# Mystery Number (Algebra?)

1. Apr 27, 2008

### LLS

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?

2. Apr 27, 2008

### Dick

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.

3. Apr 27, 2008

### LLS

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!

4. Apr 27, 2008

### Dick

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.

5. Apr 27, 2008

### LLS

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?

6. Apr 27, 2008

### Dick

{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.

7. Apr 27, 2008

### LLS

"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

8. Apr 27, 2008

### Dick

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