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put this solution on YOUR website!let h=hundreds digit, t=tens digit, and u=units digit
"three digit number which is 31 times the sum of the digits" __ 31(h+t+u)=100h+10t+u
"the units' digit is one half the sum of the other digits" __ 2u=h+t __ 2u-h=t
"If the digits are reversed, the new number obtained is 99 greater than the original number"
__ 100u+10t+h=100h+10t+u+99
__ subtracting 10t __ 100u+h=100h+u+99 __ subtracting u+h __ 99u=99h+99 __ dividing by 99 __ u=h+1 __ u-1=h
substituting __ 2u-(u-1)=t __ u+1=t
substituting __ 31((u-1)+(u+1)+u)=100(u-1)+10(u+1)+u __ 93u=111u-90 __ 90=18u __ 5=u
substituting __ (5)-1=h __ 4=h
substituting __ (5)+1=t __ 6=t
the number is 465
