The problem asks us to count how many times the digit 3 appears when listing the numbers from 236 to 635 inclusive.

1. Problem Description

The problem asks us to count how many times the digit 3 appears when listing the numbers from 236 to 635 inclusive.

2. Solution Steps

First, consider the numbers from 236 to

2

9

9. In the hundreds place, there are no 3s.

In the tens place, the numbers 230 to 239 contain 3 as the tens digit. Thus there are 10 occurrences of 3 in the tens place. From 236 to 299, we consider only 236 to 239 which means 4 of the numbers have 3 in the tens place. So there are $4 \times 1 = 4$ occurrences in the tens place.

In the units place, the numbers are of the form 2X3, where X can be 3,4,5,6,7,8,

9. There are no numbers ending in 3 from 236 to

2

9

9. The numbers are 236,237,238,

2

3

9. Next, consider the numbers from 300 to

3

9

9. In the hundreds place, there are 100 numbers starting with 3, i.e., $300, 301, \dots, 399$. So there are 100 occurrences of 3 in the hundreds place.

In the tens place, the numbers are of the form 33X. The numbers $330, 331, \dots, 339$ have 3 in the tens digit. Thus there are 10 occurrences of 3 in the tens place.

In the units place, the numbers are of the form 3X

3. The numbers $303, 313, \dots, 393$ have 3 in the units digit. Thus there are 10 occurrences of 3 in the units place.

Now consider numbers from 400 to

5

9

9. In the hundreds place, there are no 3s.

In the tens place, the numbers are of the form X3Y, where X is 4 or 5 and Y varies from 0 to

9. So the numbers $430, \dots, 439$ and $530, \dots, 539$ have 3 in the tens digit. Thus there are $10+10=20$ occurrences of 3 in the tens place.

In the units place, there are no 3s.

Now consider the numbers from 600 to

6

3

5. In the hundreds place, there are no 3s.

In the tens place, the numbers $630, \dots, 635$ have 3 in the tens digit. Thus there are $6$ occurrences of 3 in the tens place.

In the units place, the numbers $603, 613, 623, 633$ have 3 in the units place. Thus there are no occurrences of 3 in the units place.

The range is from 236 to

6

3

5. 236 to 299:

Tens: $236,237,238,239$, which are 4 times

Units: 0

300 to 399:

Hundreds: 100

Tens: 10

Units: 10

400 to 599:

Tens: $430,431,...,439$ and $530,531,...,539$. $10+10 = 20$

600 to 635:

Tens: $630,631,632,633,634,635$, so there are 6

Units: $603,613,623,633$, so there is

1.

Total: $4 + 100 + 10 + 10 + 20 + 6 + 1= 151$

There may be mistake in the numbers.

The interval is [236,635].

Consider 236 to 635

Hundreds place: 300-399: 100

Tens place: 236, 237, 238, 239, 330-339, 430-439, 530-539, 630-

6

3

5. This sums to 4 + 10 + 10 + 10 + 6 = 40

Units place: xX3, so the numbers are 243, 253, 263, 273, 283,

2

9

3. Numbers of form x3 where x from 300 to

3

9

9. 303, 313,...,

3

9

3. 403, 413,...,493; 503,513,...,593; 603,613,623,

6

3

3. So 10+1+20+ 6 =

Then $0 + (635-236+1) - (num)$ has the hundreds position be 3

Units place: 243, 253, 263, 273, 283, 293, 303, 313,...,393, 403,413,..,493,503,513,...,593,603,613,623,

6

3

3. $6+10+10+10+4=40$

100 + 40 + 40 +1 =181?

Numbers from 236 to

6

3

5. Hundreds = 300-399 =

1

0

0. Tens = 23X (4) + 33X (10) + 43X (10) + 53X (10) + 63X (6)

Units = X3 = 243, 253, 263, 273, 283,

2

9

3. 303...393 = 10, 403,413,...493=10, 503....593 =

1

0. 603,613,623,633 (4).

$100 + 4+10+10+10+6 + 6+10+10+10+4 = 180$

3. Final Answer

D. 180