Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (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?

OpenStudy (pythagoras123):

how did you get 26? O_O

OpenStudy (pythagoras123):

erm... no.

OpenStudy (inkyvoyd):

Gah, FFM, you just gave me another strange problem to think about. The sum of the fib numbers?

OpenStudy (dumbcow):

the answer is 57 but i don't have a mathematical proof for you i used a computer to evaluate all possible cases

OpenStudy (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?

OpenStudy (anonymous):

Btw what is the problem source?

OpenStudy (pythagoras123):

Homework

OpenStudy (pythagoras123):

That's why I need an explanation not just an answer

OpenStudy (anonymous):

Competitive math probably IMO?

OpenStudy (pythagoras123):

dunno, but looks like it

OpenStudy (dumbcow):

the distribution is as follows for combinations that sum up to 6,9,12 ... 1+3+7+11+13+11+7+3+1

OpenStudy (anonymous):

@anonymoustwo44: As this is combination 1+2+3 is same as 2+1+3

OpenStudy (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.

OpenStudy (anonymous):

@FoolForMath so the number 231 will be the same as the combination 321?

OpenStudy (anonymous):

they bouth divisible by 3

OpenStudy (anonymous):

so are they the same combination?

OpenStudy (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

OpenStudy (anonymous):

never mind me :))

OpenStudy (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

OpenStudy (anonymous):

3c3 + 3c1*4c1*4c1 + 4c3 + 4c3

OpenStudy (anonymous):

did i miss something? some case?maybe

OpenStudy (anonymous):

myko?

OpenStudy (anonymous):

just you have to be carefull with n=0.... wait checking...

OpenStudy (anonymous):

i only counted 3, for 3n forms {3,6,9}

OpenStudy (anonymous):

3n+1 3n+2 for n = 0 1 and 2

OpenStudy (anonymous):

never mind, you mean other thing, sry

OpenStudy (anonymous):

you right, good job

OpenStudy (anonymous):

Ishaan is correct

OpenStudy (anonymous):

Answer for this problem is 26

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!