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Mystery Number (Algebra?)

  • Thread starter LLS
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LLS
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1. Homework Statement
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. Homework 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?
 

Answers and Replies

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.
 
LLS
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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.
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!
 
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.
 
LLS
40
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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.
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?
 
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.
 
LLS
40
0
{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.
"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
 
Dick
Science Advisor
Homework Helper
26,258
618
Non-whole numbers aren't integers. You are done. It's 15.
 

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