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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 ?

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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

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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

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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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