# I have a RSA public key exponent and modulus. How can I encrypt a string using Python?

With PyCrypto, you can use the Crypto.PublicKey.RSA.construct() function. You'll need to convert the modulus to an `int`

. Here's an example (assuming big-endian):

```
from Crypto.PublicKey.RSA import construct
e = int('10001', 16)
n = int('d0eeaf...0b6602', 16) #snipped for brevity
pubkey = construct((n, e))
```

Then you can do the usual things (like encrypt) with the key:

```
pubkey.encrypt(b'abcde', None)
```

*Edit: Note that your public exponent, 10001, is mostly likely hexadecimal. This would correspond to the common public exponent 65537. I've updated the above to reflect that.*

I tried an alternative way using `Crypto.Cipher.PKCS1_OAEP`

motivated by: https://cryptobook.nakov.com/asymmetric-key-ciphers/rsa-encrypt-decrypt-examples and it just worked.

PS: There seems to be something wrong with modulus given, as modulus n must be the product of two large primes, thus should not be an even number. A tiny modification of n has been applied to make the example code runnable.

```
from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_OAEP
import binascii
e = int('10001', 16)
n = int('d0eeaf178015d0418170055351711be1e4ed1dbab956603ac04a6e7a0dca1179cf33f90294782e9db4dc24a2b1d1f2717c357f32373fb3d9fd7dce91c40b6601', 16)
# Construct a `RSAobj` with only ( n, e ), thus with only PublicKey
rsaKey = RSA.construct( ( n, e ) )
pubKey = rsaKey.publickey()
print(f"Public key: (n={hex(pubKey.n)}, e={hex(pubKey.e)})")
# Export if needed
pubKeyPEM = rsaKey.exportKey()
print(pubKeyPEM.decode('ascii'))
# Encrypt message using RSA-OAEP scheme
msg = b'Hello, world.'
encryptor = PKCS1_OAEP.new(pubKey)
encrypted = encryptor.encrypt(msg)
print("Encrypted:", binascii.hexlify(encrypted))
```