Question

The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What can be the original number?
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Answer 1

$\bf Given \:\begin {cases} \sf The \:digits\: of \:a\: two-digit\: number\: differ\; by\:\frak 3\:\sf . \\\\ \sf The\: digits \:are\: interchanged\: and \:the \:resulting\: number\: is\:\\\qquad \sf added\: to\:the\: original\: number\:,\: we \:get\:\frak 143\:\sf .\:\\\end {cases}$

Exigency To Find : The Original number .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's say that the digit at once place be x .

⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀▪︎⠀The digits of a two-digit number differ by 3.

$\qquad\longmapsto \sf Tens \: Digit \:- \: Once \: Digit \:=\: 3\:\\\\\longmapsto \sf Tens \: Digit \:- \: x \:=\: 3\:\\\\\longmapsto \sf Tens \: Digit \: \:=\: x + 3\:\\\\\longmapsto {\underline \purple{\pmb Tens \: Digit \:=\: x + 3 \:}}$

⠀⠀⠀⠀⠀Therefore ,

⠀⠀⠀⠀⠀▪︎⠀The two digit or original number number will be :

$\qquad \longmapsto \sf \:\:Two\:-\:Digit \:_{(Number )} \: = \: 10(x + 3)+x \:\\\\ \longmapsto \sf \:\:Two\:-\:Digit \:_{(Number )} \: = \: 10x + 30+x \:\\\\ \longmapsto \underline{\purple { \pmb \:\:Two\:-\:Digit \:_{(Number )} \: = \: 11x + 30}} \:$

⠀⠀⠀⠀⠀☆ On Inter - Changing the Digits :

⠀⠀⠀▪︎⠀ Once Digit number will be ( x + 3 ) .

⠀⠀⠀▪︎ ⠀Tens Digit number will be x .

⠀⠀⠀⠀⠀Therefore ,

⠀⠀⠀⠀⠀▪︎⠀The two digit number will be on inter changing the digits :

$\qquad \longmapsto \sf \:\:Two\:-\:Digit \:_{(Number )} \: = \: 10(x) + x + 3 \:\\\\ \longmapsto \sf \:\:Two\:-\:Digit \:_{(Number )} \: = \: 10x + x+3 \:\\\\ \longmapsto \pmb \:\underline{\purple{\:Two\:-\:Digit \:_{(Number )} \: = \: 11x + 3 }}\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:According \:to\: the \: Question \::}}$

⠀⠀━━━ The digits are interchanged and the resulting number is added to the original number, we get 143.

$\qquad : \implies \sf Original \:number \: + \: \;Number \:on \:Interchanging \:= 143 \:\\\\ : \implies \sf 11x + 30 \: + \: 11 x + 3 \:= 143 \:\\\\ : \implies \sf 22x + 30 \: + 3 \:= 143 \:\\\\ : \implies \sf 22x + 33 \:= 143 \:\\\\:\implies \sf 22x \:= 143 - 33\:\\\\ : \implies \sf 22x\:= 110 \:\\\\ : \implies \sf x \:= \dfrac{110}{22}\:\\\\ : \implies \bf x \:= 5\:\\\\ \qquad:\implies \pmb{\underline{\purple{\:x = 5 }} }\:\bigstar$

⠀⠀⠀Therefore,

⠀⠀⠀⠀⠀▪︎ The two-digit original is : 11x + 30 = 11 (5) + 30 = 55 + 30 = 85 .

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The \:two\: Digit \:original\: numbern\:will\:be\:\bf{85}}.}}$

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Answer 2

$\sf✵ANSWER✵$

• Step-by-step explanation:

• Let the digit in the tens place be x

And, the digit in units place be y

The original number = 10x + y

Case 1:-

The number formed by interchanging the digits = 10y + x

• As, the digits differ by 3

$\sf\color{purple}⇢ x - y = 3...(i)$

• Case 2:-

⇢ The interchanged number + Original number = 143

$\sf\color{purple}⇢ 10y + x + 10x + y = 143$

$\sf\color{purple}⇢ 11x + 11y = 143$

Dividing the whole equation by 11

$\sf\color{red}⇢ x + y = 13...(ii)$

Adding equation (i) and (ii)

$\rm⇢ x - y + x + y = 3 + 13$

$\rm⇢ 2x = 16$

$\rm⇢ x = \dfrac{16}{2} = 8$

Substituting x = 8 in equation (ii)

$⇢ x + y = 13$

$\rm⇢ 8 + y = 13$

$\rm⇢ y = 13 - 8$

$\rm\color{blue}⇢ y = 5$

The original number = 10x + y

• = 10(8) + 5

• = 80 + 5

• = 85

Therefore:-

The original number = 85