Expert: Scott A Wilson - 10/24/2009

Question
How do you  solve the following questions:
(1) what is the least possible value of a+b+C if
    a* b*b*c*c*c=3600 . a,b and c are positive integers.

(2) find (xy) if
    x+y+square root(x+y)=20
    x-y+square root(x-y)=12

Answer
1st, find the prime factors of 3600.  They are 2*2*2*2*3*3*5*5.
We must choose three of them to be c, so that means c is 2.  That leaves 2*3*3*5*5.

Next, there is only one a, so a must have a 2.  That leaves 3*3*5*5.

Now we could take b = 3, 5, or 3*5.  That would make a 50, 18, or 2.

This makes our list of choices for a, b, and c into
2, 15, 2 giving 2*15*15*2*2*2 = 30*30*4 = 900*4 = 3,600.
18, 5, 2 giving 18*5*5*2*2*2 = 18*25*8 = 18*200 = 3,600.
50, 3, 2 giving 50*3*3*2*2*2 = 50*9*8 = 50*72 = 3,600.

2a) Try x = 0, 1, 2, 3, or 4 with a value of y that gives x+y = 0, 1, 4, 9, or 16.
Once we have tried one value for x, only choose the value equal or greater for y.
For example, when x=0, we will try y=2.  It will be pointless to try x=2 and y=0.

If x=0, y=0,1,4,9,or 16.
If x=0, y=0 gives 0, y=1 gives 2, y=4 gives 4, y=9 gives 6, and y=16 gives 20, so for x=0, y=16.

If x=1, y=3,8, or 15.
If x=1, y=3 gives 6, y=8 gives 12, and y=15 gives 20, so for x=1, y=15.

If x=2, y=2,7 or 14.
If x=2, y=2 gives 7, y=7 gives 12, and 14 gives 2+14+4=20, so for x=2,y=14.

From what we've seen, x=n, y=16-n, and x+y=16, with √16=4.
Using this, we could find the values for x=3,4,5,6,7 and 8.
We could also find values for x=9 thru x=16, but they would just be repeats with x=y and y=x.

For example, if x=2, we found y=14.  If x=14, that would gives us y=2.


For x-y+√(x-y) = 12, it can be seen the y must be great than or equal to x.
If y=0, the function would be x-0+√(x-0).  This means x would have to be 0, 4, 9, or 16.
If y=0, x=1 gives 1-0+√1 = 2; x=4 gives 4-0+√4 = 6; x=9 gives 9-0+√9 = 12, and we don't have to try x=4.  The first point would be (x,y)=(9,0).

If y=1, the function would be x-1+√(x-1), so x would need to be 1,2,5,10, or 17.
So x=1 gives 1-1+0=0, x=2 gives 2-1+1=2, x=5 gives 5-1+2=6, x=10 gives 10-1+3=12, so we don't have to try x=17.  The second point is (x,y)=(10,1).

If y=2, the function would be x-2+√(x-2).  y values to try would be 2, 3, 6, or 11.
2: 2-2+0=0; 3:3-2+1=2; 6:6-2+2=6; 11:11-2+3=12, so (x,y)=(11,2).

It looks like we try y+9,y.

If y=3, x=12, we get 12-3+3=12.
If y=4, x=13, we get 13-4++3=12.
If y=5, x=14, we get 14-5+3=12.

If y=n, x=n+9, we get n - n+9 + √9 = 9 + √9 = 9 + 3 = 12.
This has an infinite number of solutions.



Something else to look into in (a) or (b) is what is y were 5.5, which means to try x=10.5.
10.5+5.5+√(10.5+5.5) = 16 + √16 = 16+4 = 20.
This means we can have solutions such that x+y=16, or y = -x + 16 for x in [0,16].

For the second one, we need x=n+9 and y=n, but n doesn't have to be an integer.
For example, if x=12.5 and y=3.5, we get 12.5 - 3.5 + √(12.5-3.5) = 9 + √9 = 9 + 3 = 12.

I've answered this question before for someone else and only given integer solutions.
This time when I answered it I was that there could be an infinite number of solutions.
For the 1st one, they are restricted to reals in [0,20] for x and y.
In the 2nd one, the only restriction is that x>=9 and y=x-9.
You could even try y=-5, then x=4 and get 4 -(-5) + √(4-(-5)) = 4+5 +√9 = 9+3 = 12.
This means y is x-9, so y>=-9.