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if m>n, then is mn divisible by 3?

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if m>n, then is mn divisible by 3? [#permalink]

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New post 26 Sep 2017, 04:51
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A
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C
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E

Difficulty:

  55% (hard)

Question Stats:

45% (01:27) correct 55% (01:47) wrong based on 11 sessions

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If m>n, then is mn divisible by 3?

1. The remainder when m+n is divided by 6 is 5.

2. The remainder when m-n is divided by 6 is 3.
[Reveal] Spoiler: OA

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if m>n, then is mn divisible by 3? [#permalink]

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New post 26 Sep 2017, 06:13
fitzpratik wrote:
If m>n, then is mn divisible by 3?

1. The remainder when m+n is divided by 6 is 5.

2. The remainder when m-n is divided by 6 is 3.


We need the value of both \(m\) & \(n\) to solve the question

Statement 1: let \(m+n=6k+5\)-----------------------(1)

from this equation we cannot find the value of \(m\) or \(n\). Hence Insufficient

Statement 2: let \(m-n=6q+3\)------------------------(2)

from this equation we cannot find the value of \(m\) or \(n\). Hence Insufficient

Combining 1 & 2: add equation 1 & 2

so \(2m=6k+6q+8\) or \(m=3(k+q)+4\)

substitute the value of \(m\) in equation 1 to get the value of \(n=3(k-q)+1\)

so \(mn=[3(k+q)+4]*[3(k-q)+1] = 9(k^2-q^2)+12(k-q)+3(k+q)+4\)

Hence on dividing \(mn\) by \(3\) remainder \(1\) will be left. Sufficient

Option C

Last edited by niks18 on 26 Sep 2017, 06:17, edited 1 time in total.

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Re: if m>n, then is mn divisible by 3? [#permalink]

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New post 26 Sep 2017, 06:16
niks18 wrote:
fitzpratik wrote:
If m>n, then is mn divisible by 3?

1. The remainder when m+n is divided by 6 is 5.

2. The remainder when m-n is divided by 6 is 3.


We need the value of both m & n to solve the question

Statement 1: let m+n=6k+5-----------------------(1)
from this equation we cannot find the value of m or n. Hence Insufficient

Statement 2: let m-n=6q+3------------------------(2)
from this equation we cannot find the value of m or n. Hence Insufficient

Combining 1 & 2: add equation 1 & 2
so 2m=6k+6q+8 or m = 3(k+q)+4
substitute the value of m in equation 1 to get the value of n = 3(k-q)+1
so mn={3(k+q)+4}*{3(k-q)+1} = 9(k^2-q^2)+12(k-q)+3(k+q)+4

Hence on dividing mn by 3 remainder 1 will be left. Sufficient

Option C

Got it! Thanx.

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if m>n, then is mn divisible by 3? [#permalink]

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New post 29 Sep 2017, 03:14
fitzpratik wrote:
If m>n, then is mn divisible by 3?

1. The remainder when m+n is divided by 6 is 5.

2. The remainder when m-n is divided by 6 is 3.

Alternative solution...

The question implicitly asks whether M or N or both are multiples of 3. So I approach it that way.

Also, observe that either of the given remainders (doesn't matter which) are odd after being divided by 6. Therefore, one of M or N must be odd, and the other must be even.

All even multiples of 3 are also multiples of 6. And all odd are multiples of 3 (duh). So let's say M may be the multiple of 6 (6m), and N may be the *ODD* multiple of 3 (3n, again note that n must be an odd number).

First consider whether M=6m. We see here that the second answer option gives a remainder of 3. The only possible way for M to be a multiple of 6 that, after being divided by 6, results in a remainder of 3, is if N=3n. However, under no circumstances will this ever result in a remainder of 5. Therefore, under no circumstances can M=6m.

So, can we be certain whether N=3n? Fist recall that N must be an even number and is NOT a multiple of 3. Here it IS possible that a remainder of 5 can be achieved. However, using the same prior logic, it's not possible to obtain a remainder of 3, after M is divided by 6, unless M=6m.

Therefore, using both A and B, we have sufficient info to show that neither M or N can be a multiple of 3. Sufficient.


...now rereading, I think my explanation is about as clear as mud....

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if m>n, then is mn divisible by 3?   [#permalink] 29 Sep 2017, 03:14
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