Problem 42. Find the alphabetic word product
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Problem Comments
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9 Comments
For the bonus question (on OSX):
words= textread('/usr/share/dict/words','%s');
for k=1:numel(words)
wordprod(k) = word_product(words{k});
end
[~,i] = min(abs(wordprod-1000000));
words{i}
wordprod(i)
% curing with word product of 1000188
For the bonus question, we can rule out any number that has a prime factor greater than 26 (z). The product found by Jason Friedman is the first number greater than 10^6 that does not have such a prime factor. The product 999856 is the largest number less than 10^6 to be part of a possible solution.
I like playing with strings and ASCII table ;)
Everett appears to miss the fact that 1000000 has no prime factors above 26, and Jason's dictionary doesn't seem to include obsolete words. 'tetty' scores a perfect 1000000.
Well, if you happen to know someone named 'BettyJ', thus a first name of Betty, last initial J, this works: word_product('BettyJ').
Could you send me the answer?
I had benn worked for 1 week but Icouldnt find answer. İf anyone solves Please send answer. Really I need it.
Nice problem
Can you give me a hint on how to program the bonus question? I used factor but that is not a good way.
If we can include multiple words in our bonus answer, the word "tea" 3 times scores perfect. n = 'teateatea'
Solution Comments
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1 Comment
easy ;)
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2 Comments
'curing' returns 1000188.
Closest 10 words are 'curing', 'Nicaragua', 'comers', 'Moors', 'rooms', 'banquet', 'Arkansas', 'sulks', 'alkaline' and 'Canaveral'.
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1 Comment
Words that make 1000,000, tetty and typey apparently, :)
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1 Comment
Very clever way of converting to a number
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1 Comment
closest I can get is 'curing', 188 away.
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1 Comment
'god' equals 420.
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1 Comment
large but working ;-)
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2 Comments
'j' corresponds to 10 according to the code above so, the word that gives exaclty 1E+6 is 'jjjjjj'
% word_product('jjjjjj')=1E+6
I don't think that 'jjjjjj' is really a word...
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