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26 Jan 2012, 09:11
Kudos
Bookmarks
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
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
vidyakmr
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
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.
General Discussion
06 Apr 2020, 10:20
Kudos
Bookmarks
vidyakmr
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
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
I know this is an old thread, but it was bumped, and I'm replying because I'm curious about it and other folks reading it now might be as well!
So, it seems like the type of problem discussed here would be one like this:
"Find the least number which, when divided by 6 and 2, leaves the remainders 3 and 1, respectively."
(The answer is 3, by the way! A good trap answer would be 9, to catch people who forget that 3 divided by 6 has a remainder of 3.)
However, the next line says:
Quote:
Then, it is always observed that
(x – a) = (z – b) = (z – c) = K (say).
(x – a) = (z – b) = (z – c) = K (say).
That isn't true in this case: 6-3 is not equal to 2-1.
Can somebody offer some clarity here? Personally, I would find the answer to this one by testing the answer choices; this is an example of a type of problem that's a lot easier on the GMAT than it is in the real world, since they HAVE to give you answer choices to work with.
