Math, asked by dipeshchadgal6266, 1 year ago

Two digit no is such that product of its digits is 18 when 63 is substracted from the number the digits interchange their place find the number?

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Answered by KarupsK
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Answered by Anonymous
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{\green {\boxed {\mathtt {☆Solution}}}}

  \rm \: let \: the \: tens \: and \: unit \: digit \: of \: the \: required \: number \: be \: x \: and \: y \: respectively \: then \\  \rm \: xy = 18 \implies \: y =  \frac{18}{x}   \\   \rm \: \purple {\: and \: (10x + y) - 63 = 10y + x }\\   \rm\implies \: 9x - 9y = 63 \implies \: x - y = 7 \:  \:  \:  \: .....(1) \\  \rm \orange{ \: putting \: y =  \frac{18}{x }  \: into \: (1) }\\  \rm \: x -  \frac{18}{x}  = 7 \\  \rm \: x {}^{2}   - 18 - 7x \implies \: x {}^{2}  - 7x - 18 \\  \rm  \implies \: x {}^{2}  - 9x + 2x - 18 = 0 \implies \: x(x - 9) + 2(x - 9) = 0 \\  \rm \implies(x - 9)(x + 2) = 0 \\  \rm \: x = 9 \: or \: x =  - 2 \:  \:  \:  \: ( but \: a \: digit \: cannot \: be \: negative) \\ \rm   \red {\:  \boxed{\therefore \: x = 9}} \\  \rm \: putting \: x = 9 \: in \: (1) we \: get \: y = 2 \\  \rm \: thus \: the \: tens \: digit \: is \: 9 \: and \: the \: unit \: digit \: is \: 2  \\  \rm hence \: the \: required \:  number \: is \: 92

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