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Question Number 157752 by Engr_Jidda last updated on 27/Oct/21

  A sum of two−digit number  is 14. The tens digit is the square  of a number which is two less than  the unit digit. Find the two−digit numbers.

$$ \\ $$$${A}\:{sum}\:{of}\:{two}−{digit}\:{number} \\ $$$${is}\:\mathrm{14}.\:{The}\:{tens}\:{digit}\:{is}\:{the}\:{square} \\ $$$${of}\:{a}\:{number}\:{which}\:{is}\:{two}\:{less}\:{than} \\ $$$${the}\:{unit}\:{digit}.\:{Find}\:{the}\:{two}−{digit}\:{numbers}. \\ $$

Answered by Rasheed.Sindhi last updated on 27/Oct/21

Required number:   tu^(−)   t+u=14⇒t=14−u  t=v^2 =u−2  t=u−2=14−u     u=8⇒t=8−2=6 (not perfect square)  The number is 68

$${Required}\:{number}:\:\:\:\overline {\mathrm{tu}} \\ $$$$\mathrm{t}+\mathrm{u}=\mathrm{14}\Rightarrow\mathrm{t}=\mathrm{14}−\mathrm{u} \\ $$$$\mathrm{t}=\mathrm{v}^{\mathrm{2}} =\mathrm{u}−\mathrm{2} \\ $$$$\mathrm{t}=\mathrm{u}−\mathrm{2}=\mathrm{14}−\mathrm{u} \\ $$$$\:\:\:\mathrm{u}=\mathrm{8}\Rightarrow\mathrm{t}=\mathrm{8}−\mathrm{2}=\mathrm{6}\:\left(\mathrm{not}\:\mathrm{perfect}\:\mathrm{square}\right) \\ $$$$\mathcal{T}{he}\:{number}\:{is}\:\mathrm{68} \\ $$

Commented by Engr_Jidda last updated on 27/Oct/21

thanks

$${thanks} \\ $$

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