Bear a two-digit number in mind.Without disclosing it, cunstruct a puzzle create two algebraic relations between the two digits of the number and solve the puzzle.

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Solution

I am a number, tell my identity, three times of me, add fifty. You still need forty to have a triple century.

Let the number be $x .$
$\begin{matrix} (3 x + 50) + 40 & = 300 3 x & = 300 - 90 3 x & = 210 x & = 70 \end{matrix}$

Hence, this is the answer.

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Bear a two-digit number in mind.Without disclosing it, cunstruct a puzzle create two algebraic relations between the two digits of the number and solve the puzzle.

I am a number, tell my identity, three times of me, add fifty. You still need forty to have a triple century.Let the number be x.(3x+50)+40=3003x=300−903x=210x=70Hence, this is the answer.

I am a number, tell my identity, three times of me, add fifty. You still need forty to have a triple century.

Let the number be $x .$
$\begin{matrix} (3 x + 50) + 40 & = 300 3 x & = 300 - 90 3 x & = 210 x & = 70 \end{matrix}$

Hence, this is the answer.