How many distinct 3 digit odd integers are there? Distinct meaning none of the digits repeat.
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
PS - you should double check this. This is just how I would go about doing it. I don't have much experience with such questions so I've simply tried my best.
To find the answer to this question, we'll break it down step by step.
First, let's consider the restrictions on the three-digit odd integers:
1. The first digit cannot be zero. (An odd integer cannot have a zero in the ones place.)
2. The third digit must be an odd number, i.e., 1, 3, 5, 7, or 9.
Now, let's analyze the possibilities for each digit:
1. The first digit can be any number from 1 to 9 (9 possibilities).
2. The second digit can be any number from 0 to 9, excluding the first digit chosen (9 possibilities minus 1).
3. The third digit can be any odd number from 1 to 9 (5 possibilities).
To find the total number of distinct three-digit odd integers, we multiply the number of possibilities for each digit together:
9 * 9 * 5 = 405.
Therefore, there are 405 distinct three-digit odd integers.