multiples of 5

[jjykim]

This problem is from the Question Bank section. I wasn't sure where to post them, but I hope it's okay to post them over here.



Which of the following is the lowest positive integer that is divisible by the first 7 positive integer multiples of 5?
140
210
1400
2100
3500

The answer choice is D, 2100, but I wasn't sure how to derive the answer.

I started by listing the first 7 multiples of 5, which are: 5, 10, 15, 20, 25, 30, and 35. Then I broke them down into their prime factors.
5 => 5
10=> 2 and 5
15=> 5 and 3
20=> 5, 2, and 2
25=> 5 and 5
30=> 5, 2, and 3
35=> 5 and 7.

I got stuck after this.
My manhattan gmat instructor suggested that we use the LCM/GCF diagram to figure out the LCM, but I wasn't sure how to apply the diagram to this situation.
Any help would be appreciated. Thanks.




- I know we

[Guest]

This problem is from the Question Bank section. I wasn't sure where to post them, but I hope it's okay to post them over here.



Which of the following is the lowest positive integer that is divisible by the first 7 positive integer multiples of 5?
140
210
1400
2100
3500

The answer choice is D, 2100, but I wasn't sure how to derive the answer.

I started by listing the first 7 multiples of 5, which are: 5, 10, 15, 20, 25, 30, and 35. Then I broke them down into their prime factors.
5 => 5
10=> 2 and 5
15=> 5 and 3
20=> 5, 2, and 2
25=> 5 and 5
30=> 5, 2, and 3
35=> 5 and 7.

I got stuck after this.
My manhattan gmat instructor suggested that we use the LCM/GCF diagram to figure out the LCM, but I wasn't sure how to apply the diagram to this situation.
Any help would be appreciated. Thanks.




- I know we

You were on the right track, because listing the prime factors for the first 7 multiples of 5 is definitely the first step.

The next step here is to remember that you're looking for the LCM of all 7 numbers. This means you are looking for the lowest multiple that 5, 10, 15, 20, 25, 30, and 35 all have in common.

I'm not sure about the LCM diagram, but here's what I did to find the LCM: First list every prime factor you found and arrange them in numerical order, without regard to how many of each appear (that will be the next step). For this question, the prime factors you see are:

2, 3, 5, 7

Now start with 2. You need to look at the prime factorizations you did for each multiple of 5 and ask, "What is the most number of times that 2 appeared as a prime factor for any one of these multiples of 5?" The most number of times 2 appeared as a prime factor for any of the multiples of 5 was for 20, where it appeared twice. This means you raise the 2 in your list to the 2nd power:

2^2, 3, 5, 7

You continue this pattern with 3, 5, and 7, each time asking "What is the most number of times this prime factor appeared for any one the multiples?" and then raise that prime factor in your list to the power of the most number of times it appeared for any one multiple--resulting in this:

2^2, 3^1, 5^2, 7^1 = 4, 3, 25, 7

Now all you need to do to find the LCM is multiply the above numbers together:

4 x 3 x 25 x 7 = 2100

Hopefully an instructor can double check all of this to make sure it is sound, or provide you with a faster way to do it, but I hope this helps you regardless.

[jjykim]

Thank you for your clear explanation !

[Guest]

Thank you for your clear explanation !

Glad to be of help. :D

[Guest]

I solve this question in less than 2 mins by keepin in mind the followin simple rules of divisibilty:-

2 divides all evens
3 if the sum of the digits in the number is multiple of 3
4 if the last two digits are divisible by 4
5 if digit ends in 5 or 0
6 if the number is divisible by both 2 and 3
7 no rule
8 if the last three digits are divisible by 8
9 if the sum of the digits in the number is multile of 9

[RonPurewal]

@guest in second post:
extremely well played. here's a virtual beer for you.

to readers of this thread, make sure that you extract the appropriate general lesson about least common multiples:
you take the highest powers of each prime number appearing in any of the factorizations.
so, for instance, because 5 is raised to the 2nd power in 25, your lcm needs to have 5 raised to the 2nd power, even though it's only raised to the 1st power in all the other factors.
etc.

[ryan.b.barton]

Why is the number 1 not included on the list of positive integer multiples of 5?

Thanks

[kishoregaurav]

140
210
1400
2100
3500

- all numbers divisible by 2,5
-140,1400,3500 not divisible by 3
- 210 not divisible by 4
- last option remains - 2100

[jnelson0612]

Why is the number 1 not included on the list of positive integer multiples of 5?

Thanks

Ryan, 1 is a factor of 5 but not a multiple of 5.

Multiples are numbers that are derived by taking our original number times consecutive integers.

The multiples of 5 are: 5 (1*5), 10 (2*5), 15 (3*5) and on.

[jnelson0612]

140
210
1400
2100
3500

- all numbers divisible by 2,5
-140,1400,3500 not divisible by 3
- 210 not divisible by 4
- last option remains - 2100

Great!

[arunangsu9362]

Go by options:

2100 has 2,3,5,7..all reqd factors

[tim]

thanks..

[CameronT327]

140
210
1400
2100
3500

- all numbers divisible by 2,5
-140,1400,3500 not divisible by 3
- 210 not divisible by 4
- last option remains - 2100

Can someone please explain to me why it matters if the number is divisible by 2, 3, or 4?

[RonPurewal]

Can someone please explain to me why it matters if the number is divisible by 2, 3, or 4?

the goal of the problem is to find a number that is divisible by all of the numbers 5, 10, 15, 20, 25, 30, and 35.

if a number isn't divisible by 2, then it won't be divisible by 10, 20, or 30.
if a number isn't divisible by 3, then it won't be divisible by 15 or 30.
if a number isn't divisible by 4, then it won't be divisible by 20.

[RonPurewal]

what's important (and unstated) here is that there are relatively straightforward tests for divisibility by these numbers. because it is QUICK to test for divisibility by THESE numbers, dividing by them first can be a 'shortcut' here.
(otherwise it would just be a needlessly indirect/roundabout way to approach the problem—which is exactly the way it would seem if you weren't aware of the divisibility tests.)