Question

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.

Answer 1

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

Answer 2

actually I see a pattern
like look at the 2 digit numbers and so on...

Answer 3

I don't see one...

Answer 4

2^1=?
2^2=?
2^3=?
2^4=?

Answer 5

Ohhhh

Answer 6

Then we have to add them up until we reach 100?

Answer 7

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

Answer 8

like which of those sums gives us 100 or more?

Answer 9

Adding to it\(2^6\) gives 124

Answer 10

so it has 6 digits

Answer 11

it's the 40th 6-digit number in base 3

Answer 12

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

Answer 13

right

Answer 14

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

Answer 15

36 numbers is a lot to write*

Answer 16

@ganeshie8 fun question for you

Answer 17

I got 124 not 128

Answer 18

http://www.wolframalpha.com/input/?i=2%2B2%5E1%2B2%5E2%2B2%5E3%2B2%5E4%2B2%5E5%2B2%5E6

Answer 19

oh okay then

Answer 20

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
\[1100011_3=?_{10} \\ 1100011_3=1\cdot 3^6+1 \cdot 3^5 + 0 \cdot 3^4+ 0 \cdot 3^3+0 \cdot 3^2+1 \cdot 3^1+1 \cdot 3^0\]

Answer 21

you definitely should check this though
it is totally possible I made a mistake in my listing there

Answer 22

alright

Answer 23

I really hope there is a shorter way :p

Answer 24

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 \(1100100_3 = \boxed{981}\).

Answer 25

I was close
I had gotten 976 :(

Answer 26

It's okay :) Thanks so much for the help!

Answer 27

and wow that one way is totally easier

Answer 28

I know 0.0