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: product and integer n [#permalink]
31 Jan 2005, 10:32
I looked at it, and broke 990 to its prime, then 11*5*3*3*2
i notice that the only prime number that shows up is 11, therefor it will be the smallest from this selection, but I think techically the smallest possible value should be 2?
Honghu can you please explain your reasoning with 189?
Antmavel wrote:
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?
another good question to be discussed and put my 2 cts views.
n to be divided by 990, n must posses all the prime factors of 990. the prime factors of 990 are 2, 3, 3, 5, and 11. the question stipulated that n is a product of all positive integers 1 to n means no single integer can repete if we need one more integer in between 1 to n. for example, here we need one more 3, we have to include 6 to fulfill the requirement of 3. similarly, to fulfill the requirement to 11, we have to stretch the n up to 11.
in case of 189, n is 9 because the prime factors of 189 are 3,3,3,and 7. we need three 3s which we can get from 1 to n=9. this n (9) also fullfill the requirement of 7 also, therefore the least possible value of n=9. n can not be 7. if n is 7, then the product of 1 to n (1x2x3x4x5x6x7) can not be divided by 189.
another good question to be discussed and put my 2 cts views.
in case of 189, n is 9 because the prime factors of 189 are 3,3,3,and 7. we need three 3s which we can get from 1 to n=9. this n (9) also fullfill the requirement of 7 also, therefore the least possible value of n=9. n can not be 7. if n is 7, then the product of 1 to n (1x2x3x4x5x6x7) can not be divided by 189.
I understand the part upto the breaking of the prime factors but I dont understand what you mean by we need 3 3s before we need 7?
ok.....for 189 you have prime factors 3x3x3x7....now for this to divide product from 1....n ....u have to have atleast go upto 7 numbers...right ?
However, in those first 7 numbers , u will only get one 3 and another three from 6 (3x2)...u need another 3 as the denomintor 3x3x3x7 has to divide the product evenly...so the product shud go all the way upto 9 to get another 3.
product = 1x2x3x4x5x6(i.e 3x2)x7x8x9 (i.e. 3x3)....u see now u have atleast three 3s and one 7...which can be divided completely by 189 (3x3x3x7)....right ? [/b]
I understand the part upto the breaking of the prime factors but I dont understand what you mean by we need 3 3s before we need 7?
ok.....for 189 you have prime factors 3x3x3x7....now for this to divide product from 1....n ....u have to have atleast go upto 7 numbers...right ? However, in those first 7 numbers , u will only get one 3 and another three from 6 (3x2)...u need another 3 as the denomintor 3x3x3x7 has to divide the product evenly...so the product shud go all the way upto 9 to get another 3.
product = 1x2x3x4x5x6(i.e 3x2)x7x8x9 (i.e. 3x3)....u see now u have atleast three 3s and one 7...which can be divided completely by 189 (3x3x3x7)....right ? [/b]
Honnestly the first explanation from MA was really great and banerjeea_98 finished the work in beauty Thanks guys, I totally understand now what I totally didn't understand 5mn ago....
Princeton and Kaplan are useless compare to your explanations