3 Remain After X Over 4

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3 Remain After X Over 4

by cunazza » Sat Feb 27, 2010 5:36 am
When the positive integer x is divided by 4, is the remainder equal to 3?

(1) When x is divided by 2, the remainder is 1

(2) x is divisible by 3

The OA is E.

I anwered correctly by solving algebrically in this way:

Stem: x/4 = q+3

(1) x/2 = t+1
(2) x/3 = z

So, if I consider Statem 1, I have 3 variables and 2 equations, same thing with Statem 2. If I consider the combination of the equations, I have 4 variables and 3 equations.
Too many unknowns and too few equations. Can anyone spot a flaw in my logic?
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by kstv » Sat Feb 27, 2010 8:42 am
(1) When x is divided by 2, the remainder is 1 so x is a odd no.
but acc. to you x/2 = t+1 or x = 2(t+1) - an even no
It is better to take x = 2a+1, this when divided by 2 will leave a remainder 1
but we cannot say whether (2a+ 1) /4 will leave a remainder 3. Not sufficient.

(2) you are using x/3 = z but it may not be necessary to use another unknown.
(1) and (2) have to be compatible so that option E is atleast a possibility.
use the eq x=2a + 1, but x is divisible by 3
so using your eq x= 3z substitute z = 2a + 1
so x=3(2a+1) = 6a + 3 but in (6a + 3)/4 the remainder is 3 only when a is even and 1 when odd - this part is not necessary. Just (6a + 3)/4 is enough to tell you it is still not Sufficient. 6a is not divisible by 4

So E.

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by outreach » Sun Feb 28, 2010 5:41 am
1.
x =3,5,7,9 etc
so when x is divided by 4 remainder wil be 3,1,3,1 etc

not suff

2.
x=3,6,9,12 etc
so when x is divided by 4 remainder wil be 3,2,1,0 etc

not suff

combined 1 and 2
x=3,15,21, etc
so when x is divided by 4 remainder wil be 3,3,0
not suff


E
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