If integer x is chosen at random from 50 to 149 inclusive, what is the

Question

If integer x is chosen at random from 50 to 149 inclusive, what is the probability that x^3 - x is a multiple of 12?

A. 1/4
B. 1/2
C. 2/3
D. 3/4
E. 4/5

CrackverbalGMAT · Accepted Answer

x^3 - x  can be written as  x(x^2 - 1) = x (x + 1)(x - 1) = (x - 1) * x * (x + 1)

Basically, we want to know if the product of 3 consecutive terms is divisible by 12 for x = 50, 51...till 149

Every set of 3 consecutive numbers will always be divisible by 3. We need to check for divisibility by 4.

When x = 50, 49 * 50 * 51. Not divisible by 4

When x = 51, 50 * 51 * 52. Is divisible by 4

When x = 52, 51 * 52 * 53. Is divisible by 4

When x = 53, 52 * 53 * 54. Is divisible by 4

When x = 54, 53 * 54 * 55. Not divisible by 4

So we see that if x is even and not divisible by 4, then  x^3 - x  is not divisible by 12.

The series is 50, 54, 58.....146

The total numbers not divisible by 12 =  \frac{146 - 50}{4} + 1 = 25

The total numbers = 149 - 50 + 1 = 100

The total numbers divisible by 12 = 100 - 25 = 75

Probability = Favorable outcomes / Total Outcomes = 75/100 = 3/4

Option D

Arun Kumar

IanStewart · Answer

x^3 - x = x(x^2 - 1) = (x-1)(x)(x + 1)

So we have a product of three consecutive integers, and this product will always be divisible by 3! = 6. So all we really care about is whether the product is divisible by 4 (since then it will be divisible by 12). Now:

• if x is odd, x-1 and x+1 are both even, so (x-1)(x)(x+1) will certainly be divisible by 4. So for the 50 odd values of x between 50 and 149 inclusive, the product will always be divisible by 12

• if x is even, (x-1)(x)(x+1) is only divisible by 4 if x itself is divisible by 4. Since every second even number is a multiple of 4, for half of the fifty even values of x between 50 and 149 inclusive, x is a multiple of 4, and we'll again get a product divisible by 12, so for 25 further values of x.

So the probability is (50 + 25)/100 = 3/4

Kinshook · Answer

Asked: If integer x is chosen at random from 50 to 149 inclusive, what is the probability that x^3 - x is a multiple of 12?

x^3 - x = x (x^2 - 1) = x (x-1) (x+1) = (x-1)x(x+1) which is always a multiple of 6 but it is a multiple of 12 
1. when x is odd
2. when x is a multiple of 4

Total possible values of x = 149 - 50 + 1 = 100
Unfavorable values of x = {50,54,58,62,...... 146} = (146-50)/4 + 1 = 25
Favorable values of x = 100 - 25 = 75

Probability that x^3 - x is a multiple of 12 = 75/100 = 3/4

IMO D

ArvindG123 · Answer

Every set of 3 consecutive numbers will always be divisible by 3. We need to check for divisibility by 4.

When x = 50, 49 * 50 * 51. Not divisible by 4

When x = 51, 50 * 51 * 52. Is divisible by 4

When x = 52, 51 * 52 * 53. Is divisible by 4

When x = 53, 52 * 53 * 54. Is divisible by 4

When x = 54, 53 * 54 * 55. Not divisible by 4

So we see that if x is even and not divisible by 4, then  x^3 - x  is not divisible by 12.

The series is 50, 54, 58.....146

The total numbers not divisible by 12 =  \frac{146 - 50}{4} + 1 = 25

Hi Arun !! Could you please explain this a bit more or link me to more information on this please ? Thanks a lot !!

EgmatQuantExpert · Answer

Solution 
  Given  
In this question, we are given that
 • Integer, x is chosen at random from 50 to 149 both inclusive

To find  
We need to determine
 • The probability that  x^3 – x  is a multiple of 12

Approach and Working out  
P(E) = favourable cases/total cases
 • Total cases = 100
• Favourable cases:
 o  x^3 – x  is a multiple of 12
o (x – 1) * x * (x + 1) = 12k
  This implies, x must be odd or a multiple of 4 
o So, favourable cases = 50 + 25 = 75

• Thus, P(E) =  \frac{75}{100} = \frac{3}{4}

Thus, option D is the correct answer.

Correct Answer: Option D

EgmatQuantExpert · Answer

Alternate Solution - Unfavourable Case Approach

Given

• x is chosen at random from 50 to 149 inclusive.

To Find

• The probability that  x^3  - x is a multiple of 12.

Approach and Working Out

• We have  x^3  – x = x ( x^2  – 1) = (x – 1)(x)(x + 1)
 o One of the 3 consecutive numbers is always a multiple of 3.
o For this to be not a multiple of 12, none of them should be a multiple of 4.

• x – 1 should be in the form 4k + 1. (For not divisible by 12)
 o x should be 4k + 2 or 50, 54, … 146.
o 25 of them.

• Remaining 75 cases the expression will be a multiple of 12.
 o Answer =  \frac{75}{100}  =  \frac{3}{4}

Correct Answer: Option D

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If integer x is chosen at random from 50 to 149 inclusive, what is the

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