In distinct odd-town, the inhabitants want to number their houses with 3-digit integers that are odd, which have all distinct digits. What is the maximum number of houses in odd-town?

Hi. I tried answering this in a previous post. I'm not entirely sure if I did this correctly but here was my best shot at it:

So 3 digit integers would be from 100 to 999.
Let's break it up:
1) 100-109 you know that there are 5 #s are odd (101,103,105,107,109) and of those #s 101 is not distinct therefore their are only 4 distinct odd #s
2) 110-119 you know that none of the #s are distinct (all 10 are non-distinct)
3) 120-129 you know that there are 5 odd numbers and of the 5, 1 # (121) is not distinct therefore there are 4 distinct odd #s
4) 130-139 you know that there are 2 non-distincts # (131,133) therefore there are 3 distinct odd #s
5) 140-149 you know is 1 non-distinct # (141), therefore 4 distinct #s
(I'm not going to explain the rest - just state)
6) 150-159 - 3 distinct
7) 160-169 - 4 distinct
8) 170-179 - 3 distinct
9) 180-189 - 4 distinct
10) 190-199 - 3 distinct

From 100-199 there are 32 distinct 3 digit odd numbers

If we were to do the 200s you would see the following:
1) 200-209 - 5 distinct
2) 210 - 219 - 4 distinct
3) 220-229 - no distinct
4) 230-239 - 4 distinct
5) 240-249 - 5 distinct
6) 250-259 - 4 distinct
7) 260-269 - 5 distinct
8) 270-279 - 4 distinct
9) 280-289 - 5 distinct
10) 290-299 - 4 distinct
Total # of distinct odd 3 digit integers = 40

If we were to do the 300s you would see the following:
1) 300-309 - 4 distinct
2) 310 - 319 - 3 distinct
3) 320-329 - 4 distinct
4) 330-339 - 0 distinct
5) 340-349 - 4 distinct
6) 350-359 - 3 distinct
7) 360-369 - 4 distinct
8) 370-379 - 3distinct
9) 380-389 - 4 distinct
10) 390-399 - 3 distinct
Total # of distinct odd 3 digit integers = 32

Do you see the pattern. If the first number of the integer is an odd number then you will have a total of 32 three-digit distinct odd integers. If the first number of the integer is even then you will have 40.

Therefore:
From 100-199 there should be 32
From 200-299 there should be 40
From 300-399 there should also be 32
From 400-499 there should also be 40
From 500-599 there should also be 32
From 600-699 there should also be 40
From 700-799 there should be 32
From 800-899 there should be 40
From 900-999 there should also be 32

Total # of odd distinct 3-digit integers is 360

the answer is 320

u have made a mistake in the addition part...
From 100-199 there should be 32
From 200-299 there should be 40
From 300-399 there should also be 32
From 400-499 there should also be 40
From 500-599 there should also be 32
From 600-699 there should also be 40
From 700-799 there should be 32
From 800-899 there should be 40
From 900-999 there should also be 32
=320

8*8*5=320....

To find the maximum number of houses in odd-town, we need to determine the largest 3-digit odd number that can be formed using distinct digits.

Step 1: Find the largest possible digit for the hundreds place. Since the number needs to be odd, the largest possible digit for the hundreds place is 9.

Step 2: Find the largest possible digit for the tens place. Since the digit for the hundreds place is already determined as 9, the largest possible digit for the tens place is 7 (since it is both odd and distinct from 9).

Step 3: Find the largest possible digit for the ones place. Since the digits for the hundreds and tens places are already determined as 9 and 7 respectively, the largest possible digit for the ones place is 5 (since it is both odd and distinct from 9 and 7).

Therefore, the largest 3-digit odd number with distinct digits is 975.

Since the houses in odd-town are numbered in ascending order, starting from the lowest odd number, the maximum number of houses in odd-town would be the same as the house number with the largest possible 3-digit odd number, which is 975.

So, the maximum number of houses in odd-town is 975.