Problem on Number Properties

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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Wed Aug 03, 2011 5:20 am

If A and B are distinct non zero numbers, is X+Y an even integer?
(1)A^x.A^y=1
(2)B^x.B^y=1

Detailed explanations would be appreciated. Don't have the OA on this one

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Source: Beat The GMAT — Data Sufficiency | Wed Aug 03, 2011 5:20 am

bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 amFollowed by: 33 members; Thanked: 118 times; GMAT Score:710

by bblast » Wed Aug 03, 2011 6:22 am

the answer should be E

if a = 1 then no matter what x,y are, the statement 1 or 2(b=1) always holds.

if a is not equal to 1, then statement 1 is sufficient. because even if a is (1.1),

a^x * a^y = 1
{above will hold only for y is always = -x} or {x and y both are 0} {in either case we know x+y = even}

however I wonder why the statement A and B are distinct is given ??

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Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

by Frankenstein » Wed Aug 03, 2011 6:39 am

Hi,
Consider a^n = 1, where n is an integer.
if a=1, n can be any number
if a = -1, n should be even
if a is not 0,1,-1, then n = 0, which is even.
Btw, any information that (x+y) is integer?
I am assuming that (x+y) is an integer
From(1):
a^(x+y) = 1. I will denote (x+y) by n.
So,
if a=1, n can be any number
if a = -1, n should be even
if a is not 0,1,-1, then n = 0, which is even.
Not sufficient
From(2):
same as above
Not sufficient

From(1) and (2):
Both a and b can't be 1. So, (x+y) has to be even.

Hence, C

Cheers!

Things are not what they appear to be... nor are they otherwise

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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 amFollowed by: 33 members; Thanked: 118 times; GMAT Score:710

by bblast » Wed Aug 03, 2011 6:47 am

I came pretty close Frank.

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