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Re: If x = (y)(y + 1) and y is a prime number less than 11, which of the f [#permalink]

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17 Aug 2016, 21:36

1

This post received KUDOS

Given y is a prime number less than 11 possible values of y are 2 3 5 7 given, x=y*(y+1) if y = 2 x=2*3=6 if y = 3 x=3*4=12 if y = 5 x=5*6=30 if y = 7 x=7*8=56

so possible values of x are 6,12,30,56

So from options A. 5x can be (5*6) B. 11x can be (11*12) C. 13x can be (13*12) D. 23x cannot be E. 57x can be (56*57) So answer option D.

A. 5(6) - ELIMINATE B. 11(12) - ELIMINATE C. 13(12) - ELIMINATE D. 23(#) --> CORRECT!! We don't have a value of x that is a consecutive integer with 23 E. 57(56) - ELIMINATE

Re: If x = (y)(y + 1) and y is a prime number less than 11, which of the f [#permalink]

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11 Sep 2017, 03:29

Ans is D::23x

x= y(y+1) , y is prime nr <11 => 2,3,5,7 only can be values of y x can be 2x3 3x4 5x6 7x8

A) 5x =>5x2x3 =6 => 5x6 consecutive B) 11x=> 11x3x4=12 => 11x12 consecutive C) 13x=> 13x3x4=12 => 13x12 or 12x13 consecutive D) 23x=>23x3x4 =12 and 5x6=30 not possible E). 57x=> 57x8x7 =56=> 56x57 consecutive

Clearly all values of x from Ato C and E are either 1 less of option or 1 more of option => Except the option D _________________

Give Kudos for correct answer and/or if you like the solution.

If x = (y)(y + 1) and y is a prime number less than 11, which of the following could not be the product of 2 consecutive integers?

a) 5x b) 11x c) 13x d) 23x e) 57x

Since x = y(y + 1) and y is a prime number less than 11, y = 2, 3, 5, or 7 and x = 6, 12, 30, or 56, respectively. As we can see, 5x, 11x, 13x, and 57x all can be the product of two consecutive integers. The only option that can’t is 23x since x is neither 22 nor 24.

Answer: D
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

If x = (y)(y + 1) and y is a prime number less than 11, which of the following could not be the product of 2 consecutive integers?

a) 5x b) 11x c) 13x d) 23x e) 57x

Since y is a prime number less than 11, there are only 4 possible vales for y. y COULD equal 2, 3, 5 or 7

This also means there are only 4 possible vales for x.

x = (y)(y + 1)

If y = 2, then x = (2)(2 + 1) = 6 If y = 3, then x = (3)(3 + 1) = 12 If y = 5, then x = (5)(5 + 1) = 30 If y = 7, then x = (7)(7 + 1) = 56

Now let's check each answer choice... a) Can 5x be the product of 2 consecutive integers? Yes. When x = 6, then 5x = 5(6) = 30 So, 5x CAN BE the product of 2 consecutive integers (5 and 6) ELIMINATE A

b) Can 11x be the product of 2 consecutive integers? Yes. When x = 12, then 11x = 11(12) So, 11x CAN BE the product of 2 consecutive integers (11 and 12) ELIMINATE B

c) 13x When x = 12, then 13x = 13(12) So, clearly, 13x CAN BE the product of 2 consecutive integers (12 and 13) ELIMINATE C

e) 57x When x = 56, then 57x = 57(56) So, clearly, 57x CAN BE the product of 2 consecutive integers (56 and 57) ELIMINATE E

By the process of elimination, the correct answer is D

GMATPrepNow What if y less than 13 would it be valid to eliminate D?

No it wouldn't.

If y = 11, then x = (11)(11 + 1) = 132

D) 23x So, 23x = (23)(132) = 3036 3036 cannot be expressed as the product of 2 consecutive integers. So, even if we allowed y to be 11, we still couldn't eliminate D

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