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x, y, and z are consecutive integers, where x < y < z. Which of the fo

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x, y, and z are consecutive integers, where x < y < z. Which of the fo [#permalink] New post   28 May 2019, 00:11
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x, y, and z are consecutive integers, where x < y < z. Which of the following must be divisible by 3 ?

I. (x + 3)yz
II. (x + 1)(y + 1)(z + 1)
III. (x + 1)(y + 2)(z + 3)

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III
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E
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Re: x, y, and z are consecutive integers, where x < y < z. Which of the fo [#permalink] New post   28 May 2019, 00:33
Bunuel wrote:
x, y, and z are consecutive integers, where x < y < z. Which of the following must be divisible by 3 ?

I. (x + 3)yz
II. (x + 1)(y + 1)(z + 1)
III. (x + 1)(y + 2)(z + 3)

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


test the relation with 1,2,3 , 3,4,5, 2,3,4
we get all given options as valid
IMO E
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Re: x, y, and z are consecutive integers, where x < y < z. Which of the fo [#permalink] New post   28 May 2019, 00:34
Bunuel wrote:
x, y, and z are consecutive integers, where x < y < z. Which of the following must be divisible by 3 ?

I. (x + 3)yz
II. (x + 1)(y + 1)(z + 1)
III. (x + 1)(y + 2)(z + 3)

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


Must be divisible? let's try to negate to quickly arrive at the soln.
x can be odd/even

Concept any three consecutive numbers must be divisible by 3.

I. Yes (X+3)(x+1)(X+2)

II. Yes (X+3)(x+1)(X+2)

III. (x+1)(x+3)(x+5), one of this terms has to be divisible by 3.

E
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Re: x, y, and z are consecutive integers, where x < y < z. Which of the fo [#permalink] New post   28 May 2019, 00:46
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Bunuel wrote:
x, y, and z are consecutive integers, where x < y < z. Which of the following must be divisible by 3 ?

I. (x + 3)yz
II. (x + 1)(y + 1)(z + 1)
III. (x + 1)(y + 2)(z + 3)

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


We know that product of any 3 consecutive integers is ALWAYS divisible by 3 (Eg: 1*2*3, 2*3*4 . . . . )
Since x, y and z are consecutive integers.
y = x+1
z = x+2

I. (x+3)yz = (x+3)(x+1)(x+2)
= (x+1)(x+2)(x+3) which is a product of 3 consecutive integers. - Divisible by 3

II. (x + 1)(y + 1)(z + 1)
= (z-2+1)(z-1+1)(z+1)
= (z-1)(z)(z+1) which is a product of 3 consecutive integers. - Divisible by 3

III. (x + 1)(y + 2)(z + 3)
= (x+1)(x+1+2)(x+2+3)
= (x+1)(x+3)(x+5)
Substitute some values of x
x ----> (x+1)(x+3)(x+5)
0 ----> 1*3*5
1 ----> 2*4*6
2 ----> 3*5*7
3 ----> 4*6*8
So, III is also divisible by 3.

IMO Option E
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Re: x, y, and z are consecutive integers, where x < y < z. Which of the fo [#permalink]
28 May 2019, 00:46
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