the sum of the digits of a two-digit numeral is 8. if the digits are reversed, the new number is 18 greater then the original number. find the original numeral.
To solve this problem, let's represent the two-digit numeral as "10a + b," where 'a' is the tens digit, and 'b' is the ones digit.
We are given two conditions:
1. The sum of the digits is 8, so we have the equation a + b = 8.
2. Reversing the digits gives us a new number that is 18 greater than the original number, so we have the equation (10b + a) = (10a + b) + 18.
Let's solve these equations simultaneously:
Equation 1: a + b = 8 (1)
Equation 2: 10b + a = 10a + b + 18 (2)
We can simplify equation 2 by subtracting 'a' and 'b' from both sides:
9b - 9a = 18 (3)
Now, let's solve equations (1) and (3) simultaneously to find the values of 'a' and 'b':
Equation (1) multiplied by 9: 9a + 9b = 72 (4)
Equation (3) minus equation (4):
9b - 9a - (9a + 9b) = 18 - 72
-18a = -54
Dividing both sides by -18:
a = 3
Now, substitute the value of 'a' (3) into equation (1):
3 + b = 8
b = 5
Therefore, the original numeral is 10a + b = 10 * 3 + 5 = 35.
Thus, the original numeral is 35.