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# On creative mathematics and modelling

I found this story on the web.

If:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Is equal to;
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Then
H+A+R+D+W+O+R+K ;
8+1+18+4+23+15+18+11=98%

K+N+O+W+L+E+D+G+E ;
11+14+15+23+12+5+4+7+5=96%

L+O+V+E;
12+15+22+5 = 54%

L+U+C+K ;
12+21+3+11 = 47%

None of them makes 100%.
Then what makes 100% ???
Is it Money?
NO !!! M+O+N+E+Y= 13+15+14+5+25=72%

NO !!! L+E+A+D+E+R+S+H+I+P= 12+5+1+4+5+18+19+8+9+16=97%

Every problem has a solution,
only if we perhaps change our “ATTITUDE”.

A+T+T+I+T+U+D+E ;
1+20+20+9+20+21+4+5 = 100%

It is therefore OUR ATTITUDE towards Life and Work that makes
OUR Life 100% Successful..

Amazing mathematics of life!

Mathematics have an amazing power in convincing because of their internal consistency. Nevertheless, we should always ask whether we use the right model. In this case, the model is arbitrary and serves only a specific narrative.

Here is a list of other words that according to this rule are 100%

Doing some maths is easy (you can do impressively complex things mechanically — see modern computer software). Doing the right maths, that is finding the right representation of your ideas in abstract level, is the hard work.

For those who are interested, the English word list can be found on python.org.gr while the code is:

```import sys

INPUTDICT = 'DictionaryE.txt'

keys = [chr(x) for x in range(ord('a'), ord('z') + 1)]
values = list(range(1, 27))
MAPPING = dict(zip(keys, values))

def word_value(word, mapping):
""" Associate a word to a value according to its letters.
>>> word_value("LOVE", MAPPING)
54
>>> word_value("KNOWLEDGE", MAPPING)
96
"""
total_value = 0
for c in word.lower():
try:
total_value += mapping[c]
except KeyError:
pass

try:
with open(inputfile, 'r') as f1:
for line in f1:
yield line.lower().strip()
except IOError:
print("Dictionary is missing", file=sys.stderr)

def word_score_association():
d = dict()
d[word] = word_value(word, MAPPING)

return d

def filter_words(dictionary, value):
return (k for k, v in dictionary.items() if v == value)

def main():
#print(word_score_association())
import doctest
doctest.testmod()
print(sorted(list(filter_words(word_score_association(), 100))))

if __name__ == '__main__':
main()
```