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Remainders.

by bblast » Wed Jan 12, 2011 9:08 pm
What number leaves a remainder of 1,2 and 4 when divided by 6,7 and 9 respectively.


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by Rahul@gurome » Wed Jan 12, 2011 10:04 pm
LCM of 6, 7, and 9 = 126
126 - 5 = 121

The correct answer is 121.
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by ankur.agrawal » Wed Jan 12, 2011 11:13 pm
Rahul@gurome wrote:LCM of 6, 7, and 9 = 126
126 - 5 = 121

The correct answer is 121.

Plz give details.
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by Rahul@gurome » Thu Jan 13, 2011 3:02 am
Ideally the question should ask for the smallest positive number.
Let the smallest positive number be N.
So N = 6*x + 1.
N = 7*y + 2.
N = 9*z + 4.
Here x, y and z are integers.
So N+5 = 6(x+1) = 7*(y+2) = 9*(z+2).
So N+5 is the L.C.M of 6, 7 and 9 which is 126.
Or N = 126 - 5 = 121.
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by nehatandon » Sat Jan 15, 2011 3:02 pm
Rahul@gurome wrote:Ideally the question should ask for the smallest positive number.
Let the smallest positive number be N.
So N = 6*x + 1.
N = 7*y + 2.
N = 9*z + 4.
Here x, y and z are integers.
So N+5 = 6(x+1) = 7*(y+2) = 9*(z+2).
So N+5 is the L.C.M of 6, 7 and 9 which is 126.
Or N = 126 - 5 = 121.
Can you please explain the part in bold ?
Thanks.
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by VivianKerr » Sat Jan 15, 2011 4:36 pm
Hi there! Even if you didn't use algebra, you could solve using logic.

A number that have a remainder of 1 when divided by 6 must be a multiple of 6, with 1 added to it.

A number that has a remainder of 2 when divided by 7, must be a multiple of 7, with 2 added to it.

A number that has a remainder of 4 when divided by 9, must be a multiple of 9, with 4 added to it.

Let's start with the lowest common multiple of 6, 7, and 9.

6 = 2 x 3
7 = 1 x 7
9 = 3 x 3

LCM: 2 x 3 x 7 x 3 = 126

Now let's look at the remainders: 1, 2, and 4. We'll need a remainder of at least 4, so let's take away 4 from the LCM.

126 - 4 = 122

When we try 122, we see that it's slightly too big.

Since 122/9 = Remainder of 5 (we want only a remainder of 4).

Let's go one down, to 121.

121 works for all 3 numbers!
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