 option where n is even and less than 9
so E
thank you for the expanation. Do you know any shortcut for picking numbers? it took me some time to check on different combinations and we cannot afford it during the test. thanks!
Re: CONSECUTIVE INTEGERS.. [#permalink]
09 Aug 2009, 21:54
2
This post received KUDOS
We can pick numbers as follows
Case 1) divide the number by 2 and play around with the numbers around it to get the sum Case 2) divide the number by 3 and play around with the numbers around it to get the sum Case 3) divide the number by 4 and play around with the numbers around it to get the sum Case 4) divide the number by 5 and play around with the numbers around it to get the sum and go on ... till you realize that sum of the n numbers exceeds 45
In the above case we see that 45 will arise when you play around with numbers in case 1,2,4 and not in 3 as it will arise in either 9+10+11+12 = 42 or 10+11+12+13 = 46
Re: CONSECUTIVE INTEGERS.. [#permalink]
09 Aug 2009, 22:29
gmanjesh wrote:
We can pick numbers as follows
Case 1) divide the number by 2 and play around with the numbers around it to get the sum Case 2) divide the number by 3 and play around with the numbers around it to get the sum Case 3) divide the number by 4 and play around with the numbers around it to get the sum Case 4) divide the number by 5 and play around with the numbers around it to get the sum and go on ... till you realize that sum of the n numbers exceeds 45
In the above case we see that 45 will arise when you play around with numbers in case 1,2,4 and not in 3 as it will arise in either 9+10+11+12 = 42 or 10+11+12+13 = 46
thank you gmanjesh. it seems like a quicker approach. +1 for you
Re: CONSECUTIVE INTEGERS.. [#permalink]
14 Aug 2009, 17:05
gmanjesh wrote:
We can pick numbers as follows
Case 1) divide the number by 2 and play around with the numbers around it to get the sum Case 2) divide the number by 3 and play around with the numbers around it to get the sum Case 3) divide the number by 4 and play around with the numbers around it to get the sum Case 4) divide the number by 5 and play around with the numbers around it to get the sum and go on ... till you realize that sum of the n numbers exceeds 45
In the above case we see that 45 will arise when you play around with numbers in case 1,2,4 and not in 3 as it will arise in either 9+10+11+12 = 42 or 10+11+12+13 = 46
sum of consecutive integers=n/2(2a+n-1) where a is first term of series in the above equation we know all the terms except a and we need to find value of n S1 and S2 do not give value of "a" directly or indirectly. even after combining them we cant find value of "a". hence both statements are insufficient. Ans=E
since we dont know whether the first term of the sequence is 1 , we cant use the formula n(n+1)/2 formula bcoz that is sum of n consecutive natural numbers starting from 1
so using the formula n/2 (a+l), where l is the last term and a is the first term we get n/2(a+l) = 45 n(a+l)=90.....A1. 10*9 18*5 6*15 2*45 30*3 now statement 1 says n is even so all the above cases are possible but only 2*45 actually satisfies the condition. then A should be good enough.
2 CASE ) n < 9 only bottom three cases are possible But again here 2*45 is the only option which gives n=2 and terms will be 22 +23. no other soluton satisfies condition A1 ...hence B is also sufficient
I KNOW THAT I AM MAKING SOME MISTAKE BUT WHERE....PLS HELP
The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is even
(2) n < 9
In how many ways can you write n consecutive integers such that their sum is 45? n = 1 -> 45 n = 2 -> 22, 23 n = 3 -> 14, 15, 16 n = 4 X n = 5 -> 7, 8, 9, 10, 11 n = 6 -> 5, 6, 7, 8, 9, 10 n = 7 X n = 8 X n = 9 -> 1, 2, 3, 4, 5, 6, 7, 8, 9
n can be 2 or 6 hence statement 1 is not sufficient. n can take many values less than 9 hence statement 2 is not sufficient. Since n can take values 2 or 6 which are even and less than 9, both statements together are not sufficient. Answer (E)
Now, the interesting thing is how to get these values of n. How do I know which values can n take and which can it not? Its pretty easy really. Follow my thought here.
"Of course n can be 1.
n can be 2 since when I divide 45 by 2, I get 22.5. So 2*22.5 is 45 so I have to find 2 consecutive integers whose mean is 22.5. The integers are obviously 22 and 23.
When I divide 45 by 3, I get 15. So I need 3 consecutive integers whose mean is 15. They are 14, 15, 16
When I divide 45 by 4, I get 11.25. Do I have 4 consecutive integers such that their mean is 11.25? No because mean of consecutive integers is always an integer or of the form n.5
When I divide 45 by 5, I get 9 so I need 5 consecutive integers whose mean is 9. They must be 7, 8, 9, 10, 11
and so on....."
Obviously, I just need to focus on getting 2 even values of n. So I check for 2, 4, 6 and I would know that answer is (E). _________________
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...