Dr. Fiendish 4

The basic question is, how to factorize?

Here is the method :

Let $abc+2ab+3ac+bc+6a+2b+3c+pqr\equiv (a+p)(b+q)(c+r)$

Then, expanding the R. H. S. of the equivalence and comparing coefficients of like terms, we get

$p=1,q=3,r=2\implies pqr=6$ . So, we can write the first equation as

$(a+1)(b+3)(c+2)=534+6=540$ .

Applying the same method, we can write the other equations as

$(a+1)(b+3)=54$

$(b+3)(c+2)=90$

$(c+2)(d+4)=140$ .

Solving we get $a=5,b=6,c=8,d=10$ .

Hence, the $5\times 6$ is the smallest size of a chocolate. Since a single number can be put in the answer box, we put the area of such a chocolate : $\boxed {30}$ square units.

\[\begin{cases} abc + 2ab + 3ac + bc + 6a + 2b + 3c {\color{red}{ +6}} &= 534 {\color{red}{ +6}} \\ ab + 3a + b + 12 {\color{green}{ -9}} &= 63 {\color{green}{ -9}} \\ bc + 2b + 3c - 4 {\color{blue}{ +10}} &= 80 {\color{blue}{ +10}} \\ cd + 4c + 2d {\color{pink}{ +8}} &= 132 {\color{pink}{ +8}} \end{cases}

\implies

\begin{cases} (a+1)(b+3)(c+2) &= 540 &\ldots {\color{red}{(1)}} \\ (a+1)(b+3) &= 54 &\ldots {\color{green}{(2)}} \\ (b+3)(c+2) &= 90 &\ldots {\color{blue}{(3)}} \\ (c+2)(d+4) &= 140 &\ldots {\color{pink}{(4)}} \end{cases} \]

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Finding $c$

$\dfrac{{\color{#D61F06}{(1)}}}{{\color{#20A900}{(2)}}} \implies (c+2) = 10 \implies \boxed{c = 8}$

Finding $a$

$\dfrac{{\color{#D61F06}{(1)}}}{{\color{#3D99F6}{(3)}}} \implies (a+1) = 6 \implies \boxed{a=5}$

Substitute in ${\color{#D61F06}{(1)}}$ to find $b$

$\begin{aligned} (a+1)(b+3)(c+2) &= 540 \\ ((5)+1)(b+3)((8)+2) &= 540 \\ b+3 &= 9 \implies \boxed{b = 6} \end{aligned}$

Substitute in $c$ in ${\color{#E81990}{(4)}}$ to find $d$

$\begin{aligned} (c+2)(d+4) &= 140 \\ ((8)+2)(d+4) &= 140 \\ d+4 &= 14 \implies \boxed{d = 10} \end{aligned}$

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The least among the three sizes is $\boxed{ab = 30} \quad bc = 48 \quad cd = 80$

Organise. $\begin{cases} abc+2ab+3ac+bc+6a+2b+3c=534\\ ab+3a+b+12-12=63-12=51\\ bc+2b+3c-4+4=80+4=84\\ \frac{2cd+8c+4d}{2}=\frac{264}{2}=132 \end{cases}$ Factorise. $\begin{cases} abc+2ab+3ac+bc+6a+2b+3c+6=(a+1)(b+3)(c+2)=540\\ ab+3a+b+3=(a+1)(b+3)=54\\ bc+2b+3c+6=(b+3)(c+2)=90\\ cd+4c+2d+8=(c+2)(d+4)=140 \end{cases}$ Now substitute $w,x,y,z$ into $a+1,b+3,c+2,d+4.$ Since $wxy=540,wx=54,y=\frac{540}{54}=10,etc.$
Similarly, we get $x=9, y=10, z=14.$
Substituting back, we have $a=5,b=6,c=8,d=10,$ so the required answer is the smallest among $ab,bc,cd,$ which is $ab=30.$