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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 x.

Then the digit at one's place =5-x (since the sum of digits is 5)

Thus the number =10x+5-x

The number formed by reversing the digits =105-x+x

According to the question,

105-x+x=10x+5-x+950-10x+x=10x+5-x+950-9x=9x+149x+9x=50-1418x=36x=3618x=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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