Looking for help not really sure how to approach problems like these and solve.

You are given the public key 3811, 47. We will use " " (space) -->00, a->01, 6-> 02..... z-> 26. Code the message using this public key cryptography "make it stop" by combining characters two at a time: (To get you started: ma-> 1301 (m->13 and a->01) Determine what gets sent over the internet: the 6 four digit numbers (you might have leading Os) that will be sent. Find the decoding exponent and us it to decode this message: You do not need that c*d is 1 mod (p-1)- 1), you need c*d is 1 mod [Icm(p-1, 2-1)] (in this case, we know c, the coding exponent is 47. We need to find d. 3247, 2301, 1728, 2608 Hint: Using Excel, I had to break down the mod arithmetic even further. If you get a #num, it is because you lost too many digits... do the computation in more steps... A*B*C*D mod n= [(A*B mod n)*(C*D mod n) mod n] You are given the public key 3811, 47. We will use " " (space) -->00, a->01, 6-> 02..... z-> 26. Code the message using this public key cryptography "make it stop" by combining characters two at a time: (To get you started: ma-> 1301 (m->13 and a->01) Determine what gets sent over the internet: the 6 four digit numbers (you might have leading Os) that will be sent. Find the decoding exponent and us it to decode this message: You do not need that c*d is 1 mod (p-1)- 1), you need c*d is 1 mod [Icm(p-1, 2-1)] (in this case, we know c, the coding exponent is 47. We need to find d. 3247, 2301, 1728, 2608 Hint: Using Excel, I had to break down the mod arithmetic even further. If you get a #num, it is because you lost too many digits... do the computation in more steps... A*B*C*D mod n= [(A*B mod n)*(C*D mod n) mod n]