probability problem!?

Question

In how many ways can three numbers be selected from the numbers 1,2...300 such that their sum is devisible by 3?

i mean i want all solutions that make the sum of the three numbers is devisible by three without remain.

Answer 1

Let us group all numbers from 1 to 300 into 3 categories.
A-all numbers of the form 3k (there will be 100 of them)
B-all numbers of the form 3k+1 (again 100 of them)
C-all numbers of the form 3k+2(100 of them)
Now the problem is to choose 3 numbers such that sum is divisible by 3...
all 3 can be picked from A -100x99x98
all 3 can be picked from B -100x99x98
all 3 can be picked from C -100x99x98
1 from A,1 from B,1 from C -100x100x100
thus, total number of ways =3x100x99x98+100x100x100=3910600

Answer 2

firstly, between 1 and 300, you can divide the numbers into three sets. One set is of the form (3n), one is (3n+1) and one is (3n+2). Basically, any number can be expressed as one form and is put into one set. In this way, you get 90 numbers in the first and third groups and 120 numbers in the second.
Now, to chose three numbers such that the sum is divisible by three, you either have to choose three from the first group, three from the second group, three from the third group or one from each group.
So, the total number of favourable cases is (90C3+90C3+120C3+{90C3*90C3*120C3}) from the above statement. And the total cases is 300C3. So now the probability is the favourable cases divided by the total cases. The final answer is some 870020294.7 although that hardly matters as long as the approach is correct.

Answer 3

First, find how many combinations of three numbers can be selected from the numbers 1 to 300. That'll be 300*299*298/(1*2*3), or 4,455,100.

Every third one of these combinations is divisible by three, so if 4,455,100 is divisible by three, the result would be your answer. Unfortunately, 4,455,100/3=1,485,033.333....

Still, every third combination is divisible by three. The first one, (1, 2, 3) is, and the last one, (298, 299, 300) is, too. Temporarily ignore the last one, and the remaining number of combinations divisible by three is 4,455,099/3=1,485,033. Add the last one back in, and the answer is 1,485,034.

Answer 4

Every number is divisible by 3 or do you mean divisible by 3 with no remainder?

Answer 5

well basically there are many possibilities like
123.
well every consecutive number on adding up is divisible by three
there fore it is an arithmatic progression
=
300(300+1)/2
=
150*301
=
45150