General Strategy for GR8767 #25

Forum for the GRE subject test in mathematics.
markisus
Posts: 6
Joined: Tue Feb 12, 2013 12:12 am

General Strategy for GR8767 #25

Postby markisus » Tue Mar 12, 2013 12:06 am

http://www.wmich.edu/mathclub/files/GR8767.pdf

How do I solve number 25?
The question is:

x and y are positive integers such that

Which of the following must also be divisible by 11?
A: 4x + 6y
B: x + y + 5
C: 9x + 4y
D: 4x - 9y
E: x + y - 1
Last edited by markisus on Tue Mar 12, 2013 12:21 am, edited 2 times in total.

kuz
Posts: 46
Joined: Wed Jan 18, 2012 3:32 am

Re: General Strategy for GR8767 #25

Postby kuz » Tue Mar 12, 2013 12:14 am

Probably the quickest way is to just multiply the original expression modulo 11 and check each of the answers.
If
3x + 7y = 0 (mod 11),
then by multiplying by 4 we find that
x + 6y = 0 (mod 11),
so adding both together shows that
4x + 2y = 0 (mod 11).
This is the same as
4x - 9y (mod 11),
so the answer is D.

markisus
Posts: 6
Joined: Tue Feb 12, 2013 12:12 am

Re: General Strategy for GR8767 #25

Postby markisus » Tue Mar 12, 2013 12:26 am

Oh nice! Thanks.




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