Consider the sequence formed by the digits of the concatenation of the positive integers, listed in order: 123456789101112131415161718192021... What is the 2011th digit of this sequence?
@Agent47
what's concatenation? lol
THis is even more lol... No idea. But from a brief google search, concatenation: the joining of two numbers by their numerals. Like in this one, the numbers are going in order.
Concatenation is basically just putting the two things next to each other. So 25 concatenated with 347 is merely 25347.
Ohhhh
Do you know what the solution is? I would like to check my answer.
For reference, I believe it should be 3.
The solution is 7
Close...I was one off. (with how I was counting) Let me see where I overcounted.
The first 7 of 707
mann KG's a genius.
Ah right, I was doing things a bit wrong. It should be 7. Specifically the first 7 of 707. Give me a minute, and I shall explain.
Wait, I think I got it. Can I post my answer and see if it's right?
sure.
Well, there are 9 single digit numbers. Then, there are 90 two digit numbers. That gives you 189 digits for 1-99. I have to get to the 2011 digit, so 2001 - 189 = 1822 and the remaining possible digits up to 2011 is 3 digit numbers. 1822/3 = 607 with a remainder of 1, so 607 + 99 = 706 which when adding 1 more digit is 7 of 707 because of the remainder. Does that make sense?
I meant 2011 near the beginning 2011 - 189
That's exactly what I was doing. Here's the explanation in my words. First, note that you have 9 single digit numbers. Then, you have concatenated 90 two digit numbers. Thus, you have a 9+2(90)=189 digits so far. We must count exactly \(2011-189=1822\) more digits. Since all the next numbers will be 3 digit numbers, we divide by 3. \[1822/3=607.3333333...\]Now, here's the tricky part. The first of these is 100, so we add 607 to 100 to get 707. However, we must also subtract 1 since we're starting at 100, and not 101. Hence, it is the digit immediately after we see \(706\). The next digit is the first in \(707\).
Alright. Thanks!
You're welcome.
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