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Question 622057: what is the three-digit number which satisfies the following conditon? The tens digit is greater than the ones digit, the sum of the digits is 9, and if the digits are reversed and the resulting three-digit number is subtracted fron the original number, the difference is 198.
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The three digits are a, b, c
then
100a + 10b + c = "the number"
:
Write an equation for each statement:
:
"The tens digit is greater than the ones digit,"
b > c
:
"the sum of the digits is 9,"
a + b + c = 9
:
" and if the digits are reversed and the resulting three-digit number is subtracted from the original number, the difference is 198."
100a + 10b + c -(100c + 10b + a) = 198
Removing the brackets changes the signs
100a + 10b + c - 100c - 10b - a = 198
Combine like terms
100a - a + 10b - 10b + c - 100c = 198
99a - 99c = 198
simplify, divide by 99
a - b = 2
a = b+2
:
Replace a with (b+2) in the sum equation
(b+2) + b + c = 9
2b + c = 9-2
2b + c = 7
From this assume b=3, then c=1 and a=5
531 is the number
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! what is the three-digit number which satisfies the following conditon? The tens digit is greater than the ones digit, the sum of the digits is 9, and if the digits are reversed and the resulting three-digit number is subtracted fron the original number, the difference is 198.
Let hundreds, tens, and units digits be H, T, and U, respectively
H + T + U = 9 ------ eq (i)
Original number: 100H + 10T + U
Reversed number: 100U + 10T + H
100H + 10T + U – (100U + 10T + H) = 198 ----- Subtracting reversed no. from original no.
100H + 10T + U – 100U – 10T – H = 198
99H – 99U = 198 ------ 99(H – U) = 99(2) ------ H – U = 2 eq (ii)
T > U
H + T + U = 9 ----- eq (i)
H – U = 2 ----- eq (ii)
T + 2U = 7 ------ Subtracting eq (ii) from eq (i) ------ eq (iii)
From eq (iii), U CANNOT BE 9, 8, 7, 6, 5, or 0
This means that U MUST either be 4, 3, 2, or 1
U = 4
Based on the fact that T + 2U = 7, T = 7 – 8, or – 1 (NOT POSSIBLE)
U = 3
Based on the fact that T + 2U = 7, T = 7 – 6, T = 1. But, T > U, so T CANNOT be 1 if U is 3
U = 2
Based on the fact that T + 2U = 7, T = 7 – 4, T = 3. This is POSSIBLE, which means that H = 2 + 2, or 4, which means that original number’s 432. Reversed, it becomes 234. Difference = 198 (TRUE)
U = 1
Based on the fact that T + 2U = 7, T = 7 – 2, or 5. This is POSSIBLE, which means that H = 2 + 1, or 3, which means that original number’s 351. Reversed, it becomes 153. Difference = 198 (TRUE).
This means that the original number can either be  or  .
Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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