A infinite sequence

Look at it this way: no matter how many 1s you have in a number of the sequence, it will always end in 11 starting with the second number in the sequence. This is the key to the problem.

Now, this means that the tens digit of 1 + 11 + 111 + ... will be the tens digit of 1 + 39 * 11 (since the tens digit cannot be affected by any other digits apart from units and tens). You have that "1" over there because, as I've pointed out, the first number in the series is 1 and not 11.

1 + 39 * 11 = 1 + 39 * 10 + 39 = 40 + 390 = 430 - tens digit 3.

why did you do 10 * 39?

I was trying to calculate 11*39 fast, by noticing that it's 39*(10 + 1) = 39*10 + 39*1 = 39*10 + 39.

you have 40 numbers that have a unit digit of 1 so 1*40 = 40 --> 4 in tens digit
you have 39 numbers that have a tens digit of 1 so 39*10 = 390 --> 9 in tens digit.

adding we get 13, so 3 is in tens digit answer is C

m&m wrote:you have 40 numbers that have a unit digit of 1 so 1*40 = 40 --> 4 in tens digit
you have 39 numbers that have a tens digit of 1 so 39*10 = 390 --> 9 in tens digit.

adding we get 13, so 3 is in tens digit answer is C

here is 3 an unit