# How to Solve Digit Problems in Algebra Part V

This is the last part of the series on How to Solve Digit Problems in Algebra. In this post, we are going to solve another digit problem involving 3-digit numbers just like in the previous post.

Problem

The sum of the digits of a 3-digit number is 10. The hundreds digit is 3 more than the tens digits. If 198 is subtracted from the number, then the digits are reversed. What is the number?

Solution and Explanation

Since this is already the fifth part of the series, we won’t be as detailed as the previous parts.

Let

$h$ = hundreds digit of the number
$t$ = tens digit of the number
$u$ = units (or ones) digit of the number.

The first sentence says the sum of the digits of the number is 10. So,
$h + t + u = 10$ –> (Equation 1).

The hundreds digit is 3 more than the tens digit.
$h = t + 3$ –> (Equation 2).

If $198$ is subtracted from the number, the digits are reversed.

$100h + 10t + u - 198 = 100u + 10t + h$.

Simplifying the preceding equation, we have

$(100h - h) + (10t - 10t) + (u - 100u) = 198$
$99h - 99u = 198$.

Dividing both sides by $99$, we have
$h - u = 2$ –> (Equation 3).

We can eliminate $u$ by adding Equation 1 and Equation 3.
$(h + t + u) + (h - u) = 10 + 2$
$2h + t = 12$ (*).

We substitute $t + 3$ from  Equation 2 to $h$ in (*)
$2(t + 3) + t = 12$
$2t + 6 + t = 12$
$3t + 6 = 12$
$3t = 6$
$t = 2$.

Substituting the value of $t$ in Equation 2,
$h = t + 3$
$h = 2 + 3$
$h = 5$.

Substituting the values of $h$ and $t$ in Equation 1,
$h + t + u = 10$
$5 + 2 + u = 10$
$7 + u = 10$
$u = 10 - 7$
$u = 3$.

So, the hundreds digit is 5, tens digit is 2, and ones digit is 3. Therefore, the number is 523.

If we check with the conditions above, we have
(1) The sum of the digits is 10. That is, 5 + 2 + 3 = 10
(2) The hundreds digit is 3 more than the tens digit. The hundred digit is 5 is 3 more than 2
(3) If 198 is subtracted from the number, the digits are reversed. That is, 523 – 198 = 325.