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4 sum
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Let ones digit number be x and tens digit number be y of the original number.
According to the question,
2x = y
Original number = y × 10 + x×1 = 2x × 10 + x
[Since 2x = y]
= 20x + x = 21x
Number after reversing digits,
= x × 10 + y × 1 = 10x + 2x
[Since 2x = y]
= 12x
Also, original number + number after reversing digits = 99
Therefore, 21x + 12x = 99
33x = 99
x = 99/33
x = 3
Hence original number = 21x = 21(3) = 63
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