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

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WE: Pharmaceuticals (Health Care)
if m>n, then is mn divisible by 3? [#permalink]

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26 Sep 2017, 05:51
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55% (hard)

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40% (00:48) correct 60% (01:47) wrong based on 10 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

_________________

Sometimes you have to burn yourself to the ground before you can rise like a phoenix from the ashes.

Dr. Pratik

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

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26 Sep 2017, 07:13
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, 07:17, edited 1 time in total.

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

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26 Sep 2017, 07:16
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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29 Sep 2017, 04:14
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, 04:14
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