84 digits (4 binary bits each = 336) except the 1st digit is a 1 so 3 binary bits "unused" .
In binary ?
10010010010011010110100100101100101001100001101111101 followed by 280 zeroes !!!
Oct 15 2003, 1:43 PM
dsjoerg
Answer has 7 votes
dsjoerg
Answer has 7 votes.
sequoianoir is mistaken. It has only 100 zeros at the end. Here's a proof:
10^100 = 2^100 * 5^100.
2^100 in binary = 1000...000 (100 zeros)
5^100 in binary ends with a 1. (because 5^n ends with a 5 for all n, therefore all 5^n are odd, therefore ends with a 1 in binary).
5^100 ends with a 1 in binary. Multiplying it by 2^100 is equivalent to sticking 100 zeros on the right-hand side of 5^100. So in binary, you get 5^100 followed by 100 zeros.
If you have the "bc" program (available in UNIX) you can see it for yourself: 100100100100110101101001001011001010011000011011111001110101100001011001001111000010011000100110011100000101111110011100010101100111001000000100011100010000100011010011111001010101010110010010000110000100010101000001011101000111100010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
