An incompetent systems administrator accidentally configured the company's encryption system to use very small keys. This RSA key modulus is only 119 bits.

-----BEGIN RSA PUBLIC KEY-----
MBYCD3AY9xf8ZmUVDBSIVPZMSQIDAQAB
-----END RSA PUBLIC KEY-----

Recover the decryption exponent (in lowercase hex).

Test Vector

This key uses a 60 bit modulus. Deriving the private key should be almost instantaneous on a modern computer if an efficient algorithm is used.

-----BEGIN RSA PUBLIC KEY-----
MA8CCA3150PxP2ofAgMBAAE=
-----END RSA PUBLIC KEY-----

They key contains

modulus: 0xdf5e743f13f6a1f
public exponent: 0x10001

From this we can derive

p, q = 0x7d453db5, 0x1c878d03

and therefore

d = 0xc8a01de0a7d799

so the solution is c8a01de0a7d799

Solution