# Proof Question [Hard]

Can someone please explain how i complete 7b)?
Original post by Matheen1

Can someone please explain how i complete 7b)?

you need to investigate the outcome of adding various values of r.... only some of the results will be compatible with the sum of the integers being a multiple of 3.
Original post by the bear
you need to investigate the outcome of adding various values of r.... only some of the results will be compatible with the sum of the integers being a multiple of 3.

How would i do this though?
Original post by the bear
you need to investigate the outcome of adding various values of r.... only some of the results will be compatible with the sum of the integers being a multiple of 3.

Amusing observation that's not necessarily directly helpful:

Spoiler

Original post by DFranklin
Amusing observation that's not necessarily directly helpful:

Spoiler

So how can i use this to solve the question?
you could draw a tree diagram like this:

each branch has a 1/3 chance. only certain end results will give a total which is a multiple of 3. so 1,1,0 can be discarded for instance.
(edited 12 months ago)
Original post by Matheen1
So how can i use this to solve the question?

As I said - it's not directly helpful (or at least proving it and then using it to solve the question isn't the most efficient approach). But it's funny because it shows that you could have any set of crazy rules for getting a total before adding a random number between 1 and 99 and it makes no difference to the actual answer.

For a short solution along these lines, suppose we have a method for getting a total N (before adding the random number). Write N = 3n + k (where k = 0,1 or 2). Suppose p(k=0) is P0, p(k=1) is P1 and p(k=2) = P2. (So P0+P1+P2=1).

You can draw a simple tree diagram (you only actually need 3 leaves if you only draw the ones that work) to find the probability that "N+ a random number between 1 and 99" is divisible by 3. You'll find it doesn't make any difference what the values of P0,P1,P2 actually are, so you don't need to worry about calculating them).
(edited 12 months ago)