Let $w, x, y$ and $z$ represent four whole numbers (not necessarily positive). If $(w-1)(x-2)(y-3)(z-4)=10$, find all possible values for $w+x+y+z$

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∻Solution by Steps∻ ‖ step 1 ⋮ Given the equation $(w-1)(x-2)(y-3)(z-4)=10$, we need to find all possible integer values for $w, x, y,$ and $z$. ‖ ‖ step 2 ⋮ We start by considering the factors of 10, which are $\pm 1, \pm 2, \pm 5, \pm 10$. ‖ ‖ step 3 ⋮ We need to find combinations of $(w-1), (x-2), (y-3),$ and $(z-4)$ that multiply to 10. ‖ ‖ step 4 ⋮ Let's consider the positive factors first: ‖ ‖ step 5 ⋮ If $(w-1)=1$, $(x-2)=1$, $(y-3)=2$, $(z-4)=5$, then $w=2$, $x=3$, $y=5$, $z=9$. ‖ ‖ step 6 ⋮ If $(w-1)=1$, $(x-2)=2$, $(y-3)=1$, $(z-4)=5$, then $w=2$, $x=4$, $y=4$, $z=9$. ‖ ‖ step 7 ⋮ If $(w-1)=1$, $(x-2)=5$, $(y-3)=1$, $(z-4)=2$, then $w=2$, $x=7$, $y=4$, $z=6$. ‖ ‖ step 8 ⋮ If $(w-1)=2$, $(x-2)=1$, $(y-3)=1$, $(z-4)=5$, then $w=3$, $x=3$, $y=4$, $z=9$. ‖ ‖ step 9 ⋮ If $(w-1)=2$, $(x-2)=5$, $(y-3)=1$, $(z-4)=1$, then $w=3$, $x=7$, $y=4$, $z=5$. ‖ ‖ step 10 ⋮ If $(w-1)=5$, $(x-2)=1$, $(y-3)=1$, $(z-4)=2$, then $w=6$, $x=3$, $y=4$, $z=6$. ‖ ‖ step 11 ⋮ If $(w-1)=5$, $(x-2)=2$, $(y-3)=1$, $(z-4)=1$, then $w=6$, $x=4$, $y=4$, $z=5$. ‖ ‖ step 12 ⋮ Now consider the negative factors: ‖ ‖ step 13 ⋮ If $(w-1)=-1$, $(x-2)=-1$, $(y-3)=-2$, $(z-4)=-5$, then $w=0$, $x=1$, $y=1$, $z=-1$. ‖ ‖ step 14 ⋮ If $(w-1)=-1$, $(x-2)=-2$, $(y-3)=-1$, $(z-4)=-5$, then $w=0$, $x=0$, $y=2$, $z=-1$. ‖ ‖ step 15 ⋮ If $(w-1)=-1$, $(x-2)=-5$, $(y-3)=-1$, $(z-4)=-2$, then $w=0$, $x=-3$, $y=2$, $z=2$. ‖ ‖ step 16 ⋮ If $(w-1)=-2$, $(x-2)=-1$, $(y-3)=-1$, $(z-4)=-5$, then $w=-1$, $x=1$, $y=2$, $z=-1$. ‖ ‖ step 17 ⋮ If $(w-1)=-2$, $(x-2)=-5$, $(y-3)=-1$, $(z-4)=-1$, then $w=-1$, $x=-3$, $y=2$, $z=3$. ‖ ‖ step 18 ⋮ If $(w-1)=-5$, $(x-2)=-1$, $(y-3)=-1$, $(z-4)=-2$, then $w=-4$, $x=1$, $y=2$, $z=2$. ‖ ‖ step 19 ⋮ If $(w-1)=-5$, $(x-2)=-2$, $(y-3)=-1$, $(z-4)=-1$, then $w=-4$, $x=0$, $y=2$, $z=3$. ‖ ‖ step 20 ⋮ Summing up all possible values of $w, x, y,$ and $z$ for each valid combination: ‖ ‖ step 21 ⋮ For $w=2, x=3, y=5, z=9$: $w+x+y+z=2+3+5+9=19$. ‖ ‖ step 22 ⋮ For $w=2, x=4, y=4, z=9$: $w+x+y+z=2+4+4+9=19$. ‖ ‖ step 23 ⋮ For $w=2, x=7, y=4, z=6$: $w+x+y+z=2+7+4+6=19$. ‖ ‖ step 24 ⋮ For $w=3, x=3, y=4, z=9$: $w+x+y+z=3+3+4+9=19$. ‖ ‖ step 25 ⋮ For $w=3, x=7, y=4, z=5$: $w+x+y+z=3+7+4+5=19$. ‖ ‖ step 26 ⋮ For $w=6, x=3, y=4, z=6$: $w+x+y+z=6+3+4+6=19$. ‖ ‖ step 27 ⋮ For $w=6, x=4, y=4, z=5$: $w+x+y+z=6+4+4+5=19$. ‖ ‖ step 28 ⋮ For $w=0, x=1, y=1, z=-1$: $w+x+y+z=0+1+1-1=1$. ‖ ‖ step 29 ⋮ For $w=0, x=0, y=2, z=-1$: $w+x+y+z=0+0+2-1=1$. ‖ ‖ step 30 ⋮ For $w=0, x=-3, y=2, z=2$: $w+x+y+z=0-3+2+2=1$. ‖ ‖ step 31 ⋮ For $w=-1, x=1, y=2, z=-1$: $w+x+y+z=-1+1+2-1=1$. ‖ ‖ step 32 ⋮ For $w=-1, x=-3, y=2, z=3$: $w+x+y+z=-1-3+2+3=1$. ‖ ‖ step 33 ⋮ For $w=-4, x=1, y=2, z=2$: $w+x+y+z=-4+1+2+2=1$. ‖ ‖ step 34 ⋮ For $w=-4, x=0, y=2, z=3$: $w+x+y+z=-4+0+2+3=1$. ‖ ∻Answer∻ ⚹ The possible values for $w+x+y+z$ are $19$ and $1$. ⚹ ∻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.⚹◊What are the possible values for $w+x+y+z$ given that $(w-1)(x-2)(y-3)(z-4)=10⍭ Generate me a similar question◊

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