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If m and n are positive integers, is m + n divisible by 4 ?

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If m and n are positive integers, is m + n divisible by 4 ? [#permalink]

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New post 27 Jul 2017, 05:33
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If m and n are positive integers, is m + n divisible by 4 ?

(1) m and n are each divisible by 2.

(2) Neither m nor n is divisible by 4.
[Reveal] Spoiler: OA

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Re: If m and n are positive integers, is m + n divisible by 4 ? [#permalink]

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New post 27 Jul 2017, 06:57
Statement 1 : if m and n take 2 and 4 then yes
But if the take 6 and 4 then No. Insufficient.

Statement 2: if 2 and 6 then yes.
If 7 and 3 , No. Insufficient.
Combining we get a positive YES.

THUS IMO answer is C.
I hope this time I have not missed anything [FACE SAVOURING DELICIOUS FOOD]

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If m and n are positive integers, is m + n divisible by 4 ? [#permalink]

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New post 27 Jul 2017, 07:11
THUS IMO answer is C.



if u combine both
10 and 6 results in yes
and
10 and 14 results in Yes

so c is the answer

Last edited by rocko911 on 20 Aug 2017, 03:18, edited 1 time in total.

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Re: If m and n are positive integers, is m + n divisible by 4 ? [#permalink]

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New post 27 Jul 2017, 07:40
rocko911 wrote:
kumarparitosh123 wrote:
Statement 1 : if m and n take 2 and 4 then yes
But if the take 6 and 4 then No. Insufficient.

Statement 2: if 2 and 6 then yes.
If 7 and 3 , No. Insufficient.
Combining we get a positive YES.

THUS IMO answer is C.
I hope this time I have not missed anything [FACE SAVOURING DELICIOUS FOOD]

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app



Unfortunately u did

its a E

if u combine both
10 and 6 results in NO
and
10 and 14 results in Yes

so E is the answer

Thanks !!!

I missed..

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If m and n are positive integers, is m + n divisible by 4 ? [#permalink]

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New post 07 Aug 2017, 06:13
according to statement I m=2p and n=2q, according to statement II p and q must be odd. adding m & n to each other we have m+n=2(p+q) and since p and p are odd, p+q is even hence m+n=4t for some 2t=p+q, therefore the answer is c.

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Re: If m and n are positive integers, is m + n divisible by 4 ? [#permalink]

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New post 07 Aug 2017, 07:53
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Statement 1 -
m and n should be divisible by 4. Hence the number should be of the form 2a = m and 2b = n
(2a + 2b)/4 = (a + b)/2
The answer depends on values of a and b
Not sufficient


Statement 2 -
m and n should not be divisible by 4. Hence the number can be m = 4x + 1/2/3 and n = 4y + 1/2/3
There are a lot of values for which this equation will/will not be satisfy.
Not sufficient

Combining statements 1 and 2
Now as m is divided by 2 and not divided by 4. Therefore the number should be m = 4x + 2 and n = 4y +2
Adding m and n we get 4x + 4y + 4 which is divisible by 4.


Please hit kudos if my answer helps.
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If m and n are positive integers, is m + n divisible by 4 ? [#permalink]

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New post 19 Aug 2017, 20:27
carcass wrote:
If m and n are positive integers, is m + n divisible by 4 ?

(1) m and n are each divisible by 2.

(2) Neither m nor n is divisible by 4.


Statement 1
- Let's say m = (2*a*b*...), which a,b,... are prime numbers.
- Let's say n = (2*x*y*...), which x,y,... are prime numbers.
- \(\frac{(m+n)}{4}\) = \(\frac{(2*a*b*...) + (2*x*y*...)}{4}\) = \(\frac{2 ( (a*b*...) + (x*y*...))}{4}\)
- Here we can see that a,b,x,y can be anything prime numbers.
- #1 If (a*b) + (x*y) = EVEN, YES, \(\frac{m+n}{4}\) divisible by 4.
- #2 If (a*b) + (x*y) = ODD, NO, \(\frac{m+n}{4}\) cannot divisible by 4.
- Hence, INSUFFICIENT.

Statement 2
- Each of m and n has only one maximum one factor of 2 so it cannot divisible by 4.
- Try to plug simple number : m = 3 and n = 5, divisible by 4, YES.
- Try to plug another random number : m = 5 and n = 9, NOT divisible by 4, NO.
- Hence, INSUFFICIENT.

Both statement
- Go back to this equation :
- \(\frac{(m+n)}{4}\) = \(\frac{(2*a*b*...) + (2*x*y*...)}{4}\) = \(\frac{2 ( (a*b*...) + (x*y*...))}{4}\)
- If m & n divisible by 2 but not by 4, so a,b,x,y MUST BE ODD.
- (ODD*ODD) + (ODD*ODD) = ODD + ODD = EVEN.
- 2 * ANY even number MUST BE DIVISIBLE BY 4.
- Hence, SUFFICIENT.
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If m and n are positive integers, is m + n divisible by 4 ?   [#permalink] 19 Aug 2017, 20:27
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