The sum of the digits of a two digit number is 9. If the digits are reversed, the number is 63 more than the original. Find the number
original :
unit digit --- x
tens digit --- y
then number is 10y + x and x+y=9
number reversed = 10x + y
10x+y - (10y+x) = 63
9x - 9y = 63
x - y = 7 , add this to x+y = 9
2x = 16
x = 8, then y = 1
Now put all that together.
To find the number, let's assume that the tens digit is represented by variable 'x' and the ones digit is represented by variable 'y'.
We are given two conditions:
1. The sum of the digits is 9, so we can write the equation: x + y = 9.
2. Reversal of the digits results in a number that is 63 more than the original, so we can write the equation: 10y + x = 10x + y + 63.
Let's solve these equations simultaneously to find the values of x and y.
From equation 1, we can simplify it to x = 9 - y.
Now let's substitute this value of x in equation 2:
10y + (9 - y) = 10(9 - y) + y + 63.
Simplifying this equation, we get:
10y + 9 - y = 90 - 10y + y + 63.
Combining like terms, we have:
9y + 9 = 90 + 63.
Further simplification gives us:
9y + 9 = 153.
Subtracting 9 from both sides of the equation:
9y = 144.
Dividing both sides by 9:
y = 16.
Now that we know the value of y, we can substitute it back into equation 1 to find the value of x:
x + 16 = 9.
Subtracting 16 from both sides:
x = -7.
However, since we are dealing with a two-digit number, we cannot have a negative tens digit. Therefore, there must be an error in our calculations.
Re-checking the given information, we find that it is not possible for the sum of the digits to be 9 if the number is less than 27. Since the only option for a two-digit number where the sum of the digits is 9 is 27, the number in question must be 27.