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If x and y are positive integers, is x divisible by 3? 1).

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If x and y are positive integers, is x divisible by 3? 1). [#permalink] New post 25 Apr 2006, 20:13
If x and y are positive integers, is x divisible by 3?

1). x-y is divisible by 3

2). x-2y is divisible by 3
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Re: divisiblity [#permalink] New post 25 Apr 2006, 21:12
joemama142000 wrote:
If x and y are positive integers, is x divisible by 3?

1). x-y is divisible by 3
2). x-2y is divisible by 3


i guess C.

i. if x = 8 and y = 5, x-y is divisible by 3. but x is not divisible by 3.
ii. if x = 8 and y = 1, x-2y is divisible by 3. still x is not divisible by 3.

if x-y and x-2y are divisible by 3, x must be divisible by 3.
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 [#permalink] New post 25 Apr 2006, 21:15
can you guys find a case for II, where x is divisble by 3?
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 [#permalink] New post 25 Apr 2006, 21:17
I would say C. 2 alone is insufficient. For example, if x=13 and y=5 then 13-2(5)/3 is an integer but x/3 is obviously not.
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 [#permalink] New post 25 Apr 2006, 21:19
Quote:
can you guys find a case for II, where x is divisble by 3?


x=12 and y=3
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 [#permalink] New post 25 Apr 2006, 21:25
thanks guys, oa is C
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 [#permalink] New post 25 Apr 2006, 21:28
joemama142000 wrote:
thanks guys, oa is C


i like your questions. i like your quarries. i like your answer timing.

you are a very positive soul..

gooooodddluck to you..
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 [#permalink] New post 25 Apr 2006, 21:42
thanks prof, i like your answers :-D
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 [#permalink] New post 28 Apr 2006, 23:25
Concur with C.

These kind of questions, is there a easier way than plugging in numbers?
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 [#permalink] New post 29 Apr 2006, 00:51
TeHCM wrote:
Concur with C.

These kind of questions, is there a easier way than plugging in numbers?


Yes it is possible, although substituting numbers might get the answer quicker.


It is evident that if x - y is divisible by 3, then "x" may or may not be divisible by 3.

Similarly it is evident for the info given in statement 2 i.e, x - 2y is divisible by 3. Then "x" may or may not be divisible by "3".

Now, when we try for "C", i.e., the information given in both the statements is true.

i.e, x - y is divisible by 3 and so is x - 2y.

For any two integers A and B, if both A and B are divisible by 3, then any combination of the two numbers say, A + B or A - B or 2A + B or 2A - B etc will also be divisible by 3.

IN this case then if one were to use B - 2A as the combination of the two numbers

i.e., x - 2y - 2(x - y) will be divisble by 3
or x - 2x is divisible by 3 or -x is divisible by 3.

So, "C" is the answer.
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 [#permalink] New post 29 Apr 2006, 04:04
Another approach!

I and II are not possible as illustrated above

So lets prove C!

Ley x-2y =3m (as its divisible by 3)
Similarly, x-y=3n

So x-2y - (x-y) = x - 2y -x + y = y
y = 3m - 3n = 3(m-n)

So y is divisible by 3

Thus as x-y is divisble by 3, x HAS to be divisible!

Hence C!
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 [#permalink] New post 29 Apr 2006, 06:05
It is C. Add both the statements, you get
2x=3(K) + 3 y, where k is some constant
2x=3(K+y)
2 is not divible by 3. Therefore x is divisible by 3
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 [#permalink] New post 29 Apr 2006, 06:46
OR if you are familar with modulo arithmetic

x-y = 0 mod 3 --(I)
x-2y = 0 mod 3 --(II)

Multiply (I) by 2 and subtract (II) from (I)
2x-2y - (x-2y) = 0 mod 3
x = 0 mod 3
Hence x must be divisible by 3.
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  [#permalink] 29 Apr 2006, 06:46
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