# Homework Help: Logic problem

1. Sep 3, 2011

### Wildcat

1. The problem statement, all variables and given/known data
All integers beginning with 1 are written down in succession. What digit is in the 1955th place?

2. Relevant equations

3. The attempt at a solutionI'm pretty sure it is 4 Wondered if someone wants to double check

2. Sep 3, 2011

### PeterO

I would suspect something a bit different.

Don't forget that the 4 digit numbers begin with 1000, 1001, 1002, ...
The 1st end sin 0
The 2nd ends in 1
the 3rd ends in 2
etc

3. Sep 4, 2011

### Lobezno

Mind posting your workings out? How did you arrive at this answer?

4. Sep 4, 2011

### Wildcat

I don't think I forgot them. There are 321 digits thru 199, then I need to go 1634 more places. I still come up with 4.

5. Sep 4, 2011

### Wildcat

1 - 1 digit
10-19. - 20 digits. So. 21
100-199 -300 digits. So 321
To get to the 1955th digit I need to go 1634 more places so 1634/4 because each number in the thousands has 4 digits. 408.5. Go to 1407 over 2 more would be the 4 in 1408

6. Sep 5, 2011

### Lobezno

Ahhh that's no fun. I thought there'd be some hardcore maths involved!

7. Sep 5, 2011

### PeterO

I think I mis-counted the 3 digit numbers - can't find my notes now - but I think I had 600 digits rather than 300.

8. Sep 5, 2011

### Wildcat

No hardcore math in this one, sorry. Just don't want to make a silly mistake.

9. Sep 5, 2011

### Staff: Mentor

There are 9 single digit numbers
There are 90 double digit numbers
There are 900 triple digit numbers

The digit with "address" 1955 will lie within the range of triple digit numbers.

The starting address of the kth three digit number in the list will be

n = <address of first 3-digit number> + (k - 1)*3

n = 1*9 + 2*90 + 1 + (I - 100)*3 .

Now, you can either find the number I by trial and error, or do something clever with 1955 to find the appropriate starting address for the number in which the 1955th digit is embedded and solve for I directly. Hint: may require integer arithmetic (or truncation or floor or ceiling operations).

10. Sep 5, 2011

### Wildcat

The original problem states that only integers that begin with 1 are written in succession, this equation would not work for this problem.

11. Sep 5, 2011

### Staff: Mentor

Ah. My apologies. I interpreted the problem statement to mean that all integers, starting with the integer 1, are written down. That is: 1,2,3,4,....