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pythagoras123
 4 years ago
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?
pythagoras123
 4 years ago
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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pythagoras123
 4 years ago
Best ResponseYou've already chosen the best response.0how did you get 26? O_O

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0Gah, FFM, you just gave me another strange problem to think about. The sum of the fib numbers?

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1the answer is 57 but i don't have a mathematical proof for you i used a computer to evaluate all possible cases

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Btw what is the problem source?

pythagoras123
 4 years ago
Best ResponseYou've already chosen the best response.0That's why I need an explanation not just an answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Competitive math probably IMO?

pythagoras123
 4 years ago
Best ResponseYou've already chosen the best response.0dunno, but looks like it

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1the distribution is as follows for combinations that sum up to 6,9,12 ... 1+3+7+11+13+11+7+3+1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@anonymoustwo44: As this is combination 1+2+3 is same as 2+1+3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@FoolForMath so the number 231 will be the same as the combination 321?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0they bouth divisible by 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so are they the same combination?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Every 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), 44 (3n+1 and 3n+2) 3c3 + 3c1*4c1*4c1 + 4c3 + 4c1 How about this? :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.03c3 + 3c1*4c1*4c1 + 4c3 + 4c3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did i miss something? some case?maybe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just you have to be carefull with n=0.... wait checking...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i only counted 3, for 3n forms {3,6,9}

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.03n+1 3n+2 for n = 0 1 and 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0never mind, you mean other thing, sry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Answer for this problem is 26
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