Python Code to solve this challenge:```import osimport collectionsimport itertools
def kasiski(text, offset):    c = 0    for i in range(0, len(text)):
        # If we find somewhere else the same character, increment counter        if text[i] == text[(i + offset) % len(text)]:            # Hint!            c += 1    return c
# Encrypted texthex = "\x0d\x13\x45\x17\x48\x09\x5e\x4b\x17\x3c\x1a\x1f\x2b\x1b\x7a\x0c\x1f\x66\x0b\x1a\x3e\x51\x0b\x41\x11\x58\x17\x4d\x55\x16\x42\x01\x52\x4b\x0f\x5a\x07\x00\x00\x07\x06\x40\x4d\x07\x5a\x07\x14\x19\x0b\x07\x5a\x4d\x03\x47\x01\x13\x43\x0b\x06\x50\x06\x13\x7a\x02\x5d\x4f\x5d\x18\x09\x41\x42\x15\x59\x48\x4d\x4f\x59\x1d\x43\x10\x15\x00\x1a\x0e\x17\x05\x51\x0d\x1f\x1b\x08\x1a\x0e\x03\x1c\x5d\x0c\x05\x15\x59\x55\x09\x0d\x0b\x41\x0e\x0e\x5b\x10\x5b\x01\x0d\x0b\x55\x17\x02\x5a\x0a\x5b\x05\x10\x0d\x52\x43\x40\x15\x46\x4a\x1d\x5f\x4a\x14\x48\x4b\x40\x5f\x55\x10\x42\x15\x14\x06\x07\x46\x01\x55\x16\x42\x48\x10\x4b\x49\x16\x07\x07\x08\x11\x18\x5b\x0d\x18\x50\x46\x5c\x43\x0a\x1c\x59\x0f\x43\x17\x58\x11\x04\x14\x48\x57\x0f\x0a\x46\x17\x48\x4a\x07\x1a\x46\x0c\x19\x12\x5a\x22\x1f\x0d\x06\x53\x43\x1b\x54\x17\x06\x1a\x0d\x1a\x50\x43\x18\x5a\x16\x07\x14\x4c\x4a\x1d\x1e"
# Generate offsets: 1 - 20 offsets = range(1,20)
for o in offsets:     print("Offset: %d\tOccurences: %d" % (o, kasiski(hex, o)))
def countletters(myfile):    """ Returns a dictionary containing a occurence frequency of each found character"""    d = collections.defaultdict(int)    myfile = open(myfile)    for line in myfile:        line = line.rstrip('\n')        for c in line:            d[c] += 1    return d
def get_letters_count(myfile):    """ Gets amount of characters in myfile """    with open(myfile) as f:        c = f.read()        return len(c)
filename = 'jq.js'  #jQuery 3.1.0 filefreqs = countletters(filename)file_size = get_letters_count(filename)
percent_freqs = {}for k,v in freqs.iteritems():    # Save ASCII code of letter and its occurence frequency     percent_freqs[ord(k)] = "{0:.8f}".format(v/float(file_size))
# For all other unoccured letters, store occurence = 0for i in xrange(0, 256):    if not i in percent_freqs:        percent_freqs[i] = "{0:.8f}".format(0)
def chi_squared(ciphertext, freqs):    d = collections.Counter(ciphertext)    res = []
    for k,v in freqs.iteritems():        c = 0        decrypted = []
        for i in ciphertext:            # Do the XOR operation            decrypted.append(chr(k ^ ord(i)))
        for l in decrypted:            expected_count = float(len(ciphertext) * float(freqs[ord(l)]))            real_count = float(d[l])
            # Avoid division by 0            if expected_count > 0:                c+= (real_count - expected_count ** 2) / expected_count
        res.append(c)
    return res
def make_columns(text, key_length):    """ Returns columns of length = key_length for text"""    blocks = []
    # Divide ciphertext into blocks of length = key_length    for i in xrange(0, len(text)/key_length):        blocks.append(list(text[key_length*i:key_length*i+key_length]))
    # Create list: [[blocks[0][0], blocks[0][1], ...], [blocks[1][0], blocks[1][1], ...]]    columns = map(list,zip(*blocks))
    # What about remaining text that doesn't fit into one block?    if len(text) % key_length:        remaining = text[key_length*(len(text)/key_length)]        for i in xrange(len(remaining)):            columns[i].append(remaining[i])    return columns
columns = make_columns(hex, 9)key = ''
for i in columns:    res = chi_squared(i, percent_freqs)    key += chr(res.index(min(res)))
#print keykey = 'dumbh4ck5'
def xor(message,key):    return ''.join(chr(ord(c)^ord(k)) for c,k in itertools.izip(message, itertools.cycle(key)))
print(xor(hex, key))
```