Let ABC be the three-digit number.(100A+10B+C)-(A+B+C)=99A+9B=9(11A+B)9(11A+B) is a perfect square.To maximize the number 100A+10B+C, we set A=9,B=1 and C=9.Therefore the answer is 919.
Let ABC be the three-digit number.
ReplyDelete(100A+10B+C)-(A+B+C)=99A+9B=9(11A+B)
9(11A+B) is a perfect square.
To maximize the number 100A+10B+C, we set
A=9,B=1 and C=9.
Therefore the answer is 919.