What Is the Largest Integer in a Sum of 31 with Restrictions?

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Homework Help Overview

The problem involves finding the largest integer in a sum of 31 composed of 8 positive integers, with the restriction that no integer can appear more than twice. Participants are exploring the implications of this restriction and the nature of the integers involved.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various combinations of integers and their sums, questioning the validity of including zero as a positive integer and exploring the implications of doubling the smallest integers. There is also a focus on how to structure the integers to maximize the largest value.

Discussion Status

The discussion is active, with participants sharing their reasoning and calculations. Some have proposed specific sets of integers that meet the criteria, while others are questioning the assumptions and exploring alternative combinations. There is no explicit consensus on the largest integer yet, but several participants are converging on the value of 15.

Contextual Notes

Participants are grappling with the definition of positive integers and the constraints of the problem, particularly regarding the maximum allowable value for the integers involved. There is uncertainty about whether non-whole numbers could be considered, which has led to confusion in the discussion.

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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?

Homework Equations



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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?
 
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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.
 
Dick said:
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!
 
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.
 
Dick said:
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?
 
{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.
 
Dick said:
{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
 
Non-whole numbers aren't integers. You are done. It's 15.
 

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