A number has two digit whose sum is 9.If 27 is added to the number its digits are reversed. Find the number.
Given :
Sum of the digits of a two-digit number = 9
When we interchange the digits, the resulting new number is greater than the original number by 27.
To do :
We have to find the original number.
Solution :
Let the two digit number be $10x+y$.
$x + y = 9$
The number formed on reversing the digits is $10y+x$.
Therefore,
$10y+x = (10x+y)+27$
$10y-y+x-10x = 27$
$9(y-x) = 27$
$y-x = 3$
$y-(9-y) = 3$
$y+y = 3+9$
$2y = 12$
$y = 6$
$x = 9-6 = 3$
The original number is $10(3)+6 = 30+6 = 36$.
The original number is 36.
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