For some of the problems (where the numbers are fairly large), you should have access, in another browser window, to a modular arithmetic calculator (google it). If the numbers aren't too large, you should be able to do the work without a calculator.

Enter an integer in the field below, then click the "Submit" button. Do not place commas in the number.

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Summary of RSA cryptography:

We have two large primes, *p* and *q* (hundreds of digits in length, unlike the small primes used in this demo).

These two primes are kept secret.
Our public modulus is *m* = *p**q*.

Factoring *m* to find *p* and *q* is computationally infeasible, because *p*, *q* and *m* are so large.

Our private modulus is *n* = (*p* − 1)(*q* − 1).

Our encryption key *e* is any integer relatively prime to *e* mod *n*.
Our decryption key *d* is the multiplicative inverse of *e* mod *n*.
Our message is an integer *M* < *m*.

Our encrypted message is *C*, where *C* = *M*^{e} mod *m*.

Although any spy can see our encrypted message *C* = *M*^{e} mod *m*, and knows the values of *e* and *m*, solving the discrete log equation *C* = *M*^{e} mod *m* to find *M* is not computationally feasible because the numbers are too large.

To decrypt the encrypted message, we raise it to the *d* power and reduce mod *m*.

Decrypted message: *C*^{d} mod *m* = (*M*^{e})^{d} mod *m* = *M*.

The public modulus *m*, the encrypted message *C* = *M*^{e} mod *m*, and the encryption key *e* are all public information.

The encrypted message cannot be deciphered without the decryption key *d*, which is secret.

In order to find the decryption key *d*, a spy would first need to find the private modulus *n*.

In order to find *n*, a spy would need to know our primes *p* and *q*, but that would require factoring *m*, which is beyond the limits of current computing technology.

The math that makes this all possible is this variation of Fermat's Little Theorem:

Let *p*, *q* be primes, let *m* = *pq*, and let *n* = (*p* − 1)(*q* − 1).

If *e* is relatively prime to *n* and *d* is the inverse of *e* mod *n*, then

(*a*^{e})^{d} mod *m* = *a* for any positive integer *a* < *m*.

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*Copyright 2015, James Wooland.
All rights reserved.
Not to be distributed for commercial purposes.*