Easy kata, ありがとう。

Ranks can't be changed ~~

this needs to be <8kyu>

1. You have a number n(let's say 10) and you need to square all numbers from 0(including) to 10(including).
   0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 --> 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

2. Let's say that we want the digit '1'. In squared array we need to sum up all digits that are '1'. (Keep in mind that if we have, for example, squared number 121, that has two digits '1', then we are adding both of them to the final sum/counter).
   0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 -->
   0, 1 + 0 + 0 + 1 + 0 + 0 + 0 + 0 + 1 + 1 = 4

I hope this helps. :)

1, 4, 9, 10, 11, 12, 13, 14, 19, 21 which squared are 1, 16, 81, 100, 121, 144, 169, 196, 361, 441

Those are the k which squared have at least a single digit 1.

Sorry, but i can't understand what you're trying to ask for.

Are you asking for counting how many "similar digits" are in each square?

i don't understand how we go from "1, 4, 9" for squaring each iteration
to "10, 11, 12, 13, 14, 19, 21" can someone explain this?