Olympiad Question

Firas

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[FONT=q_serif]Given that 5r + 4s + 3t + 6u = 100, where r ≥ s ≥ t ≥ u ≥ 0 are real numbers, find the sum of the maximum and minimum possible values of r + s + t + u. Please shed some light on how to solve this. I know you may not write the steps but can you at least explain how to do it. I have no idea how to even start - so dont ask me what i know ... please. This is not homework. This is an olympiad question, which I need to know how to do so please help me[/FONT]
[FONT=q_serif]Given that 5r + 4s + 3t + 6u = 100, where r ≥ s ≥ t ≥ u ≥ 0 are real numbers, find the sum of the maximum and minimum possible values of r + s + t + u.[/FONT]
 
[FONT=q_serif]Given that 5r + 4s + 3t + 6u = 100, where r ≥ s ≥ t ≥ u ≥ 0 are real numbers,
find the sum of the maximum and minimum possible values of r + s + t + u.[/FONT]
As shown, there is only 1 solution.

IF the 6u is a typo and should be 2u, then there are 2 solutions.

I don't know if we're allowed to give solution to an "Olympiad" problem....
I did it using a simple looper program.
 
As shown, there is only 1 solution.

IF the 6u is a typo and should be 2u, then there are 2 solutions.

I don't know if we're allowed to give solution to an "Olympiad" problem....
I did it using a simple looper program.

its an olympiad practice problem, not the actual one.
 
1 + 5 + 5 + 10 = 21
2 + 3 +6 + 10 = 21
4 + 5 + 6 +7 = 22
2 + 5 + 10 + 10 = 27
...
 
1 + 5 + 5 + 10 = 21
2 + 3 +6 + 10 = 21
4 + 5 + 6 +7 = 22
2 + 5 + 10 + 10 = 27
...
Nooooo: given condition is "r ≥ s ≥ t ≥ u ≥ 0

I originally goofed and used 9 as maximum...
there are 7 solutions:
9+7+5+2 = 23 : my only original solution (I'll go stand in the corner...)
10+5+4+3 = 22
10+8+4+1 = 23
11+6+3+2 = 22
11+6+5+1 = 23
12+7+2+1 = 22
13+5+3+1 = 22

So max = 23 and min = 22
 
Nooooo: given condition is "r ≥ s ≥ t ≥ u ≥ 0

I originally goofed and used 9 as maximum...
there are 7 solutions:
9+7+5+2 = 23 : my only original solution (I'll go stand in the corner...)
10+5+4+3 = 22
10+8+4+1 = 23
11+6+3+2 = 22
11+6+5+1 = 23
12+7+2+1 = 22
13+5+3+1 = 22

So max = 23 and min = 22

how did you know there is only 7 solutions? is listing the only way?
 
how did you know there is only 7 solutions? is listing the only way?
Listing is never away of counting as you might of missed some. Now having said that, listing is good to sometimes see how to count something.
 
Listing is never away of counting as you might of missed some. Now having said that, listing is good to sometimes see how to count something.

Generally this is true, but listing is a fine tactic for this problem as there's only finitely many solutions, so we can definitely generate an exhaustive list. Given the condition that \(\displaystyle 5r + 4s + 3t + 6u = 100\) and knowing that all four variables must be non-negative integers, we can immediately deduce that \(\displaystyle r \le 20\). Similarly, we can deduce that \(\displaystyle s \le 25\). However, the additional condition that \(\displaystyle r \ge s \ge t \ge u \ \ge 0\) means we can discard any solutions where \(\displaystyle s \ge 21\)... and so on.

I'll stop there for brevity's sake, but it should be clear how this process of logical deductions can be continued to generate a full list of every possibility. I'm not quite sure where Denis went wrong with his second attempt, but there are actually a total of 29 solutions:

  • 5(7) + 4(5) + 3(5) + 6(5) = 100
  • 5(8) + 4(6) + 3(4) + 6(4) = 100
  • 5(8) + 4(6) + 3(6) + 6(3) = 100
  • 5(9) + 4(7) + 3(3) + 6(3) = 100
  • 5(9) + 4(7) + 3(5) + 6(2) = 100
  • 5(9) + 4(7) + 3(7) + 6(1) = 100
  • 5(10) + 4(5) + 3(4) + 6(3) = 100
  • 5(10) + 4(8) + 3(2) + 6(2) = 100
  • 5(10) + 4(8) + 3(4) + 6(1) = 100
  • 5(10) + 4(8) + 3(6) + 6(0) = 100
  • 5(11) + 4(6) + 3(3) + 6(2) = 100
  • 5(11) + 4(6) + 3(5) + 6(1) = 100
  • 5(11) + 4(9) + 3(1) + 6(1) = 100
  • 5(11) + 4(9) + 3(3) + 6(0) = 100
  • 5(12) + 4(4) + 3(4) + 6(2) = 100
  • 5(12) + 4(7) + 3(2) + 6(1) = 100
  • 5(12) + 4(7) + 3(4) + 6(0) = 100
  • 5(12) + 4(10) + 3(0) + 6(0) = 100
  • 5(13) + 4(5) + 3(3) + 6(1) = 100
  • 5(13) + 4(5) + 3(5) + 6(0) = 100
  • 5(13) + 4(8) + 3(1) + 6(0) = 100
  • 5(14) + 4(3) + 3(2) + 6(2) = 100
  • 5(14) + 4(6) + 3(2) + 6(0) = 100
  • 5(15) + 4(4) + 3(1) + 6(1) = 100
  • 5(15) + 4(4) + 3(3) + 6(0) = 100
  • 5(16) + 4(2) + 3(2) + 6(1) = 100
  • 5(16) + 4(5) + 3(0) + 6(0) = 100
  • 5(17) + 4(3) + 3(1) + 6(0) = 100
  • 5(20) + 4(0) + 3(0) + 6(0) = 100

Thus, the maximum sum is 10 + 8 + 6 + 0 = 24 (or 9 + 7 + 7 + 1 = 24), and the minimum sum is 20 + 0 + 0 + 0 = 20
 
Yer keerect Ks.
Stoopid me assumed everything > 0.
(as you can tell from my list)

Going to the corner for 29 minutes....:(
 
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