How many numbers containing the digit 1.

[sutupidmath]

How many numbers containing the digit 1.Help!

Hello,

The question i am dealing with is this:Considering the integers from 1 to 10,000,000,000, answer the following two questions:
A. Is the number of integers that include the digit 1 greater or less than the number of integers that do not include the digit 1?
B. What is the difference between number (quantity) of integers that include the digit 1 and the number of integers that do not include the digit 1?


a. There are 6513211999 that include 1 as a digit, and there are 3486788001 numbers that don’t contain digit of 1.

Therefore, there are more numbers that contain the digit 1 than not.

b.3026423998



Reasoning:



After some trials, i came up with the following recurrence relation:



$$a_1=19,.. a_2=100,.. a_{2n}=(10)^{n+1}.. and... a_{2n+1}=8*\sum _{i=1}^{2n-1} a_i$$





Now the total sum of the numbers that contain digits is calculated by summing up the following:





$$a_{14}+\sum_{n=1}^{6}[a_{2n}+a_{2n+1}]=<br /> <br /> =<br /> <br /> a_{14}+\sum_{n=1}^{6}[ a_{2n}+(8*\sum _{i=1}^{11} a_i) ]$$



I don't know whether there is supposed to be a more clever and including less calculations method for doing this.

FIrst, is the answer correct? ANd second if yes, does my reasoning make sense?

THnx in advance!

 

[CRGreathouse]

Let's look at the numbers from 0 to 9,999,999,999 instead. There are ten digit positions, each of which is filled with exactly one digit (including leading zeros). There are b^10 ways to fill them with b digits. When b=10, that says that there are 10^10 numbers in the range. When b = |{0, 2, 3, 4, 5, 6, 7, 8, 9}| = 9...

Hope you don't feel too for missing that.

 

[sutupidmath]

Let's look at the numbers from 0 to 9,999,999,999 instead. There are ten digit positions, each of which is filled with exactly one digit (including leading zeros). There are b^10 ways to fill them with b digits. When b=10, that says that there are 10^10 numbers in the range. When b = |{0, 2, 3, 4, 5, 6, 7, 8, 9}| = 9...

Hope you don't feel too for missing that.

Hi, i appreciate your response.

THe deal is that, i truly believe that my method is correct, since it is even compatible with the one you just described. But, i think i might have done some mistaces in calculations, while adding stuff.

FOr example let's look at numbers between 0-9999

so according to the method you just described, we would have b^4, since there are 4 digit positions to be filled with b digits. When b=10=> 10^4=10000, and 9^4=6561

So, the number of 1'nes will be 10000-6561=3439.

Which is exactly the same result that i get if i apply the method i 'invented'.

Let's look at it.

$$a_1=19,.. a_2=100,.. a_{2n}=(10)^{n+1}.. and... a_{2n+1}=8*\sum _{i=1}^{2n-1} a_i$$


we would have n=0,1,2.

$$a_1+a_2+a_3+a_4+a_5=19+100+8*19+1000+8*[a_1+a_2+a_3]=1271+8*271=3439$$

Which is the same as above.

So, i think i should have done some mistakes while adding stuff.

But the way you pointed out is much easier and way more clever as well.

 

[matt grime]

Um, we can't comment on your reasoning because you've given none. You have just written down some sequence a_i without justification, so it's impossible to say if the a_i were correct in spirit but incorrect in practice.

 

[CRGreathouse]

So, i think i should have done some mistakes while adding stuff.

That was my conclusion, since your number was very close to mine.

But the way you pointed out is much easier and way more clever as well.

It only seemed clever because you didn't know it took me > 10 minutes to come up with it. But it is easier.