An integer \(n\) between \(1\) and \(99,\) inclusive, is to be chosen at random. What is the probability that \(n(n+1)\) will be divisible by \(3?\)
A) \(\dfrac19\)
B) \(\dfrac13\)
C) \(\dfrac12\)
D) \(\dfrac23\)
E) \(\dfrac56\)
Answer: D
Source: GMAT Prep
An integer \(n\) between \(1\) and \(99,\) inclusive, is to be chosen at random. What is the probability that \(n(n+1)\)
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Solution:
The number of multiples of 3 from 1 to 99, inclusive, is:
(99 - 3)/3 + 1 = 33
We also should see that every number that is 1 less than a multiple of 3 will also allow n(n+1) to be divisible by 3. For example, if n = 2, then n + 1 = 3, which is divisible by 3. Since there are 33 multiples of 3, there are 33 numbers that are one less than a multiple of 3. Thus, the probability that n(n+1) will be divisible by 3 is 66/99 = 2/3.
Answer: D
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