Find the one hundredth positive integer that can be written using no digits other than digits 0 and 1 in base 3. Express your answer as a base 10 integer.

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Find the one hundredth positive integer that can be written using no digits other than digits 0 and 1 in base 3. Express your answer as a base 10 integer.

Mathematics
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hmm... so we have 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000 10001 10010 10011 10100 10101 10110 10111 11000 11001 11010 11011 11100 11101 11110 11111 so we have 2 1 digits numbers and we have 2 2 digit numbers and we have 4 3 digit numbers and we have 8 4 digit numbers and we have 16 5 digit numbers trying to find a pattern so I can find the 100th number that can be written in terms of 0s and 1s only you know without writing all of them down
actually I see a pattern like look at the 2 digit numbers and so on...
I don't see one...

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2^1=? 2^2=? 2^3=? 2^4=?
Ohhhh
Then we have to add them up until we reach 100?
we want the 100th number so we have so far \[2+2^1+2^2+2^3+2^4\] \[2+2^1+2^2+2^3+2^4+2^5\] \[2+2^1+2^2+2^3+2^4+2^5+2^6\] yep we want to see which of these gives us at least 100 as a sum we want the first one that is 100 or more this will tell us the number of digits we will need
like which of those sums gives us 100 or more?
Adding to it\(2^6\) gives 124
so it has 6 digits
it's the 40th 6-digit number in base 3
well one sec remember for 2 digit numbers we had 2^1 of those and for 3 digits number we had 2^2 of those and for 4 digits numbers we had 2^3 of those we have 2^6 so we have 7 digits
right
2+2+4+8+16+32+64=128 2^6=64 There are 64 seven digit numbers that can be made up of 0's and 1's so 64-28=36 so I think we want the 36th number in the seven digits omg this is kind of hard 36 digits is a lot to write 1000000 is the first of the 7 digit numbers 1000001 is the second ... there has to be a shorter way of thinking about this one
36 numbers is a lot to write*
@ganeshie8 fun question for you
I got 124 not 128
http://www.wolframalpha.com/input/?i=2%2B2%5E1%2B2%5E2%2B2%5E3%2B2%5E4%2B2%5E5%2B2%5E6
oh okay then
1000000 is the first of the 7 digit numbers 1000001 is the second 1000010 is 3rd 1000011 is 4th 1000100 is 5th 1000101 is 6th 1000110 is 7th 1000111 is 8th 1001000 is 9th 1001001 is 10th 1001010 is 11th 1001011 is 12th 1001100 is 13th 1001101 is 14th 1001110 is 15th 1001111 is 16th 1010000 is 17th 1010001 is 18th 1010010 is 19th 1010011 is 20th 1010100 is 21th 1010101 is 22nd 1010110 is 23rd 1010111 is 24th 1011000 is 25th 1011001 is 26th 1011010 is 27th 1011011 is 28th 1011100 is 29th 1011101 is 30th 1011110 is 31st 1011111 is 32nd 1100000 is 33rd 1100001 is 34th 1100010 is 35th 1100011 is 36th I probably made a mistake in this list by if I didn't 1100011 is the 100th number of 0's 1's and this number is in base 3 and we want it in base 10 \[1100011_3=?_{10} \\ 1100011_3=1\cdot 3^6+1 \cdot 3^5 + 0 \cdot 3^4+ 0 \cdot 3^3+0 \cdot 3^2+1 \cdot 3^1+1 \cdot 3^0\]
you definitely should check this though it is totally possible I made a mistake in my listing there
alright
I really hope there is a shorter way :p
The goal is to count in base 3 using only binary digits. The \(100^{\text{th}}\) smallest positive binary integer is \(100 = 1100100_2\), so the \(100^{\text{th}}\) smallest positive integer that can be written with only the binary digits is \(1100100_3 = \boxed{981}\).
I was close I had gotten 976 :(
It's okay :) Thanks so much for the help!
and wow that one way is totally easier
I know 0.0

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