Question

Determină numărul abcd, ştiind că abcd+ abc + ab + a= 2 701.​

Answer 1

Răspuns:

$abcd+abc+ab+a=2701$

$\left(bcd+bc+b+1\right)a=2701$

$\frac{\left(bcd+bc+b+1\right)a}{bcd+bc+b+1}=\frac{2701}{bcd+bc+b+1}$

$a=\frac{2701}{bcd+bc+b+1}$

$a=\frac{2701}{bcd+bc+b+1},\left(c=-\frac{1}{d+1}\text{ or }b\neq -\frac{1}{cd+c+1}\right)\text{ and }\left(b\neq -1\text{ or }d\neq -1\text{ or }c=-\frac{1}{d+1}\right)\text{ and }\left(b\neq -1\text{ or }d\neq -1\right)$

$abcd+abc+ab+a=2701$

$abcd+abc+ab=2701-a$

$\left(acd+ac+a\right)b=2701-a$

$\frac{\left(acd+ac+a\right)b}{acd+ac+a}=\frac{2701-a}{acd+ac+a}$

$b=\frac{2701-a}{acd+ac+a}$

$b=\frac{2701-a}{a\left(cd+c+1\right)}$

$\left\{\begin{matrix}b=-\frac{a-2701}{a\left(cd+c+1\right)}\text{, }&\left(d=-1\text{ or }c\neq -\frac{1}{d+1}\right)\text{ and }a\neq 0\\b\in \mathrm{R}\text{, }&a=2701\text{ and }c=-\frac{1}{d+1}\text{ and }d\neq -1\end{matrix}\right.$

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