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 ?

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.

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it