If x is a positive integer less than 30, is x odd?

VERITAS PREP OFFICIAL SOLUTION:

While it’s possible to solve this question with a conceptual understanding, it is much easier to put some numbers to work for you. When you do employ numbers, remember that your goal is to play devil’s advocate. Your goal is to determine whether x is an odd number, so you will likely start with an odd number that satisfies statement (1). 5 works here, as 5 divided by 3, as 5/3 = 1 remainder 2. So x could be odd. Now that you’ve found an odd value of x—the answer yes to the overall question—your goal should change. You want to find an even value, because that would show that the statement is not sufficient. If you try everything you can think of and cannot find an even value of x, then you can conclude that it is sufficient. You want to play devil’s advocate to ensure that either x must be odd, or conclude that the statement is not sufficient. With that in mind, you might try 8: 8 divided by 3 provides a remainder of 2 (8/3 = 2 remainder 2). So now you have an even potential x—and the answer no to conclude that statement (1) is not sufficient.

The same process works for statement (2). 7 is an odd number that does the same, so x could still be odd, providing a yes answer. But 12 is an even number that satisfies statement (2), so you can get the answer no, and the answer is thus still maybe. Statement (2) is not sufficient. Taken together, the statements provide a bit more information, as now you know that x provides a remainder of 2 when divided by 3 and when divided by 5. You might recognize 17 as such a number, noting that 15 is the least common multiple of 3 and 5, so 17 will divide out that 15 and leave 2 remaining. Here’s where you really need to play devil’s advocate: If you chart out the values that work with each statement and look for matches between them, you may well conclude that 17 is the only such value less than 30:



But still play devil’s advocate. Is there any even number that could fit the bill? There is, but it’s very hard to find unless you remember what happens when you divide a smaller number by a larger number. 2 also works. When 2 is divided by 3, the quotient is 0 and the remainder is 2. When 2 is divided by 5, the quotient is 0, and the remainder is 5. 2 is the even counterpart, and although it may not be as readily clear as 17, if you force yourself to play devil’s advocate and consider the entire range of numbers available to you, you will often find that “catch” upon which correct answers often depend. The correct answer to this problem is E, but the authors of the question are betting that you will forget to consider 2 and therefore fall into the trap of selecting C.


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OA should be E.

1) when x is divided by 3, remainder is 2
x could be : 2,5,8,11,14,17
there are even as well as odd no.'s.
Not sufficient
2) when x is divided by 5, remiander is 2
x could be : 2,7,12,17
Again, both type of no's
Not sufficient

(1)+(2)
Even after taking both statements, we have both even and odd no's : 2, 17

Not sufficient

St 1. x could be 2 (even), 5,8 etc... NOT SUFFICIENT (multiple of 3+2) 11 14 17 20 23 26 29
St 2. x could be 2, 7, 12 NOT SUFFICIENT (multiple of 5+3) 17 22 27
St1+St2. x could be 2 or 17 NOT SUFFICIENT
Answer E

The above solutions are using a nice rule that says:

If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

For more on this rule, see our free video: https://www.gmatprepnow.com/module/gmat- ... /video/842

Cheers,
Brent

"2 also works. When 2 is divided by 3, the quotient is 0 and the remainder is 2. When 2 is divided by 5, the quotient is 0, and the remainder is 5."

I am still not clear. Can someone please explain how 2 works here? What is the remainder when 2 is divided by 5?

It is based on the rule of divisibility and definition of division.

As one of the posters above has mentioned, when you divide a smaller number by a larger number, the remainder is the smaller number itself and the quotient = 0.

Example, P=Qk+R, where P,Q,R,k are all integers and P<Q, then the remainder, R=Q and k =0.

What is the remainder and quotient when 2 is divided by 3?

As 2<3, the quotient is 0 and remainder = 2 = smaller number itself. This is true as : P=Qk+R ---> Qk+R = 0*3+2 = 2 = P.

Similarly, as 2<5, you will get remainder of 2 when 2 is divided by 5, with a quotient of 0.

This is actually the very reason why the answer to this "C-Trap" question is NOT C but is E.

Hope this helps.
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Thank you for the explanation, Engr2012 =)

If x is a positive integer less than 30, is x odd?

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

x=3q+2, where q is an non-negative integer

x could be 2,5,8,11,14

odd or even

Insufficient

(2) When x is divided by 5, the remainder is 2.

x=5k+2, where g is an non-negative integer

x could be 2, 7,12,17

odd or even

Insufficient

Combining 1 & 2

x=15z+2, then it depends on z

Answer: E

We are given that x is a positive integer less than 30 and need to determine whether x is odd.

Statement One Alone:

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

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

Statement Two Alone:

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

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

Statements One and Two Together:

From statement one, we see that x could be the following:

2, 5, 8, 11, 14, 17, 20, 23, 26, and 29

From statement two, we see that x could be the following:

2, 7, 12, 17, 22, and 27

Since x could be 2, which is even, and x could be 17, which is odd, the two statements together are still not sufficient to answer the question.

Answer: E
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