BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

least possible value

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Sun Oct 16, 2011 2:21 pm
Followed by:1 members

least possible value

by lenagmat » Wed Oct 26, 2011 4:03 am
If the product of the integers from 1 to n is divisible by 490, what is the least possible value of n?
(A) 7
(B) 14
(C) 21
(d) 28
(E) 35

Here is the answer, that I found. But I completly do not understand why it is not 7 (7<14) ?

490 = 7 x 7 x 5 x 2 = 14 x 7 x 5

Let the number be 'N'

N/490 = integer

Therefore, if the denominator is to be completely cancelled out from the numerator, N must contain 14, 7 and 5.

=> the smallest value of n must be 14.
Top
Source: — Problem Solving |
User avatar
Community Manager
Posts: 1060
Joined: Fri May 13, 2011 6:46 am
Location: Utrecht, The Netherlands
Thanked: 318 times
Followed by:52 members

by neelgandham » Wed Oct 26, 2011 4:12 am
lenagmat wrote: Here is the answer, that I found. But I completly do not understand why it is not 7 (7<14) ?
1)Let us say the answer = 7, so 1*2*3*4*5*6*7 is divisible by 490. Let us check if it really is.

490=7*7*5*2, so 1*2*3*4*5*6*7/(7*7*5*2) = (1*3*4*6/7), (dividing numerator and denominator by (7*5*2))
1*3*4*6 is not divisible by 7 => 1*2*3*4*5*6*7 is not divisible by 490(7*7*5*2).

2)Let us say the answer = 14, so 1*2*3*4*5*6*7*8*9*10*11*12*13*14 is divisible by 490. Let us check if it really is.

490=7*7*5*2, so 1*2*3*4*5*6*7*8*9*10*11*12*13*14/(7*7*5*2) = (1*3*4*6*8*9*10*11*12*13*2), (dividing numerator and denominator by (7*5*2*7)). => 1*2*3*4*5*6*7*8*9*10*11*12*13*14 is divisible by 490
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 324
Joined: Mon Jul 05, 2010 6:44 am
Location: London
Thanked: 70 times
Followed by:3 members

by kmittal82 » Wed Oct 26, 2011 4:15 am
The key here is:
"If the product of the integers from 1 to n", which is basically n!

If we use 7!, we can cancel out 7,5,2 from the denominator, which still leaves us with an extra 7 in the denominator, hence thats not the answer

Using 14, we cancel out all the terms of the denominator, hence (B)
Top
Post new topic Post Reply
• Page 1 of 1