there are 24 boxes: 5 with 100 coins, 11 with 50 and 8 with 75. 3 people want to divide in a way that everyone gets same number of boxes and same number of coins. how many solutions are there?

Question

jack1 · Answer

24 boxes
x=100
y =50
z= 75
5x +11y+8z= 24 boxes
 500 +550 +600        = 1650 coins

so each must get 8 boxes
and each must get

jack1 · Answer

each must get 550 coins

anonymous · Answer

yes, i agree

anonymous · Answer

but how many solutions are there?

jack1 · Answer

umm i honestly cant remember how to do this, sorry dude, will ask @hartnn , he's maths genius, hang on

hartnn · Answer

sorry! :P
@satellite73 @UnkleRhaukus @experimentX

jack1 · Answer

100a + 50b + 75c =550, 
100d + 50e +75f = 550, 
100g + 50h + 75i = 550, 

a+d+g=5, 
b+e+h=11, 
c+f+i=8

i just dont know where to go from here to get actual integers for answers

jack1 · Answer

gotta sec @ganeshie8 and also @whpalmer4 ... please?

whpalmer4 · Answer

I'd start by working out the various ways you can assemble 550 coins with boxes containing 100, 75 and 50 coins...

anonymous · Answer

wouldnt that take a long time?

anonymous · Answer

any ideas anyone?

whpalmer4 · Answer

Less time than you think, I think.  Remember, everyone gets the same of boxes, so the possibility tree is cut way down.

whpalmer4 · Answer

It should be clear that you can't have all 5 of the 100 coin boxes going to the same person, because then they need 3 more boxes, yet the smallest box is 50, pushing them well over the 550 coin mark.  Similarly, you can't have 4 of the 100 coin boxes going to the same person, because then they need another 150 coins from 4 boxes, but the only way to assemble that takes either 2 boxes (100+50),(75+75) or 3 boxes (3x50), so they come up short on the box count.

whpalmer4 · Answer

If you can find one arrangement that satisfies all the constraints, you should be able to construct any other possible arrangements by doing swaps starting at your first arrangement.

jack1 · Answer

100a + 50b + 75c =550,     a + b + c =8, 
100d + 50e +75f = 550,      d + e + f =8, 
100g + 50h + 75i = 550,     g + h + i =8

a + d + g = 5,  b + e + h = 11,  c + f + i = 8

jack1 · Answer

so assume numbers of a, d and g(remembering a + d + g = 5)
if a=0
b = a+2 = 2,   
c = -2 (a-3) = 6,   
so 100a + 50b + 75c =550
      0       + 100+ 450 =550 = true

if d = 3
e = d+2 = 5,   
f = -2 (d-3) = 0,   
so 100d + 50e +75f = 550,
      300   + 350 +0 = 550  = true

if g = 2
h = g+2 = 4,   
i = -2 (g-3) = 2
so 100g + 50h + 75i = 550, 
      200   + 200 + 150 = 550  = true

so use the above formulas, and play around with the 100 coin boxes, as this will have the fewest permutations (only 5 boxes)
so we've done 
0,3,2
try
0,4,1 (wont work as you wont make it to 8 boxes with 400 coins in 4 already)
0,5.0 (wont work as @whpalmer4 explained already)
1,2,2 (should work)

then u can play around with the different permutations of the ones that work

whpalmer4 · Answer

It really isn't very difficult to work out the possible combinations for the first set.

One thing worth noting is that 50, 100, and 550 are all even multiples of 25 coins, but 75 is an odd multiple.  That implies that any possible solution has to have an even number of boxes of 75 coins.  

5 boxes of 100 doesn't work because you can't get only 550 coins and still use 8 boxes.  4 boxes of 100 doesn't work for the same reason.  3 boxes of 100 allows you to complete the 550 coins/8 boxes with 5 boxes of 50 coins.  2 boxes of 100 allows you to complete the 550 coins/8 boxes with 2 boxes of 75 and 4 boxes of 100.  1 box of 100 allows you to complete the 550/8 with 4 boxes of 75 and 3 boxes of 50.  Finally, 0 boxes of 100 can be completed with 6 boxes of 75 and 2 boxes of 50.  

So, those are the possibilities for the first set of 8.  For the second and third sets, we have to work with what is left over.

3/0/5 leaves us with 2/8/6, which we can split as 1/4/3 and 1/4/3.  We could also split it as 3/0/5 + 2/2/4 + 0/6/2
2/2/4 leaves us with 3/6/7, which splits as 3/0/5 and 0/6/2 or 2/2/4 and 1/4/3
1/4/3 leaves us with 4/4/8 which splits as 2/2/4 and 2/2/4 or 3/0/5 and 1/4/3

In the end, we get the various permutations of 

{3,0,5},{1,4,3},{1,4,3}
{3,0,5},{0,6,2},{2,2,4}
{2,2,4},{2,2,4},{1,4,3}

Counting is left as an exercise for the reader.

You can also tackle this problem by viewing it as permuting all of the boxes into 3 groups of 8, then discarding all the permutations that fail to match the constraints (everyone ends up with the same number of coins, correct number of each size of box).  Here's a Mathematica expression that shows them all:

$$Select[Tuples[{x, y, z} /. 
   FindInstance[
    100 x + 75 y + 50 z == 550 && 
     x + y + z == 8 && (x > -1) && (y > -1) && (z > -1), {x, y, z} , 
    Integers, 100], 
  3], (Total[#[[All, 1 ]]] == 5 && Total[#[[All, 2]]] == 8 && 
    Total[#[[All, 3]]] == 11) &]$$

and the output:

$$\begin{array}{ccc}
 \{0,6,2\} & \{2,2,4\} & \{3,0,5\} \
 \{0,6,2\} & \{3,0,5\} & \{2,2,4\} \
 \{1,4,3\} & \{1,4,3\} & \{3,0,5\} \
 \{1,4,3\} & \{2,2,4\} & \{2,2,4\} \
 \{1,4,3\} & \{3,0,5\} & \{1,4,3\} \
 \{2,2,4\} & \{0,6,2\} & \{3,0,5\} \
 \{2,2,4\} & \{1,4,3\} & \{2,2,4\} \
 \{2,2,4\} & \{2,2,4\} & \{1,4,3\} \
 \{2,2,4\} & \{3,0,5\} & \{0,6,2\} \
 \{3,0,5\} & \{0,6,2\} & \{2,2,4\} \
 \{3,0,5\} & \{1,4,3\} & \{1,4,3\} \
 \{3,0,5\} & \{2,2,4\} & \{0,6,2\} \
\end{array}$$

If you don't grok Mathematica, FindInstance finds 100 integer solutions to the equation 100x+75y+50z=550 with the constraints that x+y+z=8 and x,y,z > -1.  Tuples makes all of the permutations of {x,y,z}, and /. is a replacement operator, plugging all the solutions found by FindInstance into the list so that Select can winnow out the ones that don't have the right number of boxes added across the row, such as {0,6,2}, {0,6,2}, {0,6,2}.

jack1 · Answer

...That is AWESOME!!!! @whpalmer4