Question

A three digit number is equal to 17 times the sum of its digits. If 198 is added to the number, the digits are interchanged. The addition of the first and third digit is 1 less than the middle digit. Find the number.

Answer 1

Hint: Assume the number to be a variable. Frame three equations using the three conditions given in the question. Solve them to get the number.

Complete step by step answer:
Let the number is $abc$.
According to the question, the number is equal 17 times the sum of its digits. So, we have:
$
   \Rightarrow 100a + 10b + c = 17\left( {a + b + c} \right) \\
   \Rightarrow 100a + 10b + c = 17a + 17b + 17c \\
   \Rightarrow 83a = 7b + 16c .....(i) \\
$
Further it is given that if 198 is added to the number, the digits are reversed. So, we have:
$
   \Rightarrow 100a + 10b + c + 198 = 100c + 10b + a \\
   \Rightarrow 99a + 198 = 99c \\
   \Rightarrow a + 2 = c \\
   \Rightarrow a = c - 2 .....(ii) \\
$
Also given that the addition of the first and third digit is 1 less than the middle digit. We have:
$ \Rightarrow a + c = b - 1$
Putting the value of $a$ from equation $(ii)$, we’ll get:
$
   \Rightarrow c - 2 + c = b - 1 \\
   \Rightarrow b = 2c - 1 .....(iii) \\
$
Putting $a = c - 2$ from equation $(ii)$ and $b = 2c - 1$ from equation $(iii)$ in equation $(i)$, we’ll get:
$
   \Rightarrow 83\left( {c - 2} \right) = 7\left( {2c - 1} \right) + 16c \\
   \Rightarrow 83c - 166 = 14c - 7 + 16c \\
   \Rightarrow 53c = 159 \\
   \Rightarrow c = \dfrac{{159}}{{53}} \\
   \Rightarrow c = 3 \\
$
Putting $c = 3$ in equation $(ii)$, we’ll get:
$
   \Rightarrow a = 3 - 2 \\
   \Rightarrow a = 1 \\
$
And putting $c = 3$ in equation $(iii)$, we’ll get:
$
   \Rightarrow b = 2\left( 3 \right) - 1 \\
   \Rightarrow b = 6 - 1 \\
   \Rightarrow b = 5 \\
$

So we have $a = 1,b = 5$ and $c = 3$.
Thus the required number is 153.

Note: The value of a three digit number $abc$ is $100a + 10b + c$.
Similarly, the value of a four digit number $abcd$ is $1000a + 100b + 10c + d$.
While writing the value of a number from its digits, every digit is multiplied by its place value.