Question

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

Answer 1

Let the tens place digit be a

And Unit place digit be b

According to first condition,

a + b = 9 -------(1)

According to second condition,

9 ( 10a + b) = 2 (10b + a)

=> 90a + 9b = 20b + 2a

=> 90a - 2a + 9b - 20b = 0

=> 88a - 11b = 0

=> 8a - b = 0 -------(2)

On adding equation 1 and 2, we get

9a = 9

=> a = 1

On Substituting the value of a in equation 1, we get

1 + b = 9

=> b = 8

Required number = 10 a + b

= 10 × 1 + 8

= 18

Answer 2

let unit place digit be' X'
tens place digit= (x-9)
original no. = 10* (x-9)+x
= 11x-9
Reversed= 1* (x-9)+10x
= 11x-9
ATQ,
9(11x-9)= 2( 11x-9)
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well it is not possible i guess or there is any mistake the tell me