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divisibility

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by cramya » Fri Mar 13, 2009 5:18 pm
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
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Re: divisibility

by sureshbala » Fri Mar 13, 2009 8:36 pm
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
So you need the LCM of the first 7 multiples of 5.

= 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
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by Anaira Mitch » Tue Dec 27, 2016 6:09 am
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.
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by [email protected] » Tue Dec 27, 2016 9:17 am
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
Contact Rich at [email protected]
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