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THE SUM OF ANY TWO SIDES SHOULD BE MORE THAN THE THIRD TO FORM A TRIANGLE

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LET THE QUADRATIC BE
AX^2+BX+C=0...SINCE THIS IS A QUADRATIC ,A IS NOT EQUAL TO ZERO..SO DIVIDING WITH A WE GET
X^2+BX/A+C/A=0
WRITE IT AS A PERFECT SQUARE USING X^2 AND X TERMS
{X^2+2*X*(B/2A)+(B/2A)^2}-(B/2A)^2+C/A=0
{X+(B/2A)}^2=(B/2A)^2-C/A=B^2/4A^2-C/A=(B^2-4AC)/4A^2
TAKING SQUARE ROOT
X+B/2A=
OR

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Find the vertex and intercepts for each quadratic function, and graph.
y = -x^2 + 2x + 3
WRITE AS A SQUARE USING X^2 AND X TERMS
Y=3-{(X^2-2*X*1+1^2)-1^2}
Y=3-(X-1)^2+1=4-(X-1)^2
SINCE (X-1)^2 IS A PERFECT SQUARE IT IS ALWAYS POSITIVE OR MINIMUM ZERO.SO WE GET A MAXIMUM VALUE FOR Y AT X=1..VALUE IS Y=4..THIS IS CALLED VERTEX.YOU CAN SEE IT IN THE GRAPH.
INTERCEPTS MEAN WHERE THE CURVE CUTS THE AXES
PUT X=0 AND FIND Y TO GET Y INTERCEPTS..WE GET Y=3 HENCE 3 IS THE Y INTERCEPT.
PUT Y=0 AND FIND X TO GET X INTERCEPTS...WE GET 0=4-(X-1)^2
(X-1)^2=4
X-1=2....OR...X-1=-2
X=3.....OR......-1...
YOU CAN SEE IT FROM GRAPH GIVEN BELOW.

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4x^2-8x > = 0
4X(X-2) > =0
EITHER X=0 OR X=2...FOR 4x^2-8x = 0..SINCE A PRODUCT IS ZERO IF EITHER OF THE 2 FACTORS IS ZERO.
FOR 4x^2-8x > 0....EITHER BOTH FACTORS SHOULD BE POSITIVE OR NEGATIVE.
X > 0 AND X > 2....OR....X < 0 AND X < 2...THAT IS.......
X > 2...............OR.......X < 0

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(2u^3-13u^2-8u+7)divided by (u-7)...YOU CAN USE SHORT DIVISION FOR THIS ...
LET U-7=0...U=7
7....|.......2........-13.......-8..........7
.....|.......0........14........7..........-7
-------------------------------------------------
.....|.......2.........1.........-1.........0
HENCE REMINDER IS ZERO
QUOTIENT IS 2U^2+U-1....
IF YOU SHOW YOUR WORKING , WE CAN TELL WHERE YOU HAVE ERRED.

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WE SHALL DEFINITELY HELP YOU.BUT YOUR PROBLEM IS NOT CLEAR.DRAW A SKETCH ,SHOWING WHAT YOU MEAN ,SCAN AND SEND ,WE SHALL ANSWER YOU IMMEDIATELY

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4x^2-8x > = 0
4X(X-2) > =0
EITHER X=0 OR X=2...FOR 4x^2-8x = 0..SINCE A PRODUCT IS ZERO IF EITHER OF THE 2 FACTORS IS ZERO.
FOR 4x^2-8x > 0....EITHER BOTH FACTORS SHOULD BE POSITIVE OR NEGATIVE.
X > 0 AND X > 2....OR....X < 0 AND X < 2...THAT IS.......
X > 2...............OR.......X < 0

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3x-x^2> 0
X(3-X)>0
FOR A PRODUCT TO BE POSITIVE EITHER BOTH FACTORS SHOULD BE POSITIVE OR BOTH SHOULD BE POSITIVE.
HENCE X > 0 AND 3 > X...THAT IS
0 < X < 3.......OR...............
X < 0 AND 3 < X....THAT IS NOT POSSIBLE
SO THE SOLUTION IS
0 < X < 3

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x^2+25<10x
x^2+25-10x <0
(X)^2+(5)^2-2*(X)*(5)<0
(X-5)^2<0
A PERFECT SQUARE IS ALWAYS POSITIVE.ITS MINIMUM VALUE IS 0.IT CAN BE NEVER NEGATIVE OR LESS THAN ZERO..HENCE THIS PROBLEM HAS NO SOLUTION

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SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK
----------------------------------------------------------------------------
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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SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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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SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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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SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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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SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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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SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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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SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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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SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------

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
----------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------
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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