diff --git a/capture_the_flag/challenges/heavy_computation/encrypt.py b/capture_the_flag/challenges/heavy_computation/encrypt.py
index e78d6a6..640f50b 100644
--- a/capture_the_flag/challenges/heavy_computation/encrypt.py
+++ b/capture_the_flag/challenges/heavy_computation/encrypt.py
@@ -1,9 +1,3 @@
-Heavy computation(400)
-Question:
-A friend of mine handed me this script and challenged me to recover the flag. However, I started running it on my school cluster and everything is burning now... Help me please!
-
-Given files:
-encrypt.py
 from Crypto.Util.number import bytes_to_long, long_to_bytes
 from Crypto.Cipher import AES
 from Crypto.Util.Padding import pad
@@ -49,76 +43,3 @@ enc = cipher.encrypt(pad(flag,16))
 
 with open("flag.enc","wb") as output_file:
     output_file.write(enc)
-flag.enc(dump)
-    0000  f4 d8 e5 a2 ac 80 6c e9  dc c1 ef 1e d5 c4 51 7c   ......l.......Q|
-    0010  e3 d8 84 1a d7 c0 77 c9  9c b0 f6 f0 ab 13 63 b0   ......w.......c.
-    0020  f9 5e 8d cd 87 ce c7 d3  88 7a 4a 68 de a9 6f 96   .^.......zJh..o.
-    0030  77 cf 1e a7 95 a0 f8 1c  be 3a 66 f0 aa 73 2c 3e   w........:f..s,>
-Solution:
-Outline:
-(1) For preparartion, we calcurate Euler's totient of N by FactorDb( http://www.factordb.com/ ):
-
-N=5*23*61*701*3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847
-
-(2) First part of the functuon "derive_key":
-
-for i in range(NB_ITERATIONS):
-    start = start ** e
-    start %= N
-It is easily simplificated by Euler's theorem.
-
-start = pow(bytes_to_long(password),pow(e,NB_ITERATIONS,phi_N), N)
-where phi_N is the totient of N,
-
-phi_N=(5-1)* (23-1)* (61-1)* (701-1)* (3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847-1)
-
-(3) Second Part of the functuon "derive_key":
-
-    key = 1
-    for i in range(NB_ITERATIONS):
-        key = key ** e
-        key %= N
-        key *= start
-        key %= N
-We can replace it with the following (using "invert" function of gmpy2).
-
-    inv_e = gmpy2.invert(e-1,phi_N)
-    key = pow(start, (pow(e,NB_ITERATIONS,phi_N)-1) * inv_e, N)
-(4) Finally, We brute force the password and get the flag!
-
-Solver:
-solve.py
-from Crypto.Util.number import bytes_to_long, long_to_bytes
-from Crypto.Cipher import AES
-from Crypto.Util.Padding import pad
-from hashlib import sha256
-import gmpy2
-
-NB_ITERATIONS = 10871177237854734092489348927
-e = 65538
-N = 14968794114523720195251887716913440457986979987674770429103169854116498198112478103466085455257317270930523061714030307370028304505577267672733143013124254253285088080041831478700041394909740024011681885623055622400205
-
-#Using FactorDB, we have N=5*23*61*701*3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847
-phi_N = (5-1)*(23-1)*(61-1)*(701-1)*(3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847-1)
-
-def derive_key(password):
-    start = bytes_to_long(password)
-    start = pow(start,pow(e,NB_ITERATIONS,phi_N), N)
-    inv_e = gmpy2.invert(e-1,phi_N)
-    key = pow(start, (pow(e,NB_ITERATIONS,phi_N)-1) * inv_e, N)
-    return sha256(long_to_bytes(key)).digest()
-
-with open('flag.enc','rb') as f:
-    flag_enc = f.read()
-    for i in range(0x20, 0x100):
-        for j in range(0x20, 0x100):
-            password = long_to_bytes(i) + long_to_bytes(j)
-            key = derive_key(password)
-            IV = b"random_and_safe!"
-            cipher = AES.new(key, AES.MODE_CBC,IV)
-            flag = cipher.decrypt(flag_enc)
-            if(flag[0:6] == b'shkCTF'):
-                 print(flag.decode('utf-8')[0:flag.decode('utf-8').find('}')+1])
-                 break
-Flag:
-shkCTF{M4ths_0v3r_p4t13Nce_b4453d1f9f5386a1846e57a3ec95678f}