|
Question 23556: 17) The sum of the digits of a three-digit number is 19. The hundreds digit is three times the tens digit. The number is 198 more than the number reversed. Find the original number.
Found 3 solutions by venugopalramana, xhia, MathTherapy:
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
----------------------------------------------------------------------------
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
--------------------------------------------------------------------------------
Answer by xhia(1)  (Show Source):
You can put this solution on YOUR website! representing x is the units, y is the tens and z is the hundreds.
eq#1 x+y+z=19
eq#2 z=3y
since "The number is 198 more than the number reversed"
so
Eq#3 100z + 10y + x=100x+10y+z+198
simplify,z-u=198
you may use substitution but it'll make your solution very long so, use elimination.
eliminate eq.1 and 3
then substitute.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! Let the hundreds, tens, and units digits be H, T, and U, respectively
Then H + T + U = 19 -------- eq (i)
Now, since the hundreds digit is 3 times the tens digit, then H = 3T
Substituting 3T for H in eq (i), we can say that 3T + T + U = 19, or 4T + U = 19 ------ eq (ii)
Original number’s value = 100(3T) + (10T) + U ----> 300T + 10T + U ----> 310T + U
Reversed number’s value = 100(U) + 10(T) + 3T ----> 100U + 10T + 3T ----> 100U + 13T
Since the original number is 198 more than when it’s reversed, then we’ll have:
Original number = Reversed number + 198, or
310T + U = 100U + 13T + 198 (notice that the middle digit remains the same although the 3-digit number has been reversed)
310T – 13T – 99U = 198
297T – 99U = 198 ------ eq (iii)
396T + 99U = 1881 ------ eq (iv)------ Multiply eq (ii) by 99
693T = 2,079 ---- Adding eqs (iii) and (iv)
T = 3
Substitute 3 for T in eq (ii) to get: 4(3) + U = 19………… U = 7
Substitute 7 for U in eq (i) to get: H + 3 + 7 = 19 ……….. H = 9
H = 9, T = 3, and U = 7. Therefore, the original number is  . When reversed, it becomes 739, which is 198 less than the original number
|
|
|
| |
