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LCM of a number

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New post 19 Mar 2019, 02:51
Dear All,

May be this is a stupid question by I am really stuck on it. Im quickly going trough MGMAT foundation of math 5th edition and on page 79 of this book there is following question:

If x is divisible by 8,12, and 45, what is the largest number that x must be divisible by? The answer shows that one needs to find LCM of a number which is a simple task. LCM is thus 360. However, if we check the answer and assume that x=720 and it is really divisible by 8,12, and 45 then 360 is not the largest number 720 must be divisible by. 720 is divisible by 720 either. what do you think?


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New post Updated on: 29 Mar 2019, 02:10
The question is equivalent to what is the lCM of 8,12, and 45
the answer is 360.
LCM means Least common multiple .
So we only find least common number
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Originally posted by Noshad on 19 Mar 2019, 09:26.
Last edited by Noshad on 29 Mar 2019, 02:10, edited 1 time in total.
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New post 19 Mar 2019, 09:26
Hi Jahangir86,

Lowest Common Multiple - LCM. For example, the lowest common multiple of 2 and 3 is 6. No number lower than 6 is divisible by both 2 and 3.
In your case, 360 is divisible by each of the numbers: 8, 12, and 45.

Learn more here:
https://gmatclub.com/forum/math-number- ... 88376.html
https://gmatclub.com/forum/gmat-math-book-87417.html

Hope this helps!
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New post 19 Mar 2019, 10:24
the answer is 360 not 720
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New post 21 Mar 2019, 08:51
I'm not sure the above replies addressed your precise question, so I'll try to explain in a different way:

The question asks: "If x is divisible by 8,12, and 45, what is the largest number that x *must* be divisible by?" This is really saying: "if the only thing you know about x is that it is divisible by 8, 12 and 45, what is the largest number you can be completely sure x is divisible by?" Of course, you're right that x could be equal to 720, and then as you say, x would be divisible not only by 360, but also by the larger number 720. But you can't be sure that x is 720, so you can't be sure that x is divisible by 720. Here x could be lots and lots of different numbers -- it might be the case that x = 360, for example, or that x = 1080, among an infinitude of possibilities.

But even though there is an infinite number of possible values of x here, no matter what x is, you can be completely certain that x is divisible by 360.
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New post 27 Mar 2019, 17:13
Jahangir86 wrote:
Dear All,

May be this is a stupid question by I am really stuck on it. Im quickly going trough MGMAT foundation of math 5th edition and on page 79 of this book there is following question:

If x is divisible by 8,12, and 45, what is the largest number that x must be divisible by? The answer shows that one needs to find LCM of a number which is a simple task. LCM is thus 360. However, if we check the answer and assume that x=720 and it is really divisible by 8,12, and 45 then 360 is not the largest number 720 must be divisible by. 720 is divisible by 720 either. what do you think?


Thanks and regards


It's not actually the math that you're having a hard time with! The math is completely correct:

- You're correct that you need to find the LCM of 8, 12, and 45;
- You've correctly calculated the LCM as 360;
- 360 is the correct answer to the question.

The only issue here is in the wording of the problem. It says 'what is the largest number that x must be divisible by.' In other words, x is some number that's divisible by 8, 12, and 45. That means x could be 360, but x could also be 720, 1080, or a lot of other numbers.

You want to know the largest number that all of those possible values are divisible by - in other words, the number that x must be divisible by. x doesn't have to be divisible by 720, since x could be 360, and 360 isn't divisible by 720. But x does have to be divisible by 360.
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New post 29 Mar 2019, 02:05
true answer is 360....
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New post 29 Mar 2019, 02:29
Coincidentally, you would not believe, I had the precise same doubt yesterday.The answer could be infinity.But I hope you got the concept behind that question!!!!
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Re: LCM of a number   [#permalink] 29 Mar 2019, 02:29
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