# How to Solve Digit Problems in Algebra Part IV

In the previous posts, Part 1, Part 2, and Part 3, we have learned how to solve digit problems involving 2-digit numbers. In this post, we are going to discuss digit problems involving 3-digit numbers. Let’s have the following problem.

Problem

The hundreds digit of a 3-digit number is twice its units digit and its tens digit is 1 more than its units digit. If 297 is subtracted from the number, then its digits are reversed. What is the number?

Solution and Explanation

Before we solve the problem, let’s recall that in the first two parts of this series that a number with tens digit $t$ and units digit $u$ can be represented by $10t + u$ and when we reverse its digits can be represented by $10u + t$. For example, the number $52$ with $t = 5$ and $u = 2$ can be represented as $10(5) + 2$ and when we reverse the digits, it becomes $25 = 10(2) + 5$.

In the same way, if we let $h$ be the hundreds digit of a number, $t$ be the tens digit, and $u$ be the units digit, then we can represent the number as $100h + 10t + u$

and with its digits reversed as $100u + 10t +h$.

In the first sentence of the problem above, it says that the hundreds digit is twice the units digit. Therefore, $h = 2t$ (*).

In the second sentence, it says that if 297 is subtracted from the number, then the digits is reversed. Putting this in equation, we have

number – 297 = number with digits reversed.

That is, $100h + 10t + u - 297 = 100u + 10t +h$.

Simplifying, we have $(100h - h) + (10t - 10t) + (u - 100u) = 297$ $99h - 99u = 297$.

Dividing both sides by 99, we have $h - u = 3$.

But from (*), h = 2u. So, substituting, we have $2u - u = u = 3$.

So, the units digit is equal to $3$.

Now, the tens digit is 1 more than the units digit, so it’s $4$.

The hundreds digit is twice the units digit, so it’s $6$.

Therefore, the number is $643$.

Check: Let’s try to subtract $297$ and see if the digits are reversed. $643 - 297 = 346$. 