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Three different integers from 1 to 11 are selected. In how many of these combinations of 3 numbers are their sums a multiple of 3?

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how did you get 26? O_O
erm... no.
Gah, FFM, you just gave me another strange problem to think about. The sum of the fib numbers?

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Other answers:

the answer is 57 but i don't have a mathematical proof for you i used a computer to evaluate all possible cases
@FoolForMath... Can I ask what combination of 3numbers is. does that mean order doesn't matter? or for instance, in this cause would 132 be same as 123?
Btw what is the problem source?
That's why I need an explanation not just an answer
Competitive math probably IMO?
dunno, but looks like it
the distribution is as follows for combinations that sum up to 6,9,12 ... 1+3+7+11+13+11+7+3+1
@anonymoustwo44: As this is combination 1+2+3 is same as 2+1+3
you can try like that: the sum of numbers a,b,c is divisible by 3 if the sum of their digits is divisible by 3.
@FoolForMath so the number 231 will be the same as the combination 321?
they bouth divisible by 3
so are they the same combination?
what do you mean? the maximum sum to check in this case is: 11+10+9 = 30 (digits sum =3) you never even get to 123 or 321
never mind me :))
Every number can be expressed as 3n, 3n+1,3n+2 Case1. 3n,3n,3n Case2. 3n, 3n+1,3n+2 Case3. 3n+2,3n+2,3n+2 Case 4. 3n+1,3n+1,3n+1 In the set {1,2,..,11} 3 (3n forms), 4-4 (3n+1 and 3n+2) 3c3 + 3c1*4c1*4c1 + 4c3 + 4c1 How about this? :D
3c3 + 3c1*4c1*4c1 + 4c3 + 4c3
did i miss something? some case?maybe
just you have to be carefull with n=0.... wait checking...
i only counted 3, for 3n forms {3,6,9}
3n+1 3n+2 for n = 0 1 and 2
never mind, you mean other thing, sry
you right, good job
Ishaan is correct
Answer for this problem is 26

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