Write numbers from 100 to 365 in a sequence like 100101102103104105.......…

To count how many times a particular digit occurs across a range of consecutive integers, split the range into place values (hundreds, tens, units) and count that digit's occurrences independently in each place value, then add the counts.

1. Split the range 100 to 365 into three parts: 100-199, 200-299, and 300-365.

2. Hundreds digit: across the whole range the hundreds digit only takes the values 1, 2 or 3, so the digit 9 never appears in the hundreds place. Contribution = 0.

3. Tens digit: in 100-199 the tens digit equals 9 for the ten numbers 190 to 199, giving 10 occurrences; likewise 290 to 299 gives another 10 occurrences in 200-299. In 300-365 the tens digit only ranges from 3 to 6, so it never equals 9, giving 0 occurrences. Tens-place total = 10 + 10 + 0 = 20.

4. Units digit: in 100-199 the units digit equals 9 once every ten numbers (109, 119, 129, ..., 199), giving 10 occurrences; likewise 10 more occurrences in 200-299. In 300-365 the numbers with units digit 9 are 309, 319, 329, 339, 349 and 359, giving 6 occurrences (369 falls outside the given range). Units-place total = 10 + 10 + 6 = 26.

5. Add the three place-value totals: 0 (hundreds) + 20 (tens) + 26 (units) = 46.

Cross-check: the complete block 100-199 contributes 10 (tens) + 10 (units) = 20 occurrences, the complete block 200-299 also contributes 20 occurrences, and the partial block 300-365 contributes only 6 occurrences (units place only, since it is not a complete hundred block). Adding 20 + 20 + 6 confirms the total of 46.