Which of the following is the lowest positive integer that

Question

Which of the following is the lowest positive integer that is divisible by the first 7 positive integer multiples of 5?

A. 140 
B. 210 
C. 1400 
D. 2100 
E. 3500

Bunuel · Accepted Answer

The first 7 positive multiples of 5 are 5, 10, 15, 20, 25, 30 and 35. The least common multiple of these numbers is 2,100.

Answer: D.

Or, a shortcut approach:  
Discard 140, 1,400 and 3,500 because they are not divisible by 3.
Discard 210 because it's not divisible by 4.
Only option D remains.

hitman5532 · Answer

Why are we looking at divisibility by 3 and 4?

Bunuel · Answer

The least common multiple of 5, 10,  15 ,  20 , 25, 30 and 35 must be a multiple of 3 since 15 is a multiple of 3 and it must a multiple of 4 since 20 is a multiple of 4.

Hope it's clear.

hitman5532 · Answer

I knew I was missing something very simple. Thank you

shanmugamgsn · Answer

first 7 positive multiples of 5 are 5, 10, 15, 20, 25, 30 and 35.

5 = 1* 5
10 = 2 * 5
15 = 3 * 5
20 = 4 * 5 = \(2^2 * 5\)
25 = 5 * 5 = \(5 ^ 5\)
30 = 6 * 5= \(2*3*5\)
35 = 7* 5

So, LCM would be 1 * \(2^2\) * 3 * \(5^5\) * 7 = 2100

actually i made mistake while computing first after going through Bunuel soln i got above one :-D

ENGRTOMBA2018 · Answer

One way to look at it this :

1st 7 multiples of 5: 
5 =1*5
10=2*5
15=3*5
20=4*5=2^2*5
25=5^2
30=2*3*5
35=5*7

Always try to break down the numbers into their corresponding prime factors.

For LCM (least common multiple) take the highest powers of 2,3,5,7   which gives us : 2^2 *3*5^2*7 = 2100. Hence D is the correct answer.

mbaapp1234 · Answer

First write out in a single column the first seven positive multiples of 5:

5
10
15
20
25
30
35

Then in the next column write out their prime factors:

5
2*5
3*5
2^2*5
5^2
2*3*5
7*5

For an integer to be divisible by each of these prime groups, it must have all of these primes at their highest power. So it must have at least 5^2 for instance because 5^2 is 25 and this integer must be divisible by 25.

We can narrow the list down to 5^2*7*2^2*3

100*7*3

100*21 = 2100.

EMPOWERgmatRichC · Answer

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

Crater · Answer

First 7 integer multiples of 5 are:
5, 10, 15, 20, 25, 30, 35

For a number to be divisible by all 7 numbers, take LCM:
This can be reduced to:
 5*LCM(1,2,3,4,5,6,7)=5*2*LCM(1,1,3,2,5,3,7)= 5*2 * 2*5 *3*7=2100

Hence, option (D)

bumpbot · Answer

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Which of the following is the lowest positive integer that

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