y = x(x + 1)(x + 2) where x is an even integer. Which of the following

Question

y = x(x + 1)(x + 2)  where x is an even integer. Which of the following must be true?

I. y is divisible by 6

II. y is divisible by 12

III. y is divisible by 18

A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above

chetan2u · Answer

Hi 
  Numbers are even * odd * even. So following properties 
1) one even will be multiple of 2 and one of 4... ATLEAST 
2) ATLEAST one will be multiple of 3..
3) so the product will be multiple of ATLEAST 2*4*3=24

Now 6 and 12 are factors of 24, so I and II MUST be true.
18 is NOT a factor of 24... So not a MUST

B

Luckisnoexcuse · Answer

Although it does not make any difference, just want to highlight that since it is mentioned that x is an even integer which includes 0 as well and the value of y can be 0 which is considered divisible by all integers but none of the integers are considered divisible by "0"

Abhishek009 · Answer

Let's plug in some numbers and check -

Let x = 2 ; So,  y = 2*3*4  ( Divisible by 6 & 12 )
Let x = 4 ; So,  y = 4*5*6  ( Divisible by 6 & 12 )
Let x = 6 ; So,  y = 6*7*8  ( Divisible by 6 & 12 )

Thus, in each case y is divisible by 6 & 12, so the correct answer must be (B)

sashiim20 · Answer

y = x(x + 1)(x + 2)

y  is product of  3  consecutive integers. Hence  y  must be divisible by  6 .

Given  x  is even integer. Hence  y  must be divisible by  12 .

Let  x = 2

y = 2(2+1)(2+2) = 2*3*4 = 24

24  is divisible by  6  and  12

24  is not divisible by  18

I and II only . Answer (B)...

shahul. · Answer

0 is even and an integer too. IF x=0 option E becomes the answer. please clear the doubt Bunuel

Bunuel · Answer

0 is divisible by every integer except 0 itself: 0/integer = 0 = integer.

shahul. · Answer

Thanks bunuel , but if x =0, y will become zero and option d should be the answer. But b is the OA. Why is that ?  Bunuel

Bunuel · Answer

Notice that the question asks "which of the following statements  must  be true", not "which of the following statements  could  be true". While y  could  be divisible by any integer, it  must  be divisible only by some of the numbers.

AkshdeepS · Answer

The only reason I marked (E) is that x can be "0". Point noted, 0 is divisible by every number except "0".

rishabhmishra · Answer

IMO B  By plug in numbers we know these are 3 consecutive integers where middle no. is odd and 1st and last no. is even so take any example 2,3,4 or 4,5,6 or 14,15,16 we can say only I and II situation is true.

RahulPaul · Answer

But if x is -2 (even integer); it would give 0 and wouldn't the answer be E then

chetan2u · Answer

0 is divisible by every number..
12*0=0, 5*0=0

Bunuel · Answer

Zero is divisible by EVERY integer except 0 itself ( \frac{0}{x} = 0 , so 0 is a divisible by every number, x).

ZERO:

1. Zero is an  INTEGER .

2. Zero is an  EVEN  integer.

3. Zero is  neither positive nor negative  (the only one of this kind)

4. Zero is  divisible by EVERY integer except 0 itself  ( \frac{0}{x} = 0 , so 0 is a divisible by every number, x).

5. Zero is a  multiple of EVERY integer  ( x*0 = 0 , so 0 is a multiple of any number, x)

6. Zero is  NOT a prime number  (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed:  anything/0 is undefined .

8. Any non-zero number  to the power of 0 equals 1  ( x^0 = 1 )

9.  0^0  case is  NOT tested on the GMAT .

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined.  You CANNOT take 0 to the negative power .

12.  0! = 1! = 1 .

Hope it helps.

y = x(x + 1)(x + 2) where x is an even integer. Which of the following

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y = x(x + 1)(x + 2) where x is an even integer. Which of the following

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