SOLUTION: 19) A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!!

Question 23580: 19) A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!!
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!!
LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T
WE ARE GIVEN THAT T-H=1.................I
SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1
52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T
104H+52=100H+10(H+1)=110H+10
110H-104H=52-10=42
6H=42
H=7
HENCE T=H+1=7+1=8
THE NUMBER IS 780

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