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
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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\((C1)C(C+1)\) should be divisible by \(12.\)BTGmoderatorDC wrote: ↑Sun Sep 12, 2021 10:03 pmIf 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
Question is: How many of the integers from \(20\) to \(99\) are either \(ODD\) or Divisible by \(4.\)
\(ODD=\dfrac{9921}{2}+1=40\)
Divisible by \(4= \dfrac{9620}{4}+1=20\)
Total\(=9920+1=80\)
\(P= \dfrac{\text{Favorable}}{\text{Total}}=\dfrac{40+20}{80}=\dfrac{60}{80}=\dfrac{3}{4}\)
Therefore, C