The digits of a four-digit multiple of 5 are written in reverse order to give a second four-digit number. Subtracting the second from the first gives 1359. Find one such original number.

Step-by-step solution

1. 1

   Write both numbers in place-value form. Let the original be $\overline{abcd}=1000a+100b+10c+d$; its reverse is $\overline{dcba}=1000d+100c+10b+a$. Subtracting,

   $N-N_{\text{rev}}=999(a-d)+90(b-c).$

   Every problem of this type reduces to that one identity.

2. 2

   Pin down the units digit. $N$ is a multiple of $5$, so $d\in\{0,5\}$. But the reverse must also be a four-digit number, so its leading digit $d$ cannot be $0$. Hence

   $d=5.$

3. 3

   Solve the identity for the digit differences. We need $999(a-d)+90(b-c)=1359$. Trying $a-d=1$ gives $90(b-c)=1359-999=360$, so $b-c=4$ — a valid pair of digit differences. Trying $a-d=2$ gives $90(b-c)=1359-1998=-639$, which is not a multiple of $90$; larger values overshoot further. So $a-d=1$ and $b-c=4$ is forced.

4. 4

   Assemble the digits. With $d=5$ and $a=d+1=6$, and $b-c=4$ with both digits in $0..9$, the possibilities are $(b,c)=(4,0),(5,1),(6,2),(7,3),(8,4),(9,5)$:

   $6405,\ 6515,\ 6625,\ 6735,\ 6845,\ 6955.$

5. 5

   Check one and note the pattern. $6405-5046=1359$ ✓, and $6955-5596=1359$ ✓. A brute-force scan of all four-digit multiples of $5$ confirms these six are the complete list; any one of them answers the question.