pythagoras123
  • pythagoras123
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?
Mathematics
jamiebookeater
  • jamiebookeater
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pythagoras123
  • pythagoras123
how did you get 26? O_O
pythagoras123
  • pythagoras123
erm... no.
inkyvoyd
  • inkyvoyd
Gah, FFM, you just gave me another strange problem to think about. The sum of the fib numbers?

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dumbcow
  • dumbcow
the answer is 57 but i don't have a mathematical proof for you i used a computer to evaluate all possible cases
anonymous
  • anonymous
@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
  • anonymous
Btw what is the problem source?
pythagoras123
  • pythagoras123
Homework
pythagoras123
  • pythagoras123
That's why I need an explanation not just an answer
anonymous
  • anonymous
Competitive math probably IMO?
pythagoras123
  • pythagoras123
dunno, but looks like it
dumbcow
  • dumbcow
the distribution is as follows for combinations that sum up to 6,9,12 ... 1+3+7+11+13+11+7+3+1
anonymous
  • anonymous
@anonymoustwo44: As this is combination 1+2+3 is same as 2+1+3
anonymous
  • anonymous
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.
anonymous
  • anonymous
@FoolForMath so the number 231 will be the same as the combination 321?
anonymous
  • anonymous
they bouth divisible by 3
anonymous
  • anonymous
so are they the same combination?
anonymous
  • anonymous
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
anonymous
  • anonymous
never mind me :))
anonymous
  • anonymous
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
anonymous
  • anonymous
3c3 + 3c1*4c1*4c1 + 4c3 + 4c3
anonymous
  • anonymous
did i miss something? some case?maybe
anonymous
  • anonymous
myko?
anonymous
  • anonymous
just you have to be carefull with n=0.... wait checking...
anonymous
  • anonymous
i only counted 3, for 3n forms {3,6,9}
anonymous
  • anonymous
3n+1 3n+2 for n = 0 1 and 2
anonymous
  • anonymous
never mind, you mean other thing, sry
anonymous
  • anonymous
you right, good job
anonymous
  • anonymous
Ishaan is correct
anonymous
  • anonymous
Answer for this problem is 26

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