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# Are both of the integers X and Y divisible by 3? (1) X+Y and

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Manager
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Are both of the integers X and Y divisible by 3? (1) X+Y and [#permalink]  22 May 2004, 06:31
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Are both of the integers X and Y divisible by 3?

(1) X+Y and X-Y are divisible by 3
(2) X^2 and Y^2 are divisible by 9
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Manager
Joined: 10 Mar 2004
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Can be solved by each individually so ....(d)

Please note that it is already given that X and Y are intergers. If it was not specifically given then (2) may not hold good.

cheers
Manager
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Can you explain (1) ? I am trying to solve the equation but I am not so sure it's the right method.

If x+y is divisible by 3 => x+y = 3n , when n = integer -(a)
If x-y is divisible by 3 => x-y = 3m, when m = integer -(b)

From (a) and (b) => x = 3(m+n)/2 and y = 3(n-m)/2
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Manager
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becoolja wrote:
Can you explain (1) ? I am trying to solve the equation but I am not so sure it's the right method.

If x+y is divisible by 3 => x+y = 3n , when n = integer -(a)
If x-y is divisible by 3 => x-y = 3m, when m = integer -(b)

From (a) and (b) => x = 3(m+n)/2 and y = 3(n-m)/2

SORRYYYYYYYYYYYYYYYYYYY

was wrong above.

Cannot be solved by (1)

I hope it can be solved by (2) so (b)
Manager
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Maybe, we can wait other idea....Anyone please
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Senior Manager
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Re: DS_10_15 [#permalink]  22 May 2004, 10:32
1. x+y = 0 (3) x-y = 0 (mod 3)
or x = y = 0 (3) sufficient

2. x^2 = 0 (9), x = 0 (3)
sufficient.

D it is.
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becoolja wrote:
Can you explain (1) ? I am trying to solve the equation but I am not so sure it's the right method.

If x+y is divisible by 3 => x+y = 3n , when n = integer -(a)
If x-y is divisible by 3 => x-y = 3m, when m = integer -(b)

From (a) and (b) => x = 3(m+n)/2 and y = 3(n-m)/2

Since x and y are integers then (m+n)/2 and (m-n)/2 have to either integers
or k/3.
For no integers pairs (m,n) (m+n)/2 or (m-n)/2 yields k/3 Hence
(m+n)/2 and (m-n)/2 have to be integers. If they are integers then both x and y are divisible by 3
Manager
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Thanks for anandnk's explanation.
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Manager
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Anyway, if n=3 and m= 3, is your explanation still applicable? Quite headache
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Senior Manager
Joined: 07 Oct 2003
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anandnk wrote:
becoolja wrote:
Can you explain (1) ? I am trying to solve the equation but I am not so sure it's the right method.

If x+y is divisible by 3 => x+y = 3n , when n = integer -(a)
If x-y is divisible by 3 => x-y = 3m, when m = integer -(b)

From (a) and (b) => x = 3(m+n)/2 and y = 3(n-m)/2

Since x and y are integers then (m+n)/2 and (m-n)/2 have to either integers
or k/3.
For no integers pairs (m,n) (m+n)/2 or (m-n)/2 yields k/3 Hence
(m+n)/2 and (m-n)/2 have to be integers. If they are integers then both x and y are divisible by 3

Got a technical question here,
is 0 divisible by 3?
Senior Manager
Joined: 02 Mar 2004
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Of course, yes.

0 = 3*0

or 0 is a multiple of any integer!
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