Question

5. A two digit number is such that the product of the digits is 14. When 45 is added to the

number, then the digits interchange their places. Find the number
I need what is the answer and the procedure in solving this​

Answer 1

Answer:

Hence, the required number is 27

Answer 2

AnswEr :

⠀⠀⠀⌬⠀Let , Digit at Unit's Place : A

⠀⠀⠀⌬⠀And , Digit at Ten's Place : B

⠀⠀⠀⌬⠀Then , Original Number : ( 10 B + A )

⠀⠀⠀⌬⠀Also , Interchanged Number : (10 A + B )

⠀⠀⠀⠀⠀⠀⠀(I) The Product of the Digit's of Original number is 14 .

$\\:\implies \sf Unit\:Digit\:\times Tens \:Digit \:=\:14\:$

$:\implies \sf B \:\times A \:=\:14\:$

$:\implies \sf B \:\times A \:=\:14\: \qquad \qquad \Bigg\lgroup \: Eq^n\:(i)\:\Bigg\rgroup$

⠀⠀⠀⠀⠀⠀⠀(II) If 45 is added to the number, then the digits interchange their places.

$\\\longrightarrow \sf Original \:Number \:+\:45\:=\:Interchanged\:Number \:\:$

$\longrightarrow \sf (\:10B\:+\:A\:) \:+\:45\:=\:(\:10A\:+\:B\:) \:\:$

$\longrightarrow \sf (\:10B\:+\:A\:) \:-\:(\:10A\:+\:B\:)\:=\:- 45 \:\:$

$\longrightarrow \sf \:10B\:+\:A\: \:-\:10A\:-\:B\:=\:- 45 \:\:$

$\longrightarrow \sf \:9B\:-\:9A\:=\:- 45 \:\:$

$\longrightarrow \sf \:B\:-\:A\:=\:-5 \:\:$

$\longrightarrow \sf \:B\:=\:-5 + A \:\:\qquad \qquad \Bigg\lgroup \: Eq^n\:(ii)\:\Bigg\rgroup$

✇ Substituting Value of B from Eq.(ii) in Eq.(i) , we get —

$\\\dashrightarrow \sf B \:\times A \:=\:14\:$

$\dashrightarrow \sf (\:-5\:+\:A\:) \:\times A \:=\:14\:$

$\dashrightarrow \sf (\:-5\:+\:A\:) \:\times A \:=\:14\:$

$\dashrightarrow \sf \:-5A\:+\:A^2\: \:=\:14\:$

$\dashrightarrow \sf \:A^2\:-\:5A\:-14 \:=\:0\:$

$\dashrightarrow \sf \:A^2\:-\:7A\:+ 2A\:-14 \:=\:0\:$

$\dashrightarrow \sf A\:(\:A\:-\:7\:)\:+\:2\:(\: A \:-7 ) \:=\:0\:$

$\dashrightarrow \sf \:(\:A\:+\:2\:)\:(\: A \:-7 ) \:=\:0\:$

$\dashrightarrow \sf \:A \:=\:-2\:,\:7\:$

• ATQ , Original Number cannot be in '-ve' .

$\\\dashrightarrow \sf \:A \:=\:7\:\qquad \qquad \Bigg\lgroup \: \pmb{\sf{-\:2\:}}\: rejected \:\Bigg\rgroup$

$\dashrightarrow \pmb{\sf \:A \:=\:7\:}$

$\qquad \dag \underline {\sf Now\:,\: By \:Using \:Eq^n\:(ii)\:\::\:}$

$\dashrightarrow \sf \:B\:=\:-5 \:+\: A\:$

$\dashrightarrow \sf \:B\:=\:-5 \:+\: 7\:$

$\dashrightarrow \pmb{\sf \:B\:=\:2\:}$

Therefore,

• Digit at Unit's Place , A = 7 &
• Digit at Ten's Place, B = 2 .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \bigstar \:\underline {\pmb{\sf Now\;,\:\:Original \:Number \:\::\:}}$

$\twoheadrightarrow \:\sf Original \:_{\:(\:Number \:)\:}=\: \:(\:10B \:+\;A \:)\:\\\\\\ \twoheadrightarrow \:\sf Original \:_{\:(\:Number \:)\:}\:=\: \:10(2) \:+\;7 \:\\\\\\ \twoheadrightarrow \:\sf Original \:_{\:(\:Number \:)}\:=\: \:20\:+\:7\:\\\\\\ \twoheadrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:Original \:_{\:(\:Number \:)}\:=\:27\:}}}}}\:\bigstar \:$

$\therefore \:\underline {\sf Hence, \:Original \:Number \:is\:\pmb{\sf 27\:}.\:}$