# Thread: To find the number in series?

1. ## To find the number in series?

Q) Find the 28383 term of the series

12345678910111213..........

Any help would be appreciated?

Thanks,

2. ## Series

Hello a69356
Originally Posted by a69356
Q) Find the 28383 term of the series

12345678910111213..........

Any help would be appreciated?

Thanks,
I assume that the series is

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2, 1, 3, ...

where the natural numbers are written as separate digits.

In which case, let's do a little counting:

The first 9 terms represent the natural numbers 1...9

The next 180 terms represent the 90 natural numbers with two digits in their representation: 10...99

So, if we add up so far, the (9+180)th term is 9, representing the units digit of the number 99. That's the 189th term.

There are then 900 3-digit numbers from 100...999. That will take up the next 2700 terms of the series. So the 2889th term is 9, the units digit of 999.

Next, there are 9000 4-digit numbers from 1000...9999. That's 36000 single-digit terms. So the 38889th term is 9, the units digit of 9999. The 28383rd term is somewhere in this block.

Now 28383-2889 = 25494. So we want the 25494th digit in the block representing the numbers 1000...9999. Let's find out whether it's a units, tens, hundreds or thousands digit.

The thousands digits are represented by the terms whose positions are 1, 5, 9, 13, ... in this block; i.e. the terms whose positions leave a remainder 1 when divided by 4

The positions of the hundreds digits leave remainder 2 when divided by 4, and so on.

Now 25494 = 6373 x 4 + 2. So the 25494th term represents a hundreds digit.

The first 4000 terms represent the numbers from 1000...1999; the next 4000 the numbers from 2000...2999 and so on.

6 x 4000 = 24000 and 25494 - 24000 = 1494

So we want the hundreds digit of the number represented by the 1494th term in the block 7000...7999.

The 100 numbers 7000...7099 (i.e. those with 0 as their hundreds digit) will take 4 x 100 = 400 terms. Then the next 100 numbers, 7100...7199, another 400, and so on. 3 x 400 = 1200 and 4 x 400 = 1600, so we're in the 4th block of 100; i.e. the numbers from 7300...7399, with a hundreds digit of 3.

So I reckon (if I haven't counted wrong at any point) that the 28383rd term in the sequence is 3.

3. Thanks a lot for explaining it so thoroughly