A three-digit number has 0 as its tens digit, and twice its hundreds digit is 2 more than its units digit. Swapping two of its digits produces a number that is 198 larger. Find the number.

Step-by-step solution

1. 1

   Encode the shape of the number. Let the digits be $a$ (hundreds), $b$ (tens) and $c$ (units). The tens digit is given as $0$, so $b=0$ and the number is

   $N=100a+0+c=100a+c$

2. 2

   Turn the second sentence into an equation. Twice the hundreds digit is $2$ more than the units digit:

   $2a=c+2$

3. 3

   Work out which swap can raise the value. Swapping the hundreds and tens digits would move $a$ into the tens place and make the number smaller, and the same is true of swapping the tens and units. Only exchanging the hundreds and units digits can increase it, giving $100c+a$:

   $100c+a=(100a+c)+198$

   $99c-99a=198\ \Longrightarrow\ c-a=2$

4. 4

   Solve the two-equation system. From the two conditions, $c=2a-2$ and $c=a+2$. Setting them equal:

   $2a-2=a+2\ \Longrightarrow\ a=4$

   $c=a+2=6$

5. 5

   Assemble and check both conditions. The number is $N=406$. Twice the hundreds digit: $2\times 4=8$, and the units digit plus $2$ is $6+2=8$ $\\checkmark$. The swap:

   $604-406=198\quad\checkmark$

   Both conditions hold, so the number is $406$.