I need answer pls

272

Thanks so much .i will appreciate if you explain how it was gotten

Didn't he just tell you?

yes he did and thanks to him . I will like to know how it was gotten

ohhhhh.....

I get it, you not only want the answer, you want us to actually DO your homework for you.

Do you think God would be pleased with that Darius?

I am here to learn please. I don't have home work

154 = 1 mod 3
101 = 2 mod 3
46 = 1 mod 3
22 = 1 mod 3
19 = 1 mod 3
17 = 2 mod 3
16 = 1 mod 3
13 = 1 mod 3
10 = 1 mod 3

Total = 398 = 2 mod 3

A + B + C = 398
2B + B + C = 398
3B + C = 398
C = 2 mod 3

Groups of 3 are going to be:
0 mod 3 (1 + 1 + 1),
1 mod 3 (1 + 1 + 2), or
2 mod 3 (1 + 2 + 2)

Since C = 2 mod 3, it must contain 101 and 17.

That leaves 154, 46, 22, 19, 16, 13, and 10.

Since A = 2B, 154 must be in A or C.
If 154 is in A, then A is at least 154+13+10 = 177 and B is at most 46+22+19=87.

Therefore, A is greater than twice B.
Therefore, 154 must be in C.
Therefore, C is 154+101+17 = 272.

No need to evaluate the C(9,3)*C(6,3)*C(3,3) = 1680 ways of distributing the nuggets among A, B, and C.

Thanks so much man

Thanks markr very helpful thing you shared here.