Is the positive integer x an even number?

Question : Is the positive integer x an even number?

Statement 1: If x is divided by 3, the remainder is 2.
i.e. x = 3a+2 (Possible values of x = 5, 8, 11, 14, 17, 20, 23, 26, 29...etc.)
@a = Even, x = even+2 = Even
@a = Odd, x = Odd+2 = Odd. Hence,
NOT SUFFICIENT


Statement 2: If x is divided by 5, the remainder is 2.
i.e. x = 5b+2 (Possible values of x = 7, 12, 17, 22, 27, 32, 37...etc.)
@b = Even, x = even+2 = Even
@b = Odd, x = Odd+2 = Odd. Hence,
NOT SUFFICIENT

Combining the two statements
Statement 1 says (Possible values of x = 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38...etc.)
Statement 2 says (Possible values of x = 7, 12, 17, 22, 27, 32, 37...etc.)

i.e. x may still be 17, 37, 52, 67... etc. i.e. x may still be Even as well as odd. Hence,
NOT SUFFICIENT

Answer: Option E

From A :
\(X =3(m)+2\)
For m=0,1,2,3,4.... X =2,5,8.....

X can be even or odd .Option A is not sufficient.

From B:
\(X=5(m)+2\)
For m=0,1,2,3,4.... X =2,7,12....
X can be even or odd .Option B is not sufficient.

Combining both the options together :
\(X=15m+2\)
For m=0,1,2,3,4.... X =2,17,47,62....
X can be even or odd .

Option E is the correct answer.

VERITAS PREP OFFICIAL SOLUTION:

Question Type: Yes/No. The question asks: “Is x an even number?”

Given information in the question stem or diagram: x is a positive integer.

Statement 1: If x is divided by 3, the remainder is 2. This statement can be easily addressed by considering numbers and Playing Devil’s Advocate. First attempt to get an even number from Statement 1: The number 8 gives you a remainder of 2 when divided by 3. This is an even number so you have a “yes” answer. Next choose a number exactly 3 higher or lower than 8: 11 is an odd number and also gives a remainder of 2 when divided by 3. So 11 works but gives you a “no” answer. This statement is not sufficient, so eliminate choices A and D.

Statement 2: If x is divided by 5, the remainder is 2. This statement can be treated in the same way as Statement 1. First get a “yes” answer. If x = 12, then 12/5 = 2 remainder 2. Next choose a number exactly 5 above or below 12. If x = 7, then 7/5 = 1 remainder 2. So this works as well and gives you a “no” answer. This statement is also not sufficient, so eliminate choice B.

Together: Taken together you know that x must be 2 greater than a multiple of both 5 and 3. So x must yield a remainder of 2 when divided by 15. So 17 would work. 17/5 = 3 remainder 2, and 17/3 = 5 remainder 2. 17 is an odd number so you have a “no” answer. Now add 15 to 17 to get 32. 32 is an even number and will give you a “yes” answer. 32/5 = 6 remainder 2 and 32/3 = 10 remainder 2. So 32 satisfies both statements as well. Note: Number picking is of course not required to answer this question. With a good conceptual understanding of the number line, and quotient/remainder problems in particular, you can see quite clearly that the answer will be E in this problem. As is often the case, Arithmetic problems in Data Sufficiency can be quickly and reliably answered if you have a deep conceptual understanding of the subject matter.
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x=3q+2
x=5r+2

3q=5r, q/r=5/3, so x can be even or odd

E

Given => x is a positive integer
WE need to examine it fro its even/odd property
Statement 1
here x=3k+2 for some k
x=2
x=5
From above two test cases we can say that this statement is insufficient
Statement 2
x=5m+2 for some m
x=2
x=7
from the above two test cases we can say that this statement is insufficient.
Combing the two statements
we can say that x=2+15p for some p
x=2
x=17
clearly from the above two test cases we can say that this statement is insufficient.
Hence E
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Plug in ad check

FROM STATEMENT - I ( INSUFFICIENT )

x = 5 , x/3 = remainder 2 ( x = Odd )
x = 8 , x/3 = remainder 2 ( x = Even )

Thus x can be even/Odd..

FROM STATEMENT - II ( INSUFFICIENT )

x = 7 , x/5 = remainder 2 ( x = Odd )
x = 12 , x/5 = remainder 2 ( x = Even )

Thus x can be even/Odd..

FROM STATEMENT - I & II ( INSUFFICIENT )

Since, we do not have a unique value of x we can not state whether x = Even/Odd

Hence, Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed, answer will be (E)..
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Hi Bunuel, thanks for careful explanation on the problem.
It took me around 6~ minutes to get options satisfying both (1) & (2) and yield in "E". Your formula for finding LCM of quotients and adding the remainder is really useful and can save plenty of time in real exam conditions. I'm studying for GMAT using MGM 6th edition course however it doesn't provide for this approach. Would you please give a link to a source or explain more how to tackle similar problems in case we have different remainders (2 & 3 for instance) or if one quotient is a multiple of another one.

Thanks in advance!

Check the links below for more:

Divisibility and Remainders on the GMAT

Divisibility Tips

Theory on remainders problems
Tips on remainders

Units digits, exponents, remainders problems

DS remainders problems
PS remainders problems

DS divisibility problems
PS divisibility problems
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stmt-1: If x is divided by 3, the remainder is 2
x = 5 (odd) / 3 , remainder is 2
x = 2 (Even) / 3 , remainder is 2

stmt-1: If x is divided by 5, the remainder is 2
x = 7 (odd) / 5 , remainder is 2
x = 2 (Even) / 5 , remainder is 2

Stmt-1 + stmt-2: remainder is 2 when divided by 15
x = 17(odd) / 15 , remainder is 2
x = 2 (Even) / 15 , remainder is 2

Answer E

We need to determine whether x is an even number.

Statement One Alone:

If x is divided by 3, the remainder is 2.

Statement one alone is not sufficient to answer the question. For example, x could be 5, which is an odd number, or x could be 8, which is an even number.

Statement Two Alone:

If x is divided by 5, the remainder is 2.

Statement two alone is not sufficient to answer the question. For example, x could be 7, which is an odd number, or x could be 12, which is an even number.

Statements One and Two Together:

Using the statements together, we see that the first number that fulfills both statements is n = 2, which is an even number.

To determine the next value in the list, we can use the least common multiple of 3 and 5, which is 3 x 5 = 15. Thus, the next number that could be x is 2 + (5 x 3) = 17, which is an odd number. Thus, the two statements together are still not sufficient to answer the question.

Answer: E
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#1If x is divided by 3, the remainder is 2.
x can be 2 ,5 ; insufficient
#2
If x is divided by 5, the remainder is 2
x can 2,7 insufficient
from 1 &2
x can be 2,17,32 where x is divided by 15
insufficient
OPTION E

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