Question

A three-digit number is equal to 17 times the sum of its digits. If the digits are reversed the new number is 198 more than the original number. The sum of the extreme digits is 1 less than the middle digit. Find the original number.

Answer 1

Hint: Systems with three equations and three variables can also be solved using the Addition/Subtraction method.
Pick any two pairs of equations in the system. Then use addition and subtraction to eliminate The Same Variable from both pairs of equations. This leaves two equations with two variables-one equation from each pair. Solve this system using the Addition/Subtraction method. Then plug the solution back into one of the original three equations to solve for the remaining variable.

Complete step by step solution:
Let the number (n) be $100x + 10y + z$ that is digit at units place is z, digit at tens place is y and digit at hundreds place is x.
Now,
According to the question,
The three-digit number is equal to 17 times the sum of its digits.
$
  \therefore This \Rightarrow 100x + 10y + z = 17(x + y + z) \\
   \Rightarrow 100x + 10y + z = 17x + 17y + 17z \\
   \Rightarrow 100x + 10y + z - (17x + 17y + 17z) = 0 \\
   \Rightarrow 100x + 10y + z - 17x - 17y - 17z = 0 \\
   \Rightarrow 83x - 7y - 16z = 0...........(1) \\
$
Again,
If the digits are reversed the new number is 198 more than the original number.
So, the reversed number will be $ = 100z + 10y + x$
$
  \therefore This \Rightarrow 100z + 10y + x = 198 + (100x + 10y + z) \\
   \Rightarrow 100z + 10y + x - (100x + 10y + z) = 198 \\
   \Rightarrow 100z + 10y + x - 100x - 10y - z = 198 \\
   \Rightarrow 99z - 99x = 198 \\
   \Rightarrow z - x = 2..............(2) \\
$


And,
The sum of the extreme digits of the original number is 1 less than the middle digit
$
  \therefore This \Rightarrow x + z = y - 1 \\
   \Rightarrow x - y + z = - 1................(3) \\
$
Now, we have to solve these three equations to get the desired value.
So,
At first taking equation 1 and equation 2
$\therefore $\[equation{\text{ }}1 + ((equation{\text{ 2)}} \times {\text{16)}}\]
$
  \therefore this \Rightarrow (83x - 7y - 16z) + 16 \times (z - x) = 0 + 16 \times 2 \\
   \Rightarrow 83x - 7y - 16z + 16z - 16x = 32 \\
   \Rightarrow 67x - 7y = 32.............(4) \\
$

Now , taking equation 2 and 3
$\therefore $\[equation{\text{ 2}} - (equation{\text{ 3)}}\]
$
  \therefore this \Rightarrow (z - x) - (x - y + z) = 2 - ( - 1) \\
   \Rightarrow z - x - x + y - z = 3 \\
   \Rightarrow - 2x + y = 3 \\
   \Rightarrow 2x - y = - 3...............(5) \\
$
And taking equation 4 and equation 5
$\therefore $\[equation{\text{ 4}} - {\text{(7}} \times (equation{\text{ 5)}}\]
$
  \therefore this \Rightarrow (67x - 7y) - (7 \times (2x - y)) = 32 - (7 \times ( - 3)) \\
   \Rightarrow (67x - 7y) - (14x - 7y) = 32 - ( - 21) \\
   \Rightarrow 67x - 7y - 14x + 7y = 32 + 21 \\
   \Rightarrow 53x = 53 \\
   \Rightarrow x = 1 \\
$
Now on putting x = 1 in equation 5 we will get y = 5
And on putting x = 1 in equation 2 we will get z = 3

Thus the required value of variables are x = 1, y = 5 and z = 3

Thus the required number is =$100x + 10y + z$=153

Note: Always cross-check the answer by putting values and checking the conditions.