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Divisibility by 3

Given a three...

Question

Given a three digit number $x + 5 + y$ where $x$ is the digit at hundreds place and $y$ is the digit at ones. If the number $x + 5 + y$ is divisible by $9$ , find least positive value $x + y$ ?

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Solution

By divisibility test of 9 sum of digits must be divisible by 9
hence x+y+5=y
where y is a real number
minimum value of y possible is 9
hence
x+y+5=9
x+y=4

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