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
2 answers:
You want to multiply two consecutive numbers, and see when you get a multiple of three.
Since two consecutive numbers share no divisors (except 1), either n or n+1 must be a multiple of 3.
So, your experiments succeeds if n is itself a muliple of three, or if lies immediately before one. So, 2 numbers out of 3 are ok.
Let me try to visualize the situation: I'll list some integers. Below, I'll write if they are multiple of three (M) or not (N). Above, I will write if they will cause the experiment to succeed (S), i.e. n(n+1) will be dibisible by 3, or to fail (F), i.e. n(n+1) will not be a multiple of three. We have
So, since there are exactly 99 numbers between 1 and 99, and 2/3 of the numbers work, the probability that you will choose a number n such that n(n+1) is divisible by 3 is 2/3.
You might be interested in
Answer:
Step-by-step explanation:
First let's multiply the 1st equation by 10.
So, we can see that the equations have the same coefficients and that implies they are equal.
So the equation has no solutions for.