document.write( "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!! \n" ); document.write( "

Algebra.Com's Answer #12335 by venugopalramana(3286) $\"\"$ $\"About$

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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!!
\n" ); document.write( "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
\n" ); document.write( "WE ARE GIVEN THAT T-H=1.................I
\n" ); document.write( "SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1
\n" ); document.write( "52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T
\n" ); document.write( "104H+52=100H+10(H+1)=110H+10
\n" ); document.write( "110H-104H=52-10=42
\n" ); document.write( "6H=42
\n" ); document.write( "H=7
\n" ); document.write( "HENCE T=H+1=7+1=8
\n" ); document.write( "THE NUMBER IS 780\r
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