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 Post subject: Book XIV, Q# 29Posted: Thu Feb 02, 2012 5:44 pm

Joined: Thu Feb 02, 2012 5:21 pm
Posts: 3
In this problem, I understand why statement 2 is not sufficient (there are to many possible options). But why the answer is A?? I don´t understand why statement 1 alone is sufficient. I tried to substitute the integer r with real numbers and, this is what happens:

60/3=20, r=0
61/3=20, r=1
62/3=20, r=2
.......

So, since there are 3 different reminders depending the value of r, why the answer is saying that the reminder is always 0??? Where am I lost?

Thanks!!!

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 Post subject: Re: Book XIV, Q# 29Posted: Fri Feb 03, 2012 4:04 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
From (1)

the remainder will always be 0

r = 1, 6r = 6 6/3 = 2 R0
r = 2, 6r = 12, 12/3 = 4 R0
r = 3, 6r = 18, 18/3 = 6 R0
etc.

from (2), r could be a fraction (and 6r could still be an integer)

examples:
r = 1/2, 6r = 3, 3/3 = 1 R0
r = 4/3, 6r = 8, 8/3 = 2 R2

since we can get different remainders (2) is not sufficient.

Veritas Help

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