Question:
The sum of the digits of a two-digit number is 15. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number.
Solution:
Let the tens place digit be a and the units place digit be b.
Then, the number is (10a + b).
Given:
a + b = 15 ... (1)
When the digits are interchanged the number will be (10 b + a).
Given:
10a + b + 9 = 10 b + a
∴ 10a - a + b - 10b = -9
9a - 9b = -9
a - b = -1 ... (2)
Adding equations (1) and (2)
Using a = 7 in equation (2):
7 - b = -1
∴ b = 8
Original number $=10 a+b=10 \times 7+8=78$