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There are one thousand. Starting at 2000 and going to 2999
-48
Select all the numbers that $4221462$ is divisible by.
1 x 10 x 10 x 10 x 9 = 9000 1 = it must be 2 10's = any digits 9 = any digit BUT zero
Let me first re-phrase your question: What is the number of (positive) integers less than 10000 (5 digits) and greater than 999 (3 digits)? The greatest 4 digit integer would be 9999. The greatest 3 digit integer would be 999. Let's do some subtraction: 9999 - 999 = 9000 This works because as we count up from 999, each positive integer encountered satisfies your requirements until reaching 10000.
Three of them.
18 positive integers and 36 integers (negative and positive)
90
44
There are 10 one digit positive integers (0 - 9) and 9 one digit negative integers (-9 to -1) making 19 in all.
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the range of three-digit integers is from 100 to 999. Therefore, there are 300 positive three-digit integers that are divisible by neither 2 nor 3.1 day ago
To find the number of three-digit positive integers with digits whose product is 24, we can break down 24 into its prime factors: 2 x 2 x 2 x 3. The possible combinations for the three digits are (2, 2, 6), (2, 3, 4), and (2, 4, 3). These can be arranged in 3! ways each, giving a total of 3 x 3! = 18 three-digit positive integers.
There are one thousand. Starting at 2000 and going to 2999
the answer would be 900
4
there are 999 - 100 + 1 = 900 positive triple digit positive integers, between 100 and 999.(e.g. there are 102 - 100 + 1 = 3 triple digit integers between 100 and 102,namely 100, 101 and 102.)multiply that by 2 to take in consideration of the negative integers,you have 1800 triple digit integers.