Problem Of The Week #456 February 22nd 2021

[anemone]

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 $(x+20)(y+20)(z+20)$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[anemone]

Congratulations to the following members for their correct solution, which you can find below:

1. Theia
2. kaliprasad

Solution from Theia :

Spoiler

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 \(L = (x + 20)(y + 20)(z + 20) = 6748.\)