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HCF and LCM question apporach how to derive

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HCF and LCM question apporach how to derive [#permalink] New post 26 Jan 2012, 08:11
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
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Re: HCF and LCM question apporach how to derive [#permalink] New post 26 Jan 2012, 21:37
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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] New post 11 Dec 2013, 05:18
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Re: HCF and LCM question apporach how to derive   [#permalink] 11 Dec 2013, 05:18
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