Asksia AI LOGO


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
Sign in to unlock the answer
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
The possible values for w+x+y+zw+x+y+z are 1919 and 11.
Key Concept
Solving Diophantine Equations
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.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages
Strong algorithms that better know you
Early access to new release features
Study Other Question