The digits of a four-digit number that is a multiple of 5 are written in reverse order, producing a second four-digit number. Subtracting the second number from the first gives 1458. Find one such original number.

Step-by-step solution

1. 1

   Write both numbers in expanded place-value form. Let the digits be $a,b,c,d$, so the original number is

   $N=1000a+100b+10c+d,\qquad a\neq 0$

   and the reversed number is

   $N'=1000d+100c+10b+a$

   Turning the digit statement into arithmetic is the entire trick; everything after this is routine.

2. 2

   Force the last digit using both conditions. Divisibility by $5$ leaves $d\in\{0,5\}$. But the reversed number must also be a four-digit number, so its leading digit $d$ cannot be $0$. Hence

   $d=5$

   This is the step that decides the problem: with $d=0$ the reversal would collapse to at most three digits.

3. 3

   Subtract and simplify.

   $N-N'=1000a+100b+10c+5-(5000+100c+10b+a)$

   $=999a+90b-90c-4995$

   Setting this equal to $1458$ and moving the constant across:

   $999a+90(b-c)=1458+4995=6453$

4. 4

   Search over the single digit $a$, allowing $b-c$ to be negative. The term $90(b-c)$ can range from $-810$ to $810$, so $999a$ must land within $810$ of $6453$: only $a=7$ qualifies ($999\cdot 7=6993$, and $999\cdot 6=5994$ is $459$ short, which is not a multiple of $90$). With $a=7$:

   $6993+90(b-c)=6453\quad\Longrightarrow\quad 90(b-c)=-540\quad\Longrightarrow\quad b-c=-6$

   Note that $b-c$ is negative here. Testing only nonnegative differences is the trap that makes this problem look unsolvable.

5. 5

   List the numbers and pick one. Digits with $b-c=-6$ are $(b,c)\in\{(0,6),(1,7),(2,8),(3,9)\}$, giving

   $7065,\quad 7175,\quad 7285,\quad 7395$

   Any of them answers the question; the smallest is

   $\boxed{7065}$

6. 6

   Verify. $7065$ ends in $5$, so it is divisible by $5$. Reversing gives the four-digit number $5607$, and

   $7065-5607=1458$

   as required. The same method with a right-hand side of $1359$ gives $999a+90(b-c)=6354$, i.e. $a=6$ and $b-c=4$, whose smallest solution is $6405$.