When the positive integer x is divided by 4, is the

I still think that C is correct.

1) When x/3 is divided by 2, the remainder is 1.
It means that x is divisible by 3 because x/3 has to be an integer.
So,
x=3, r=3
x=9, r=1
insufficient

2) x is divisible by 5.

So,
x=5, r=1
x=15, r=3
insufficient

1)&2)
write out x that satisfy 1st condition and find two the closest numbers that satisfy 2nd condition also:

x=3,9,15,21,27,43,49,55

So,
x=15, r=3
x=55, r=3
sufficient


Give two examples that satisfy both conditions and give different remainders.

E

you have 25 and 55, they meet both conditions

and 25/4 ---> remainder is 1

55/4 ---> remainder is 3

25 is not divisible by 3 and therefore first condition is not satisfied or maybe you know what is remainder when 8.333333 is divided by 2

hey walker, statement 1 does not say that x has to be divisible by 3, it say x/3 divide by 2 ---> x/6. This is another typical question by chineseburn, where the wording is unnecessary complicated and designed to play "gotcha".

(1) When 25/3 is divided by 2, the remainder is 1.

it has no sense, as remainder is applicable only for integers.

I disagree that 1st question simply means x/6 and can say that in real GMAT the condition would state as:

(1) X is divisible by 3 and when x/3 is divided by 2, the remainder is 1.

But if this is the case, x can not equal to 55 either.

Thanks! Silly mistake! should be:

x=3,9,15,21,27,33,39,45

So,
x=15, r=3
x=45, r=1

insufficient.

E

I think C is correct.

I think the examples of 25, 55 and 65 do not satisfy (1), i.e. 25/3, 55/3 and 65/3 divided by 2 do not lead to remainder 1.

Agreed with walker.

**** I got C too..

anyways..i realized my mistake...

stem x=4N+3 i.e x=3, 7, 11, 15 etc

1) x=6N+1 x=1, 7, 13, 19, 25...

clearly Insuff

2) x is divible by 5.., x=5 remainder 1..x=10 remainder 2...x=15 remainder 3...

insuff..

together x can be 15, 55...

HERE is what i did wrong..i accidently ended up dividing by 3..the question clearly asks x/4...GRRRRRR

I agree with OA=e..remainder can be 1 or 3..

In my earlier post above,
I assumed that the integer x is being divided by 6. (Otherwise you must assume that x/3 is an integer). Indeed, the division algorithm (actually a theorem that relates to all integers positive, zero and negative with the constraint that d ≠ 0 ) states that for two integers a and d, with d ≠ 0, there are unique integers q and r such that a = qd + r and 0 ≤ r < |d| ,
The integer
-q is called the quotient
- r is called the remainder
-d is called the divisor
-a is called the dividend
This question needs to be more clear as it is the case in the GMAT.
The set of "qualified" integers you get will depend on your choice of "x/3" or "x/6" since an integer can be divisible by 3 without being divisible by 6. If you choose to go with "x/6" then you'll have the "qualified" integer 55. 55 is divisible by 5 and when you divide 55 by 6 the remainder is 1. So the two conditions are satisfied. Finally when you divide 55 by 4 the remainder is 3. The "qualified'' integer 25 will give a remainder of 1 when divided by 4.
Yet, no matter what option you go with, the answer will still be E.
Very nice discussion overall!