What is the value of x?

Question

(1) (x)(x + 1) = (2013)(2014)

(2) x is odd

Bunuel · Accepted Answer

VERITAS PREP OFFICIAL SOLUTION:

C. This problem is a classic example of the "Why Are You Here" strategy. Clearly statement 2 is not sufficient on its own, so why was it written?

In statement 1, the "obvious" answer for x is that x = 2013 and (x + 1) would then equal 2014. Which looks pretty sufficient. But there's one additional, not as obvious possibility: x = -2014 and (x + 1) = -2013. Since negative-times-negative is positive, that would give the same result. So statement 1 looks pretty sufficient but it is not. Statement 2 provides that little clue by emphatically stating that x is odd. That should get you thinking "how could x not be odd?" and of course that would be if x were -2014 and x + 1 were -2013. With both statements together, that negative-negative possibility is off the table, so the correct answer is C.

chetan2u · Answer

ans C.... 
1) statement one tells us that x is either -2014 or 2013....insfficient
2) statement 2 tells us that it is odd..insufficient
 combined it tells us that it is 2013

TeamGMATIFY · Answer

This is classic question in which you can be trapped easily.
The questions seems easy but there is a little catch involved here.

Statement 1 : (x)(x + 1) = (2013)(2014)
If directly seems that the equation means x = 2013 and (x + 1) 2014
But x can be - 2014 and (x+1) can be - 2013
Hence x has two values.
INSUFFICIENT

Statement 2 : x is odd.
This in itself does not tell us anything.

Combining statement 1 and 2 : 
We are left with just one value of x: 2013
SUFFICIENT

Option C

0akshay0 · Answer

This link help will help one understand the "Why Are You Here" strategy https://youtu.be/eY_MIpioA8M

analytica233 · Answer

x = ? (1) x(x + 1) = 2013.2014 <=> x^2 + x - 2013.2014 = 0 <=> (x + 2014)(x - 2013) = 0 --> x = -2014 or x = 2013. Insufficient. (2) x is odd. x can be any odd number. Insufficient. (1) + (2) --> x = 2013. Sufficient. Answer (C).

lakshya14 · Answer

Hi, what if I prime factorize 2013 = 3 x 671 and multiply either of the values in 2014, lets say 3.
Then it becomes, 671(2014x3 + 3),
 = 671(6045)
 = 671(60444 + 1)
 Then, similarly with retaining 3 only as both are odd, will give 2 values of x.

Bunuel · Answer

What are two odd values of x which satisfy (x)(x + 1) = (2013)(2014)?

(x)(x + 1) = (2013)(2014) has two solutions for x, one is even (x = -2014) and another is odd (x = 2013).

BrentGMATPrepNow · Answer

Target question :    What is the value of x?

Statement 1 : (x)(x + 1) = (2013)(2014)  
At first glance, statement 1 looks sufficient. However, there are two possible values of x: 
 Case a : x = 2013 and x+1 = 2014. In this case, the answer to the target question is  x = 2013 
 Case b : x = -2014 and x+1 = -2013. In this case, the answer to the target question is  x = -2014 
Since we can’t answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : x is odd 
Definitely NOT SUFFICIENT

Statements 1 and 2 combined    
Statement 1 indirectly tells us that EITHER x = 2013 OR x = -2014
Statement 2 tells us that x is odd
In order for both statements to be true, it must be the case that  x = 2013 
Since we can answer the  target question  with certainty, the combined statements are SUFFICIENT

Answer: C

bumpbot · Answer

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What is the value of x?

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