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# HCF and LCM question apporach how to derive

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Intern
Joined: 26 Jan 2012
Posts: 1
HCF and LCM question apporach how to derive [#permalink]

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26 Jan 2012, 09:11
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Find the LEAST NUMBER
which when divided by x, y
and z leaves the remainders
a, b and c respectively.
Then, it is always observed that
(x – a) = (z – b) = (z – c) = K (say).
∴ Required number
= (L.C.M. of x, y and z) – K.

How and why
(L.C.M. of x, y and z) +K it is not the result
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8077
Location: Pune, India
Re: HCF and LCM question apporach how to derive [#permalink]

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26 Jan 2012, 22:37
2
KUDOS
Expert's post
vidyakmr wrote:
Find the LEAST NUMBER
which when divided by x, y
and z leaves the remainders
a, b and c respectively.
Then, it is always observed that
(x – a) = (y – b) = (z – c) = K (say).
∴ Required number
= (L.C.M. of x, y and z) – K.

How and why
(L.C.M. of x, y and z) +K it is not the result

Tip: If a concept seems tricky, plug in some numbers. If you want to understand the 'why' and 'how' in Quant, it is not difficult. Just reason out with some numbers. Let me explain.

"Find the LEAST NUMBER which when divided by x, y and z leaves the remainders a, b and c respectively. "

Case 1:

"Find the LEAST NUMBER which when divided by 2, 3 and 5 leaves the remainder 1 in each case."

Now, LCM of 2, 3 and 5 is 30. Here, the required number is 30+1 since 30 is divisible by each of 2, 3 and 5 so when 31 is divided by these numbers, it will leave a remainder of 1 in each case.

Case 2:

"Find the LEAST NUMBER which when divided by 2, 3 and 5 leaves the remainders 1, 2 and 4 respectively." (this is the version you have given)

Note here that there is a common remainder here. It is -1. When we say a number divided by 2 leaves a remainder of 1 (i.e. it is 1 more than a multiple of 2), we can also say it leaves a remainder of -1 (i.e. it is 1 less than a multiple of 2). When we say a number divided by 3 leaves a remainder of 2, we can also say it leaves a remainder of -1. Similarly, when we say a number divided by 5 leaves a remainder of 4, we can also say it leaves a remainder of -1.

Mind you, here $$(x-a) = (y-b) = (z-c) = (2-1) = (3-2) = (5-4) = 1$$

So, just like in the case above, we find the LCM and ADD the common remainder (-1) here,
LCM of (2, 3, 5) = 30 + (-1) since -1 is the common remainder.
Required number = 29 = LCM - K

29 is the number which is 1 less than a multiple of 2 , 3 and 5. So when divided by these numbers, it will leave a remainder of -1 in each case or we can say, it will leave remainders 1, 2 and 4 respectively.

There can also be a general case where there is no common remainder, positive or negative. I would suggest you to go through these posts to understand all these concepts in detail.

http://www.veritasprep.com/blog/2011/04 ... unraveled/
http://www.veritasprep.com/blog/2011/04 ... y-applied/
http://www.veritasprep.com/blog/2011/05 ... emainders/
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Re: HCF and LCM question apporach how to derive [#permalink]

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31 Mar 2017, 07:19
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: HCF and LCM question apporach how to derive   [#permalink] 31 Mar 2017, 07:19
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