There are 900 four-digit positive integers with the thousands digit being 2. Why? Because you have 10 choices for the hundreds digit (0-9), 10 choices for the tens digit (0-9), and 1 choice for the units digit (0-9), making a total of 10 * 10 * 1 = 1000. But since we are looking for positive integers, we subtract the 100 integers that start with 0, leaving us with 900 four-digit positive integers with 2 in the thousands place.
The numbers are 51, 53, 55 and 57.
The numbers, when summed, are 75, 77, 79 and 81.
5*5*4*4 = 400
-2, -1, 0, 1
21, 22, 23
There are 200 positive four digit integers that have 1 as their first digit and 2 or 5 as their last digit. There are 9000 positive four digit numbers, 1000 through 9999. 1000 of them have 1 as the first digit, 1000 through 1999. 200 of them have 2 or 5 as their last digit, 1002, 1005, 1012, 1015, ... 1992, and 1995.
Four random, positive, one-digit integers.
There are 320 such numbers.
If you allow non integers, then your smallest four-digit number would be .0001. If you only allow integers, then your smallest four-digit number would be 1000. Your largest four-digit number would be 9999.
The first four positive integers of 13 are : 26, 39, 52, 65
1.382 OR 0.000 or 2831
The positive integers up to 4 are: 1, 2, 3, and 4. This is a total of four positive integers.
The final result is positive.
Four 1-digit integers.
1,3,5,7
This is impossible with positive integers. However, four numbers separated by a difference of 1 with a sum of -92 are: -21.5 -22.5 -23.5 -24.5
The thousands digit in 395341 is 3, hence the number four greater than 3 would be 3 + 4 = 7