Question

# A number consists of two digits whose sum is five. When the digits are reversed the number becomes greater by nine. Find the number.

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Solution

## We have to find a two-digit number whose sum of digits is $5$ and when the digits are reversed the number becomes greater by $9$.Let the digit at ten's place be $\mathrm{x}$.Then the digit at one's place $=5-\mathrm{x}$ (since the sum of digits is $5$)Thus the number $=10\mathrm{x}+\left(5-\mathrm{x}\right)$The number formed by reversing the digits $=10\left(5-\mathrm{x}\right)+\mathrm{x}$According to the question,$10\left(5-\mathrm{x}\right)+\mathrm{x}=10\mathrm{x}+\left(5-\mathrm{x}\right)+9\phantom{\rule{0ex}{0ex}}⇒50-10\mathrm{x}+\mathrm{x}=10\mathrm{x}+5-\mathrm{x}+9\phantom{\rule{0ex}{0ex}}⇒50-9\mathrm{x}=9\mathrm{x}+14\phantom{\rule{0ex}{0ex}}⇒9\mathrm{x}+9\mathrm{x}=50-14\phantom{\rule{0ex}{0ex}}⇒18\mathrm{x}=36\phantom{\rule{0ex}{0ex}}⇒\mathrm{x}=\frac{36}{18}\phantom{\rule{0ex}{0ex}}⇒\mathrm{x}=2$Therefore the digit at ten's place is $2$.The digit at one's place $=5-2=3$Then the number $=10×2+3=20+3=23$Therefore the required number is $23$.

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