Which of the following is the lowest positive integer that is divisible by the first 7 positive integer multiples of 5?
3500
2100
1400
210
140
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
divisibility
This topic has expert replies
-
anishprabhu
- Senior | Next Rank: 100 Posts
- Posts: 46
- Joined: Sun Feb 15, 2009 3:09 pm
- Thanked: 1 times
- GMAT Score:720
-
cramya
- Legendary Member
- Posts: 2467
- Joined: Thu Aug 28, 2008 6:14 pm
- Thanked: 331 times
- Followed by:11 members
2100
First 5 positive multiples of 5 are
5,10,15,20,25,30,35
Break these down in to primes u would need 2^2 5^2 3 and 7 minimum.
140 1400,3500 eliminate not divisible by 3 since sum of digits not divisible by 3
210->3*5*2*7
Missing a 5 and 2 so if u add this u need atleast 2100
First 5 positive multiples of 5 are
5,10,15,20,25,30,35
Break these down in to primes u would need 2^2 5^2 3 and 7 minimum.
140 1400,3500 eliminate not divisible by 3 since sum of digits not divisible by 3
210->3*5*2*7
Missing a 5 and 2 so if u add this u need atleast 2100
-
sureshbala
- Master | Next Rank: 500 Posts
- Posts: 319
- Joined: Wed Feb 04, 2009 10:32 am
- Location: Delhi
- Thanked: 84 times
- Followed by:9 members
So you need the LCM of the first 7 multiples of 5.anishprabhu wrote:Which of the following is the lowest positive integer that is divisible by the first 7 positive integer multiples of 5?
3500
2100
1400
210
140
= 5 x LCM(1x2x3x4x5x6x7)
Since 1,2 and 3 are factors of 6, LCM(1,2,3,4,5,6,7) = LCM(4,5,6,7)=420
Hence the required number = 5x420 = 2100
-
Anaira Mitch
- Master | Next Rank: 500 Posts
- Posts: 235
- Joined: Wed Oct 26, 2016 9:21 pm
- Thanked: 3 times
- Followed by:5 members
The first 7 integer multiples of 5 are 5, 10, 15, 20, 25, 30, and 35. The question is asking for the least common multiple (LCM) of these 7 numbers. Let's construct the prime box of the LCM.
In order for the LCM to be divisible by 5, one 5 must be in the prime box.
In order for the LCM to be divisible by 10, a 5 (already in) and a 2 must be in the prime box.
In order for the LCM to be divisible by 15, a 5 (already in) and a 3 must be in the prime box.
In order for the LCM to be divisible by 20, a 5 (already in), a 2 (already in), and a second 2 must be in the prime box.
In order for the LCM to be divisible by 25, a 5 (already in) and a second 5 must be in the prime box.
In order for the LCM to be divisible by 30, a 5 (already in), a 2 (already in) and a 3 (already in) must be in the prime box.
In order for the LCM to be divisible by 35, a 5 (already in) and a 7 must be in the prime box. Thus, the prime box of the LCM contains a 5, 2, 3, 2, 5, and 7. The value of the LCM is the product of these prime factors, 2100.
The correct answer is D.
In order for the LCM to be divisible by 5, one 5 must be in the prime box.
In order for the LCM to be divisible by 10, a 5 (already in) and a 2 must be in the prime box.
In order for the LCM to be divisible by 15, a 5 (already in) and a 3 must be in the prime box.
In order for the LCM to be divisible by 20, a 5 (already in), a 2 (already in), and a second 2 must be in the prime box.
In order for the LCM to be divisible by 25, a 5 (already in) and a second 5 must be in the prime box.
In order for the LCM to be divisible by 30, a 5 (already in), a 2 (already in) and a 3 (already in) must be in the prime box.
In order for the LCM to be divisible by 35, a 5 (already in) and a 7 must be in the prime box. Thus, the prime box of the LCM contains a 5, 2, 3, 2, 5, and 7. The value of the LCM is the product of these prime factors, 2100.
The correct answer is D.
GMAT/MBA Expert
-
[email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi All,
This question has a few built-in Number Properties that you can take advantage of (and you can use the 5 answer choices to your advantage):
We're looking for the SMALLEST positive integer that's divisible by 5, 10, 15, 20, 25, 30 and 35.
30 is divisible by 5, 10 and 15, so we don't have to 'find' any of those numbers in the final answer... we just have to make sure that 30 is divisible into it. Since 30 is divisible by 3, the final answer MUST also be divisible by 3. Using the 'rule of 3', you can quickly eliminate Answers A, C and E.
Between the final two answers 2100 and 210, you should be able to easily see that 210 is NOT divisible by 25.
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
This question has a few built-in Number Properties that you can take advantage of (and you can use the 5 answer choices to your advantage):
We're looking for the SMALLEST positive integer that's divisible by 5, 10, 15, 20, 25, 30 and 35.
30 is divisible by 5, 10 and 15, so we don't have to 'find' any of those numbers in the final answer... we just have to make sure that 30 is divisible into it. Since 30 is divisible by 3, the final answer MUST also be divisible by 3. Using the 'rule of 3', you can quickly eliminate Answers A, C and E.
Between the final two answers 2100 and 210, you should be able to easily see that 210 is NOT divisible by 25.
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
