Problem Of The Week #456 February 22nd 2021

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anemone

MHB POTW Director
Staff member
Here is this week's POTW:

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If $x,\,y$ and $z$ are real numbers satisfying

\$(x+1)(y+1)(z+1)=3\$$x+2)(y+2)(z+2)=-2\\(x+3)(y+3)(z+3)=-1 find the value of (x+20)(y+20)(z+20). ----- Remember to read the POTW submission guidelines to find out how to submit your answers! anemone MHB POTW Director Staff member Congratulations to the following members for their correct solution, which you can find below: 1. Theia 2. kaliprasad Solution from Theia : We have \(\displaystyle \begin{cases}(x + 1)(y + 1)(z + 1) = 3 \\ (x + 2)(y + 2)(z + 2) = -2 \\ (x + 3)(y + 3)(z + 3) = -1 \end{cases}$$

and we want to calculate $$L = (x + 20)(y + 20)(z + 20).$$
Let's write

$$\displaystyle \begin{cases}x = a - 1 \\ y = b - 1 \\ z = c - 1.\end{cases}$$

After substituting to the original equation and expanding we have

$$\displaystyle \begin{cases}abc = 3 \\ abc + ab + ac + bc + a + b + c + 1 = -2 \\ abc + 2(ab + ac + bc) + 4(a + b + c) + 8 = -1.\end{cases}$$

Now substitute $$abc = 3$$ into two other equations and write

$$\displaystyle \begin{cases}ab + ac + bc = q \\ a + b + c = r.\end{cases}$$

Hence we have obtained simultaneous equations

$$\displaystyle \begin{cases}q + r = -6 \\ 2q + 4r = -12, \end{cases}$$

whose solution is $$q = -6, r = 0.$$ The expression $$L$$ is in terms of $$a,b,c$$:

\displaystyle \begin{align*}L &= (a + 19)(b + 19)(c + 19) \\ &= abc + 19(ab + ac + bc) + 19^2(a + b + c) + 19^3 \\ &= 3 + 19\cdot q + 19^2\cdot r + 19^3 \\ &= 3 - 6\cdot 19 + 19^2\cdot 0 + 19^3 \\ &= 6748.\end{align*}

Hence $$L = (x + 20)(y + 20)(z + 20) = 6748.$$

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