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If a googol (1 followed by 100 zeros) was written in binary notation, how many digits would it have?

Question #39902. Asked by jigaboo_sue.

fosse4
Answer has 5 votes
fosse4

Answer has 5 votes.
333 bits )binary digits

Oct 15 2003, 12:20 PM
sequoianoir
Answer has 7 votes
Currently Best Answer
sequoianoir
20 year member
2091 replies

Answer has 7 votes.

Currently voted the best answer.
1249AD2594C37D0000000000000000000000000000000000000000000000000000000000000000000000

In hexadecimal to be precise!

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

Jun 21 2007, 8:52 AM
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