The sum of two digits number and the number obtained by reverse the digit is 66 .if the digit of numbers differ by2 , find the number
Answers
Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number can be written as 10x + y in the expanded form.
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x.
According to the given condition.
(10x + y) + (10y + x) = 66
11(x + y) = 66
x + y = 6 ... (1)
You are also given that the digits differ by 2. Therefore,
either x – y = 2 ... (2)
or y – x = 2 ... (3)
If x – y = 2, then solving (1) and (2) by elimination, you get x = 4 and y = 2. In this case, the number is 42.
If y – x = 2, then solving (1) and (3) by elimination, you get x = 2 and y = 4. In this case, the number is 24.
Thus, there are two such numbers 42 and 24.
Answer:
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