Question

The sum of digits of a two digit number is 9. Also nine times this number is twice the number obtained by revering the order of the digits. Find the number

Answer 1

A n s w e r

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G i v e n

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• The sum of digits of a two digit number is 9

• Nine times the original number is twice the number obtained by revering the order of the digits

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F i n d

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• The number

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S o l u t i o n

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• Let the ones digit be "y"

• Let the tens digit be "x"

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$\underline{\bold{\texttt{Original number :}}}$

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➠ 10x + y ⚊⚊⚊⚊ ⓵

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$\underline{\bold{\texttt{Reversed number :}}}$

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➠ 10y + x

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$\underline{\bold{\texttt{9 times the original number :}}}$

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➜ 9(10x + y) ⚊⚊⚊⚊ ⓶

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$\underline{\bold{\texttt{2 times the reversed number :}}}$

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➜ 2(10y + x) ⚊⚊⚊⚊ ⓷

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Given that , The sum of digits of a two digit number is 9

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So ,

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➜ x + y = 9 ⚊⚊⚊⚊ ⓸

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Also given that , nine times the original number is twice the number obtained by revering the order of the digits

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Thus ,

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Equation ⓶ = Equation ⓷

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➜ 9(10x + y) = 2(10y + x)

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➜ 90x + 9y = 20y + 2x

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➜ 90x - 2x = 20y - 9y

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➜ 88x = 11y

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⟮ Dividing the above equation by 11 ⟯

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➜ $\sf \dfrac { 88x} { 11 } = \dfrac { 11y } { 11 }$

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➜ 8x = y ⚊⚊⚊⚊ ⓹

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⟮ Putting y = 8x from ⓹ to ⓸ ⟯

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➜ x + y = 9

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➜ x + 8x = 9

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➜ 9x = 9

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➜ x = 1 ⚊⚊⚊⚊ ⓺

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• Hence the tens digit is 1

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⟮ Putting x = 1 from ⓺ to ⓸ ⟯

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➜ x + y = 9

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➜ 1 + y = 9

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➜ y = 9 - 1

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➜ y = 8 ⚊⚊⚊⚊ ⓻

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• Hence the ones digit is 8

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⟮ Putting x = 1 & y = 8 from ⓺ & ⓻ to ⓵ ⟯

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➜ 10x + y

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➜ 10(1) + 8

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➜ 10 + 8

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➨ 18

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• Hence the original number is 18