## If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?

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### If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?

by BTGmoderatorDC » Sun Sep 12, 2021 10:03 pm

00:00

A

B

C

D

E

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If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?

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

OA C

Source: Manhattan Prep

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### Re: If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12

by swerve » Tue Sep 14, 2021 7:00 am
BTGmoderatorDC wrote:
Sun Sep 12, 2021 10:03 pm
If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?

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

OA C

Source: Manhattan Prep
$$(C-1)C(C+1)$$ should be divisible by $$12.$$

Question is: How many of the integers from $$20$$ to $$99$$ are either $$ODD$$ or Divisible by $$4.$$

$$ODD=\dfrac{99-21}{2}+1=40$$
Divisible by $$4= \dfrac{96-20}{4}+1=20$$
Total$$=99-20+1=80$$

$$P= \dfrac{\text{Favorable}}{\text{Total}}=\dfrac{40+20}{80}=\dfrac{60}{80}=\dfrac{3}{4}$$

Therefore, C

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