Crypto 2

Computer Science Level pending

Answer is: 010100010111000101001110010100011001 010100001000001111010101010011000001001110000100010011, 000110001111010101010010 001000010101001110000100010010000101000100 000001001110000100 010100001000001001010010010100011001.

I think its a bit too hard to be placed as second...

Hint: Computers

~thefourseasons


The answer is 20430.

Aloysius Ng by Aloysius Ng , 20, Singapore

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2 solutions

Laurent Shorts
Apr 28, 2016 
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t = []
offset = ord('A') - 1
for s in  ['010100010111000101001110010100011001','010100001000001111010101010011000001001110000100010011','000110001111010101010010','001000010101001110000100010010000101000100','000001001110000100','010100001000001001010010010100011001']:
        for i in xrange(0, len(s), 6):
                t += [ chr( int(s[i:i+6], 2) + offset ) ]
        t += [ ' ' ]
print "".join(t)

T W E N T Y T H O U S A N D S F O U R H U N D R E D A N D T H I R T Y \boxed{\mathrm{TWENTY\ THOUSANDS\ FOUR\ HUNDRED\ AND\ THIRTY}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Aloysius Ng
Oct 7, 2014

Change the numbers from binary to base-10 and into letters. They are all in bunches of 6 digits.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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