SOLUTION: help please!! /The units digit of a two-digit number is 5 more than the tens digit. If the sum of the digits is one-thir of the number, find the number.

Question 23560: help please!! /The units digit of a two-digit number is 5 more than the tens digit. If the sum of the digits is one-thir of the number, find the number.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK
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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!!
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
SEE THE FOLLOWING EXAMPLE.
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LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U
the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!!
TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H
UNITS DIGIT EXCEEDS TENS DIGIT BY U-T
THESE ARE SAME .HENCE
T-H=U-T...OR 2T=U+H..................................................I
WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES
UTH...ITS VALUE = 100U+10T+H
NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U
=99U-99H=99(U-H)
THIS IS EQUAL TO 198
HENCE 99(U-H)=198
U-H=198/99=2.........................................................II
U+H=2T..................FROM....I
ADDING,WE GET
2U=2T+2.....DIVIDING BY 2,WE GET
U=T+1...OR........T=U-1......................AND
H=U-2...........FROM.......II
SO THE ORIGINAL NUMBER IS
HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE
U=9..........789.............987-789=198
U=8...........678............876-678=198...ETC
U=7..........567
U=6..........456
U=5............345
U=4............234
U=3............123
U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198
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