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Mystery Number (Algebra?)

  1. Apr 27, 2008 #1

    LLS

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    1. The problem statement, all variables and given/known data
    The sum of 8 positive integers is 31. If no individual integer value can appear more than twice in the list of 8 integers, what is the greatest possible value that one of the integers can have?

    2. Relevant equations

    ?

    3. The attempt at a solution
    My answer is 16. I did a lot of calculating; a lot of trial and error.

    I doubled 3 of the lowest possible numbers, which totaled 6. I used 6 numbers; there are 2 left.

    0+0=0
    1+1=2
    2+2=4

    16+6=22 (7 numbers have been used)

    22+9=31 (all 8 numbers have been used)

    17 would be too high, even if the lowest possible numbers are used. 16 is the highest possible number.

    How do I formulate the equation to solve this problem?

    Is 16 the correct answer?
     
  2. jcsd
  3. Apr 27, 2008 #2

    Dick

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    Your logic gets a little confusing but doubling the lowest numbers sounds like a good start. What is your final set of 8 numbers? Before you answer make sure it doesn't have 0 in it. 0 isn't positive.
     
  4. Apr 27, 2008 #3

    LLS

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    Ouch-I counted 0 as positive. My final sets are:

    1+1=2
    2+2=4
    3+3=6
    4+4=8

    Total = 20

    15 is too high
    14 is too high
    13 is too high
    12 is too high

    11 would work, but that leaves 1 number short

    can't be 10 (used 1), 9 (used 2), 8 (used 3)

    7 would be the highest

    If that's right, what's the equation?

    And thanks!
     
  5. Apr 27, 2008 #4

    Dick

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    You are confusing me again. Using your logic I would write the list of numbers as {1,1,2,2,3,3,4,x}, picking the first seven numbers to be as small as possible, so the last can be as large as possible. What's x if the sum is 31? It's really not worth writing an equation for this, it's more of a logic problem.
     
  6. Apr 27, 2008 #5

    LLS

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    I'm confused now-sorry

    If I add 1+1+2+2+3+3+4+x, it totals 16 - Those numbers are the lowest possible choices.

    16+x=31

    The largest number would then be 15, if the above equation is correct.

    Is 15 the correct choice?
     
  7. Apr 27, 2008 #6

    Dick

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    {1,1,2,2,3,3,4,15} does add up to 31. And every number is positive and none are repeated more that twice. Can you think of a way to pick 8 numbers that might have a larger max? Do YOU think it's correct? My opinion doesn't count.
     
  8. Apr 27, 2008 #7

    LLS

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    "Can you think of a way to pick 8 numbers that might have a larger max?"

    I guess that the numbers don't have to be whole numbers. If that is what's true, I am really lost as to how to solve this problem.

    Going by the numbers above, 15 would be the largest. If non whole numbers are are possible, I really need help.

    Thank you
     
  9. Apr 27, 2008 #8

    Dick

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    Non-whole numbers aren't integers. You are done. It's 15.
     
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