Number Theory

 

How many three-digit numbers are equal to five times the sum of their digits?    

 

100a + 10b + c  =  5 • (a + b + c)    

100a + 10b + c  =  5a + 5b + 5c     

 

  95a + 5b – 4c  =  0     

 

                  95a  =  4c – 5b    

 

Stipulating that "a" cannot be zero; because, if so, the number would be a two-digit number.    

 

If a = 1 then   95 is too large to equal the most that 4c – 5b can be, namely 36 – 45, i.e., – 9    

If a = 2 then 190 is too large to equal the most that 4c – 5b can be, namely 36 – 45, i.e., – 9    

If a = 3 then 285 is too large, etc., and so on for the rest of the digits that "a" might equal.    

 

I think this little demonstration shows that there is no solution.    

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