Trollathon #2.4 : The Encrypted Message #2
00111100000010000100011101111010000100100011110 11001111110011101000000101100110111010000
Refer to the 4th element.
The answer is 56.
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1 solution
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Now, the given e and N and the long binary sequence suggests the problem is encrypted with the RSA encryption. Now, after decrypting (using the link above), we find that the message is actually "A123456", which looks like an oeis entry. (especially when the problem states refer to the 4th element)
So, we find the 4th element of A123456 in oeis, which is 5 6 .
For those who don't know oeis, here's the link.