sum of digits of two digit number is 9. If we interchange the digit than sum of original number and new number is 99. Find the original number
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Let the unit place digit of a two-digit number be x.
Therefore, the tens place digit = 9-x
\because∵ 2-digit number = 10 x tens place digit + unit place digit
\therefore∴ Original number = 10(9-x)+x
According to the question, New number
= Original number + 27
\Rightarrow10x+\left(9-x\right)=10\left(9-x\right)+x+27⇒10x+(9−x)=10(9−x)+x+27
\Rightarrow10+9-x=90-10x+x+27⇒10+9−x=90−10x+x+27
\Rightarrow9x+9=117-9x⇒9x+9=117−9x
\Rightarrow9x+9x=117-9⇒9x+9x=117−9
\Rightarrow18x=108⇒18x=108
\Rightarrow x=\frac{108}{18}=6⇒x=
18
108
=6
Hence, the 2-digit number = 10(9-x)+x = 10(9-6)+6 = 10 x 3 + 6 = 30 + 6 = 36
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