Need help getting started - whole numbers using only 0 and 1?

Question

Just a disclaimer; this is for homework and I don't want an answer. I'm just not sure how to start out on this problem...help! How many whole numbers are there that have between 1 and at 10 digits and that can be written using only the digits 0 and 1?
I'm assuming I need to use multiplication to find all the possibilities, but I'm not sure where to begin. Any help is appreciate...thanks!

Answer 1

I'm going to start by assuming that numbers starting with 0, other than zero itself, don't count. Normally, since there are two choices for any individual digit, the number of numbers possible for any number of digits (d) is 2^d.

However, since 0 can't be the starting digit for any but the one-digit numbers, your choices are cut in half, which would be 2^(d-1).

Thus, your answer would be the sum of:
2 + 2^1 + 2^2 +...+ 2^9

Interestingly enough, you'll notice that this answer ends up being 2^10.

Answer 2

METHOD 1:
Count the number of ways to get 1 digit numbers, 2 digit numbers, 3 digit numbers, etc. See a pattern and calculate.

For example:
1 digit numbers (0, 1) --> 2 ways
2 digit numbers (10, 11) --> 2 ways
3 digit numbers (100, 101, 110, 111) --> 4 ways
4 digit numbers (1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111) --> 8 ways.

Because zero is a whole number and it is the only number that is allowed to have a leading zero, it causes an exception in the pattern. Otherwise the pattern would be 1 + 2 + 4 + 8 + 16 + ... + 256 + 512. This comes out to 1023, but if you add zero, you get 1024.

METHOD 2:
Let's use that same list of numbers, but put in the leading zeroes.
0000000000
0000000001
0000000010
0000000011
0000000100
0000000101
0000000110
0000000111
...
1111111011
1111111100
1111111101
1111111110
1111111111

The first digit can be zero or one (2 choices).
The second digit can be zero or one (2 choices)
The third digit can be zero or one (2 choices)
...
etc.

To figure out how many numbers you can form this way, just multiply it out. The answer is 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2, which can also be written as 2^10 (2 to the 10th power).

You can calculate this by hand or plug it into your calculator, but the answer is:
2^10 = 1024.

So you can use either method to get to your answer.

(BTW, this all assumes you are using the definition of "whole number" meaning the non-negative integers {0, 1, 2, 3, etc.}. Some textbooks use whole numbers as a synonym for natural numbers {1, 2, 3, etc.} and some other sources use whole number as a synonym for integers {..., -2, -1, 0, 1, 2, ...}. Obviously if you exclude 0, you get an answer of 1023.)

Answer 3

If you are using only 0 and 1, you are talking about the binary number system. If you were restricted to 2 digits

0 = 00
1 = 01
2 = 10
3 = 11

If zero is considered a whole number, there are 4 numbers, else 3 and 2^2 - 1 is the answer

With ten digits, 2^10 - 1???

Answer 4

Simplify the problem... Look for all 10-digit numbers having only 0's and 1's, with leading zeros allowed (so you would write 1000 as 0000001000).

You have 10 positions and two possibilities at each position. Methinks it would be the same number of possibilities as you would have with 10-digit binary numbers.

Answer 5

a million- For the first digit, it will be purely a million(otherwise that is going to likely be 9 digits, you won't be able to commence with 0) and for the left we may be able to apply 2^9 diverse combos. 2-2^10 for each digit, there are 2 opportunities and there are ten digits. 3-2^0+2^a million+2^2+2^3+...+2^9. We repeat what we did in decision one for 9-digit, 8-digit, 7-digit,... , a million-digit numbers. i desire it facilitates. in case you do not realize, be at liberty to ask. :)

Answer 6

Hi. The series goes: 0,1,10,11,100,101,110,111,1000... Notice a pattern?

Answer 7

yeahhh, it looks like you need to find out what 1111111111 converts to from binary to decimal (i think the last answer guy said that). here's a link to check your work.

http://acc6.its.brooklyn.cuny.edu/~gurwitz/core5/nav2tool.html