SOLUTION: Word problem:- A three digit number is equal to 17 times the sum of its digits. If 198 is added to the original number, the digits get interchanged. The addition of the first & t

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Question 1024485: Word problem:-
A three digit number is equal to 17 times the sum of its digits. If 198 is added to the original number,
the digits get interchanged. The addition of the first & the third digit is 1 less than the middle digit.
Find the original number. (steps required)

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Assign three variables and translate the description into equations.

h hundred
t tens
u one or units

The number: .

.

Simplify and solve the system of equations.

A start on that:


-

-


-
Revised system of equations:
Substitution method might be most comfortable from here.


---Continuing---
Use the last or third equation and solve it for t in terms of the other variables.


Substitute this into the first "equals 0" equation and simplify.




A revised, shorter system not including variable t, is this:

and the "99" equation is simplifiable making a system:

Solve THIS system for h and u.

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Word problem:-
A three digit number is equal to 17 times the sum of its digits. If 198 is added to the original number,
the digits get interchanged. The addition of the first & the third digit is 1 less than the middle digit.
Find the original number. (steps required)
Number: .
Don't you think you need to make an attempt? It's not that difficult to try something!

After your E-Mail, here's the solution:

Let the hundreds, tens, and units digits, be H, T, and U, respectively. From clue 1, we get:
100H + 10T + U = 17(H + T + U)		
100H + 10T + U = 17H + 17T + 17U		
83H - 7T - 16U = 0 -------- eq (i)	
		
From clue 2, we get:
100H + 10T + U + 198 = 100U + 10T + H		
99H - 99U = - 198		
99(H – U) = 99(- 2)  ------- Dividing by GCF, 99		
H – U = - 2 -------- eq (ii)
 
Finally, clue 3 tells us that: 
H + U = T - 1		
H - T + U = - 1 ------ eq (iii)		

- 7H + 7T - 7U = 7 ---- Multiplying eq (iii) by - 7		
 83H - 7T - 16U = 0 --- eq (i)		
 76H - 23U = 7 -------- Adding eqs (iii) & (i) ------- eq (iv)

    H –   U = - 2 ----- eq (ii)
- 23H + 23U =  46 ----- Multiplying eq (ii) by – 23 ------- eq (v) 		
  76H – 23U =   7 ----- eq (iv)
        53H = 53 ------ Adding eqs (v) & (iv)
          H, or hundreds digit = , or 1		

1 - U = - 2 -------- Substituting 1 for H in eq (ii)		
  - U = - 2 - 1		
  - U = - 3
    U, or units digit = , or 3
		
1 – T + 3 = - 1 ------- Substituting 1 for H, and 3 for U in eq (iii)		
    4 – T = - 1
      - T = - 1 – 4
      - T = - 5
        T, or tens digit = , or 5

Number:

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