Question

A number consists of two digit when the number is divised by the sum of its digit the quationt is 9

Answer 1

Answer:

$\bold{81}$

Step-by-step explanation:

$Let the number be \ 10x + y. \\ \\ Then the sum of its digits will be \ x + y. \\ \\ \\ When \ 10x + y\ is divided by \ x + y,\ we get 9 as quotient. \\ \\ \therefore\ 10x + y = 9(x + y) \\ \\ = 10x + y = 9x + 9y \\ \\ \\ 10x - 9x = 9y - y \\ \\ = x = 8y$

$\\ \\ \\ Here, it is clear from this that, if a two digit number divided by the sum of its digits gives 9 as quotient, then the digit at tens place of the number will be 8 times that at ones place of the number. \\ \\ \\ Here, both \ x\ and \ y\ are one digit numbers, as they are digits of \ 10x + y. \\ \\ If \ y = 1,\ x = 8y = 8 \times 1 = 8 \\ \\ If \ y = 2,\ x = 8y = 8 \times 2 = 16.\ Here, \ x\ becomes a two digit number. So this is not possible.$

$\\ \\ If \ y \geq 2,\ x > 10.\ So it is not possible when \ y \ge 2. \\ \\ \\ So there is only one possibility which satisfy this property, and that is, when \\ y = 1. \\ \\ \\ \therefore\ The two digit number is \ \bold{81}.$

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