let

a3 = bet the thousand's digit

a2 = bet the hundred's digit

a1 = bet the ten's digit

a0 = bet the one's digit

thus the original number can be written as a3*10^3 + a2*10^2 + a1*10 + a0

the hundred's digit is one more than four times the ten's digit

a2 = 1 + 4a1

the ten's digit is half of the thousand's digit

a1 = a3/2

and the one's digit is two less than the thousand's digit

a0 = a3 - 2

if the thousand's and hundred's digits are interchanged, the resulting number is 2700 more than the original

a2*10^3 + a3*10^2 + a1*10 + a0 = 2700 + a3*10^3 + a2*10^2 + a1*10 + a0

a2*10^3 + a3*10^2 = 2700 + a3*10^3 + a2*10^2

900(a2 - a3) = 2700

a2 - a3 = 3

by solving the system of equations

a2=1+4*a1

a1=a3/2

a0=a3-2

a2-a3=3

we have

a2 = 1 + 4a1

a1 = a3/2 => a3 = 2a1

a0 = a3 - 2 => a0 = 2a1 - 2

a2 - a3 = 3 => 1 + 4a1 - 2a1 = 3 => a1 = 1

a1 = 1

a2 = 1 +4a1 = 1 + 4 = 5

a2 = 5

a3 = 2a1 = 2

a3 = 2

a0 = 2a1 - 2 = 0

a0 = 0

2*10^2 + 5*10^2 + 1*10 + 0 = 2510

indeed 5210 = 2510 + 2700

the original number is 2510.