mmcooley33 wrote:
If 1<n<99, what is the probability that n(n + 1) is perfectly divisible by 3 ?
We are given that an integer n is to be selected at random from 1 to 99 inclusive, and we must determine the probability that n(n+1) will be divisible by 3.
Since probability = favorable outcomes/total outcomes and we know that the total number of outcomes is 99, because there are 99 integers from 1 to 99 inclusive, we need to determine the number of values of n such that n(n+1) is divisible by 3.
First, we can determine the number of values of n that are divisible by 3, that is, the number of multiples of 3 that are between 3 and 99 inclusive. To calculate this, we can use the formula:
(Largest multiple of 3 – smallest multiple of 3)/3 + 1
(99 - 3)/3 + 1
96/3 + 1
32 + 1 = 33
Thus, there are 33 multiples of 3 between 1 and 99 inclusive. That is, the value of n can be any one of these 33 multiples of 3 so that n(n + 1) will be divisible by 3.
Similarly, if n + 1 is a multiple of 3, n(n + 1) will be also be divisible by 3. Since we know that there are 33 values of n that are multiples of 3, there must be another 33 values of n such that n + 1 is a multiple of 3. Let’s expand on this idea:
When n = 2, n + 1 = 3, and thus n(n+1) is a multiple of 3.
When n = 23, n + 1 = 24, and thus n(n+1) is a multiple of 3.
When n = 98, n + 1 = 99, and thus n(n+1) is a multiple of 3.
We can see that there are 33 values of n that are multiples of 3, and 33 more values of n for n + 1 to be multiples of 3. Thus, the probability of selecting a value of n so that n(n+1) is a multiple of 3 is:
66/99 = 2/3
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