an interesting math problem that has to do with flipping number I saw this question years ago and tried to solve it but I couldn't solve it , so maybe someone can help me.

a and b are 4 digits numbers such that:
  a = flipped(b)

For example:
a=1234 b=4321
a=4512 b=2154
Find all a's and b's where:
a=3b+4 or a=4b+3 ( b=3a+4 or b=4a+3 )

For $a=4b+3$, note that the first digit of $a$ must be at least $4$ and the first digit of $b$ cannot be above $2$ or the multiply will carry.  Since $a$ is odd the first digit of $b$ must be $1$.  Then the last digit of $a$ is $1$ and the last digit of $4b$ is $8$, meaning the last digit of $b$ is $2$ or $7$.  Only $7$ is greater than $4$.  This means that $4b$ has to carry $3$ into the thousands place and the second digit of $b$ is at least $7$.  This gives $$\begin {align} 1cd7&\\ \underline {\times \quad 4}&\\
7\_\_8& \\ \underline {+ \quad 3}&\\7dc1\end {align}$$
with $c=7,8,9$. The tens digit of the product must be even to be a multiple of $4$, which means $c$ is odd.  If $c=7, d=1,6$ to make the tens digit of the product $6$ with the carry of $2$ coming in.  Neither works.  If $c=9, d=4,9$ and neither works again.  We conclude there is no solution with $a=4b+3$.  You can take a similar approach with $a=3b+4$ but it looks like a little more work.
