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Math
Posted about 1 month ago

Let w,x,yw, x, y and zz represent four whole numbers (not necessarily positive). If (w1)(x2)(y3)(z4)=10(w-1)(x-2)(y-3)(z-4)=10, find all possible values for w+x+y+zw+x+y+z
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Answer from Sia
Posted about 1 month ago
Solution by Steps
step 1
Given the equation (w1)(x2)(y3)(z4)=10(w-1)(x-2)(y-3)(z-4)=10, we need to find all possible integer values for w,x,y,w, x, y, and zz
step 2
We start by considering the factors of 10, which are ±1,±2,±5,±10\pm 1, \pm 2, \pm 5, \pm 10
step 3
We need to find combinations of (w1),(x2),(y3),(w-1), (x-2), (y-3), and (z4)(z-4) that multiply to 10
step 4
Let's consider the positive factors first:
step 5
If (w1)=1(w-1)=1, (x2)=1(x-2)=1, (y3)=2(y-3)=2, (z4)=5(z-4)=5, then w=2w=2, x=3x=3, y=5y=5, z=9z=9
step 6
If (w1)=1(w-1)=1, (x2)=2(x-2)=2, (y3)=1(y-3)=1, (z4)=5(z-4)=5, then w=2w=2, x=4x=4, y=4y=4, z=9z=9
step 7
If (w1)=1(w-1)=1, (x2)=5(x-2)=5, (y3)=1(y-3)=1, (z4)=2(z-4)=2, then w=2w=2, x=7x=7, y=4y=4, z=6z=6
step 8
If (w1)=2(w-1)=2, (x2)=1(x-2)=1, (y3)=1(y-3)=1, (z4)=5(z-4)=5, then w=3w=3, x=3x=3, y=4y=4, z=9z=9
step 9
If (w1)=2(w-1)=2, (x2)=5(x-2)=5, (y3)=1(y-3)=1, (z4)=1(z-4)=1, then w=3w=3, x=7x=7, y=4y=4, z=5z=5
step 10
If (w1)=5(w-1)=5, (x2)=1(x-2)=1, (y3)=1(y-3)=1, (z4)=2(z-4)=2, then w=6w=6, x=3x=3, y=4y=4, z=6z=6
step 11
If (w1)=5(w-1)=5, (x2)=2(x-2)=2, (y3)=1(y-3)=1, (z4)=1(z-4)=1, then w=6w=6, x=4x=4, y=4y=4, z=5z=5
step 12
Now consider the negative factors:
step 13
If (w1)=1(w-1)=-1, (x2)=1(x-2)=-1, (y3)=2(y-3)=-2, (z4)=5(z-4)=-5, then w=0w=0, x=1x=1, y=1y=1, z=1z=-1
step 14
If (w1)=1(w-1)=-1, (x2)=2(x-2)=-2, (y3)=1(y-3)=-1, (z4)=5(z-4)=-5, then w=0w=0, x=0x=0, y=2y=2, z=1z=-1
step 15
If (w1)=1(w-1)=-1, (x2)=5(x-2)=-5, (y3)=1(y-3)=-1, (z4)=2(z-4)=-2, then w=0w=0, x=3x=-3, y=2y=2, z=2z=2
step 16
If (w1)=2(w-1)=-2, (x2)=1(x-2)=-1, (y3)=1(y-3)=-1, (z4)=5(z-4)=-5, then w=1w=-1, x=1x=1, y=2y=2, z=1z=-1
step 17
If (w1)=2(w-1)=-2, (x2)=5(x-2)=-5, (y3)=1(y-3)=-1, (z4)=1(z-4)=-1, then w=1w=-1, x=3x=-3, y=2y=2, z=3z=3
step 18
If (w1)=5(w-1)=-5, (x2)=1(x-2)=-1, (y3)=1(y-3)=-1, (z4)=2(z-4)=-2, then w=4w=-4, x=1x=1, y=2y=2, z=2z=2
step 19
If (w1)=5(w-1)=-5, (x2)=2(x-2)=-2, (y3)=1(y-3)=-1, (z4)=1(z-4)=-1, then w=4w=-4, x=0x=0, y=2y=2, z=3z=3
step 20
Summing up all possible values of w,x,y,w, x, y, and zz for each valid combination:
step 21
For w=2,x=3,y=5,z=9w=2, x=3, y=5, z=9: w+x+y+z=2+3+5+9=19w+x+y+z=2+3+5+9=19
step 22
For w=2,x=4,y=4,z=9w=2, x=4, y=4, z=9: w+x+y+z=2+4+4+9=19w+x+y+z=2+4+4+9=19
step 23
For w=2,x=7,y=4,z=6w=2, x=7, y=4, z=6: w+x+y+z=2+7+4+6=19w+x+y+z=2+7+4+6=19
step 24
For w=3,x=3,y=4,z=9w=3, x=3, y=4, z=9: w+x+y+z=3+3+4+9=19w+x+y+z=3+3+4+9=19
step 25
For w=3,x=7,y=4,z=5w=3, x=7, y=4, z=5: w+x+y+z=3+7+4+5=19w+x+y+z=3+7+4+5=19
step 26
For w=6,x=3,y=4,z=6w=6, x=3, y=4, z=6: w+x+y+z=6+3+4+6=19w+x+y+z=6+3+4+6=19
step 27
For w=6,x=4,y=4,z=5w=6, x=4, y=4, z=5: w+x+y+z=6+4+4+5=19w+x+y+z=6+4+4+5=19
step 28
For w=0,x=1,y=1,z=1w=0, x=1, y=1, z=-1: w+x+y+z=0+1+11=1w+x+y+z=0+1+1-1=1
step 29
For w=0,x=0,y=2,z=1w=0, x=0, y=2, z=-1: w+x+y+z=0+0+21=1w+x+y+z=0+0+2-1=1
step 30
For w=0,x=3,y=2,z=2w=0, x=-3, y=2, z=2: w+x+y+z=03+2+2=1w+x+y+z=0-3+2+2=1
step 31
For w=1,x=1,y=2,z=1w=-1, x=1, y=2, z=-1: w+x+y+z=1+1+21=1w+x+y+z=-1+1+2-1=1
step 32
For w=1,x=3,y=2,z=3w=-1, x=-3, y=2, z=3: w+x+y+z=13+2+3=1w+x+y+z=-1-3+2+3=1
step 33
For w=4,x=1,y=2,z=2w=-4, x=1, y=2, z=2: w+x+y+z=4+1+2+2=1w+x+y+z=-4+1+2+2=1
step 34
For w=4,x=0,y=2,z=3w=-4, x=0, y=2, z=3: w+x+y+z=4+0+2+3=1w+x+y+z=-4+0+2+3=1
Answer
The possible values for w+x+y+zw+x+y+z are 1919 and 11.
Key Concept
Solving Diophantine Equations
Explanation
This problem involves finding integer solutions to a polynomial equation, which is a type of Diophantine equation. By considering the factors of the constant term, we can systematically find all possible combinations of the variables that satisfy the equation.

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