1. ## Who made the least?

There are two groups of people and each group has three people in them.

In the first group, there are Alan, Bert, and Craig.
In the second group, there are Xavier, Yun, and Zack.

All six people made some money, Alan made $a, Bert made$b, Craig made $c, Xavier made$x, Yun made y, and zack made $z. We know: a<b<c and x<y<z c<z abc= xyz a+b+c=x+y+z Who made the least amount and why? I can use ab>zy to show that b>z, but I get stuck here. Can anyone help me out? 2. Originally Posted by Redbeard There are two groups of people and each group has three people in them. In the first group, there are Alan, Bert, and Craig. In the second group, there are Xavier, Yun, and Zack. All six people made some money, Alan made$a, Bert made $b, Craig made$c, Xavier made $x, Yun made y, and zack made$z.

We know:
a<b<c and x<y<z
c<z
abc= xyz
a+b+c=x+y+z
Who made the least amount and why?
I can use ab>zy to show that b>z, but I get stuck here. Can anyone help me out?
It's between a and x.

TAKE #1
a = x (yz)/(bc)
We need to decide if (yz)/(bc) is greater than 1 or less than 1 in order to find the magnitude of a relative to x.
(yz)/(bc) = (y/b)(z/c) > y/b since z/c > 1

TAKE #2
On the other hand:
a = x + y + z - b - c
We need to decide us y + z - b - c is positive or negative.
Since b > z (by your work), we have:

y + z - b - c < y + z - z - c = y - c
If y - c is negative, then we are done: a is less than x since y + z - b - c would be less than a negative number, thus negative, thus x + negative = a.

Let's see how y relates to c.

We know that:
b + y < z + c
y-c < z - b < 0
So y - c < 0 --> y < c

So a is the least amount.

I hope this makes sense.

3. Oops I screwed up in the OP. What i meant to say was:

I can use ab>xy to show that b>x. Clearly b cannot be >z because b<c<z.

But I do see where I need to go with the problem, though I'm not quite sure yet how this changes Take#2.

4. Originally Posted by Redbeard
...

a<b<c and x<y<z

c<z

abc= xyz

a+b+c=x+y+z
If the composite abc = xyz, then xyz is a re-arrangement of the factors of abc.
Since they are equal it implies that we are working with whole pennies.
a = $p_1 \times p_2 \times p_3$
b = $p_4 \times p_5 \times p_6$
c = $p_7 \times p_8 \times p_9$
(a,b,c could have more or less than 3 factors each)

The products are equal and the sums are equal and the values of each are given relative to the others.

That almost suggests that there is a unique solution.

5. There is no unique solution.
Code:

SUMS         PRODUCTS
a  b  c  x  y  z a+b+c x+y+z   a*b*c  x*y*z
01  9 10  2  3 15    20    20      90     90
02  8  9  3  4 12    19    19     144    144
03  8 10  4  5 12    21    21     240    240
04  9 10  5  6 12    23    23     360    360
05  9 14  6  7 15    28    28     630    630
06 10 14  7  8 15    30    30     840    840
07 15 18  9 10 21    40    40    1890   1890
08 12 15  9 10 16    35    35    1440   1440
09 15 16 10 12 18    40    40    2160   2160
10 18 22 11 15 24    50    50    3960   3960
11 18 20 12 15 22    49    49    3960   3960
12 20 21 14 15 24    53    53    5040   5040
13 21 24 14 18 26    58    58    6552   6552
14 18 20 15 16 21    52    52    5040   5040
15 20 24 16 18 25    59    59    7200   7200
16 25 27 18 20 30    68    68   10800  10800
18 25 28 20 21 30    71    71   12600  12600
20 27 28 21 24 30    75    75   15120  15120
...