Practice question for interview - digits

There are 9 one-digit numbers and 90 two-digit numbers.
The rest are three-digit numbers.

This is more difficult. Not a visualization.

I get 1,000,051 employees.

1 to 9
20 to 29
30 to 39
40 to 49
50 to 59
60 to 69
70 to 79
80 to 89
90 to 99 is 189 digits

Then 100 to 999 is 270 digits
1000 to 9999 is 360 digits
10000 to 99999 is 450 digits
100,000 to 999,999 is 540 digits
1,000,000 to 1,000,029 is 210 digits

189+270+360+450+540+210 = 2019

1,000,030 to 1,000,051 = 21

Total is 2040.  

Ends in 1 because no employee number 0

This one is tricky and you need to write down the groups to see.

thanks, let me see.

I made an error in the last row.

Up to 1,000,029 is 2019 digits.

then 1,000,030 is 7, 1,000,031 is 7 1,000,032 is 7 and 2019 +21 = 2040

So the number is 1,000,032 (not 051).

Total is 1,000,032

Thanks, John. I'll check it. Had to call it quits for the day :) thanks for all the help today.

If there are about 190 digits for the first 100, then the first 1000 should have a lot more than 270 digits, I would think.

100-109 --> 30 digits
110-119 --> 30 digits
...
190-199 --> 30 digits
______________________
100-199 -->10 * 30 = 300 digits
200-299 = 300 digits
900-999 = 300 digits
_________________________
100-999 = 9 * 300 = 2700 digits

since only 2040 digits are needed, and if there were 1000 employes, we would need over 2700 digits, then N < 1000.

Hope you can take it from here.

According to the example provided, digits are not unique and re-used, so that changes how we calculate.

Yes, it's a bit like buying numbers to stick on the front of their lockers.
It's a shame they didn't use leading "0"s :)

The answers weren't right. But it's ok. If I get this on the test, I'll skip it.

>> Between 100 and 999 each employee number uses three digits at most that's 899 x 3 digits which = 2697
Off by one: If there are 999 employees, then (999 - 100 + 1) = 900 employees will have 3 digit numbers, in which case there are 900 x 3 digits = 2700 digits.

From this we conclude that 100 < N < 999.

>> The answers weren't right. But it's ok.

It looks to me like d-glitch is giving you the correct answer.
What answer are expecting?

>> If I get this on the test, I'll skip it.
If you do not have time to answer all the questions, then skip the hardest ones, I guess. But, the point of this partitioning exercise is to see how well you think. I advise to show the partitioning that you learned here and come up with an answer even if it turns out to be wrong.

>> It's a shame they didn't use leading "0"s :)
In a real company, they probably would do that, or just start the numbering scheme with 10000 (reserved for the CEO). :)

I put the answer in the test and I get incorrect. Let me try again.

Please give us the link for this online test whether it be practice or for credit.
BTW, this problem is quite a bit harder than your previous problem. As you can see, this problem even got some experts close but not quite on target. One of the pitfalls is to realize that when counting from 0 or from 100, you have to take special care because we are used to counting from 1.

You can solve this by brute force in two columns of an Excel spreadsheet:
  Put a 1 in Col 1, Rows 1-9, and a 2 in Col 1, Rows 10-99, and a 3 in Col 1, Rows 100-999.  
  Then do a running total in Col 2 and look for 2040.

This takes less than 2 minutes and confirms my answer.

You can also solve this programmatically with for-loops or recursion.

yes, 716 is correct.  I just double checked.

Interesting because the practice hint says count ALL digits even if re-used.

Helped me solve this one
How many digits are needed to assign numbers to 283 employees using the numbers 1 through 283?

741.

That is correct.

Yeah, this thread and your explanations helped. Thanks