Put the natural numbers in a row from small to large, and form a series of numbers 12345 ~ 91011! 99100101 ~. What is the position of 0 on the bit of natural number 100?

Put the natural numbers in a row from small to large, and form a series of numbers 12345 ~ 91011! 99100101 ~. What is the position of 0 on the bit of natural number 100?

192 formula 9 + 20 * (x-1) = 189
It is required that 0 in 100 bits should be 189 + 3 = 192
Note: X means that there are 10 tens from 1 to 100, like 20 to 30
This problem only needs to divide a series of numbers into 1, 2, 3... 91, 92, 93
98 99 100
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There is a ladder classroom with 25 rows of seats. The height of the first row from the classroom floor is 17cm. The height difference between the front and back rows of the first 16 rows is 8cm. From the 17th row, the height difference between the front and back rows is 10cm (including the height difference between the 16th and 17th rows). Find the height of the last row from the classroom floor

The problem of finding the nth term of arithmetic sequence
It is divided into two parts
The first part, Prime Minister 17, the tolerance is 8, a total of 16 items, then the height of the 16th row from the classroom floor is
17+（16-1）x8=137cm.
The second part, Prime Minister 137, tolerance 10, a total of 10 items (because the first item is calculated from the 16th row), then the height of the last row from the classroom floor is
137+（10-1）x10=227cm.

If the sequence {an} satisfies: for any n ∈ n +, there are only finite positive integers m such that am

Sequence an = n & sup2;, namely an = 1,4,9,16,25
From the meaning of the question, (A5) * = satisfy an in the sequence an