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Math Help - Can this be made intoa formula or algorithem?

  1. #1
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    Smile Can this be made intoa formula or algorithem?

    I would like a formula for the following:

    Lets say I have a number x that is equal to 240

    I would like a formula from the following information for all variations that will give me x or greater.

    The following will be true:

    There will always be numbers 1 through 6

    You will know how many of each there are (1 six, 1 five, 1 four, 16 threes, 2 twos and 3 ones as an example). there may be 0 or n but most likely n will not be over 50.

    For each group above (like the 2 two's) , each occurrence can have a factor of 0,1,2,3 or 4. So for instance….the 2 two’s could be 1 four and 1 three. The 16 three’s could be 5 fours, 5 threes, 3 twos, 1 one and 1 zero. You then multiply every 0-4 by the grouping 1-6 and then by the occurrence and add them together.

    For example the group of twos of 4 and 3 would be 4*2*1 + 3*2*1 = 14 the group of 3’s would be 4*3*5 + 3*3*5 + 2*3*3 + 1*3*1 + 0*3*1 = 126

    Summing all the groupings 1-6 together needs to be >= x

    How can I come up with every combination that will give 240 or higher?

    One such combination is

    1 6 which is a 4 = 24

    1 5 which is a 4 = 20

    1 4 which is a 4 = 16

    16 3’s which are 1-1,2-2s, 6-3s,7-4s = 153

    2 2’s which are both 4s = 16

    3 1’s which are 1-3 and 2-4s = 11

    Add together and you get 240

    i would like a way to figure out every possible combination for 240 and higher without writing a program to get every single combination and then testing for 240 or higher.

    can anyone help. i hope i posted this in correct subject.
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  2. #2
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    was this posted under the appropriate header?

    sorry if not!
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  3. #3
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    actually,

    i was incorrect about the groupings 1-6

    the groupings will only be 1-3
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