The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3.
The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.
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Let x be the ones digit and y be the tens digit. Then
Two digit number before reversing = 10y + x
Two digit number after reversing = 10x + y
As per the question (10y + x) + (10x + y) = 121
⇒11x + 11y = 121
⇒x + y = 11 …….(i)
Since the digits differ by 3, so
x – y = 3 ……….(ii)
Adding (i) and (ii), we get
2x = 14
⇒ x = 7
Putting x = 7 in (i), we get
7 + y = 11
⇒ y = 4
Changing the role of x and y, x = 4 and y = 7
Hence, the two-digit number is 74 or 47.
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