When a 4-digit number is multiplied by 1.5, the order is reversed. What is the number?
I did several things. Here is my work.
$1.5(1000a+100b+10c+d)=1000d+100c+10b+a$
$1500a+150b+15c+1.5d=1000d+100c+10b+a$
$1499a+140b=998.5d+85c$
I know $d$ must be even, or we get a decimal. Also, $d\neq0$
Now up to this point I cannot deduce the relationship of $a$ and $d$. I wrote a program to find the number but I want an intuitive approach to teach this to students.
You already have $$1499a+140b=998.5d+85c$$ Multiplying the both sides by $2$ gives $$2\times 1499a+2^3\times 5\times 7b=1997d+2\times 5\times 17c$$
Setting $d=2d'$ where $d'=1,2,3,4$ and dividing the both sides by $2$ give
$$1499a+2^2\times 5\times 7b=1997\times d'+5\times 17c\tag1$$
Note here that we have to have $a\le 6$ since $7000\times 1.5=10500\gt 9999$. (Fimpellizieri has already pointed that out in a comment.)
So, in mod $5$, $$a\equiv 3d'\implies (d',a)=(1,3),(2,1),(2,6),(3,4),(4,2)$$
For $(d',a)=(1,3)$, $(1)\iff 28b=17c-500\lt 0$
For $(d',a)=(2,1)$, $(1)\iff 17c=28b-499\lt 0$
For $(d',a)=(2,6)$, $(1)\iff 28b=17c-1000\lt 0$
For $(d',a)=(3,4)$, $(1)\iff 28b=17c-1$. Since $c\equiv 1\pmod 4$, we have $c=1,5,9$ giving $b=\frac{4}{7},3,\frac{38}{7}$ respectively.
For $(d',a)=(4,2)$. $(1)\iff 17c=28b-998\lt 0$
Hence, $\color{red}{(a,b,c,d)=(4,3,5,6)}$ is the only solution.