|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2
Kudos [?]:
2
[0], given: 0
|
Are both of the integers X and Y divisible by 3? (1) X+Y and [#permalink]
 22 May 2004, 07:31
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
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
_________________
Exceed your goals and then Proceed to Succeed!!
|
|
|
|
|
|
|
Manager
Joined: 10 Mar 2004
Posts: 69
Location: Dallas,TX
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2
Kudos [?]:
2
[0], given: 0
|
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
_________________
Exceed your goals and then Proceed to Succeed!!
|
|
|
|
|
|
Manager
Joined: 10 Mar 2004
Posts: 69
Location: Dallas,TX
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2
Kudos [?]:
2
[0], given: 0
|
Maybe, we can wait other idea....Anyone please 
_________________
Exceed your goals and then Proceed to Succeed!!
|
|
|
|
|
|
Senior Manager
Joined: 02 Mar 2004
Posts: 372
Location: There
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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.
|
|
|
|
|
|
SVP
Joined: 30 Oct 2003
Posts: 1963
Location: NewJersey USA
Followers: 3
Kudos [?]:
25
[0], given: 0
|
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
Just extend your analysis further...
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
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2
Kudos [?]:
2
[0], given: 0
|
Thanks for anandnk's explanation.
_________________
Exceed your goals and then Proceed to Succeed!!
|
|
|
|
|
|
Manager
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2
Kudos [?]:
2
[0], given: 0
|
Anyway, if n=3 and m= 3, is your explanation still applicable? Quite headache 
_________________
Exceed your goals and then Proceed to Succeed!!
|
|
|
|
|
|
Senior Manager
Joined: 07 Oct 2003
Posts: 380
Location: Manhattan
Followers: 1
Kudos [?]:
2
[0], given: 0
|
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 Just extend your analysis further... 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
Posts: 372
Location: There
Followers: 1
Kudos [?]:
0
[0], given: 0
|
Of course, yes.
0 = 3*0
or 0 is a multiple of any integer!
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
If X,Y r integers, is X*Y divisible by 3 ? 1) (X+Y)^2
|
allabout |
7 |
08 Jan 2006, 13:06 |
|
|
|
|
if x and y are integers is xy + 1 divisible by 3 ? 1. when x
|
winsome |
3 |
17 Mar 2007, 21:20 |
|
|
|
|
If x and y are integers, is xy + 1 divisible by 3? 1) when x
|
rao |
4 |
01 Aug 2008, 17:17 |
|
|
1
|
|
If x and y are integers, is xy divisible by 3 ? 1. (x+y)^2
|
bigfernhead |
9 |
27 Apr 2009, 15:51 |
|
|
1
|
 |
If x, y and k are integers, is xy divisible by 3? (1) y =
|
shrive555 |
7 |
24 Oct 2010, 12:34 |
|
|
|
|
|
|
close
