Find digits $x$, $y$, $z$, and $p$ such that $(xyzp)_{10}=4\cdot (pzyx)_{10}$

Question

Given that equation $xyzp=4\cdot pzyx$ is valid, how do I solve showing my work clearly? By trial and error the valid answers are $x=8,y=7,z=1,p=2$. These are the only conditions provided.

Answer 1

Here's a start (assuming I have the right interpretation of the problem in the comments).

1. $p$ must be even (why?)

2. $p\lt3$ (why?)

3. So $x$ is 8 or 9 (why?)

4. But $x$ can't be 9 (why?)

Answer 2

p z y x

× 4

x y z p x × 4 = p in unit place. similarly p × 4 = x

4 digit number × 4 gives 4 digit answer, p < 3 But p = 4 × x is even so p = 2 and x = 8 so it becomes 2 z y 8

× 4

8 y z 2 it shows that there is no carry forward in z × 4 so z must be 1 2 1 y 8

× 4

8 y 1 2 so you can calsukate y 8 × 4 = 32 ----- 3 c/f. y × 4 + 3 = 1 in unit place so y = 7

x = 8,y = 7,z = 1,p = 2