3000 and 6000 both have 3 0's, so they are out. The numbers between them all start with 3, 4, or 5, followed by 3 digits. There are 9 possible digits for the hundreds place, which leaves 8 for the tens and 7 for the units digit. The answer is therefore 3 x 9 x 8 x 7 = 1512
Answer 1
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There are 4 possible first digits, 9 possible second digits, 8 possible third digits and 7 possible 4 digits, making a total of 4 x 9 x 8 x 7 = 2016.
There are 2001 numbers.
This question cannot be answered sensibly. There are infinitely many numbers that are not factors of 3000. For example, all integers greater than 3000 belong to the set of "not factors of 3000". To that infinite set you can add all integers from 1501 to 2999 (both inclusive). There are also a lot more numbers smaller than 1500 that are not factors.
The largest 4 digit number that can be rounded off to 3000 is 3499. If 3499 is rounded to nearest thousand then we get 3000 but if we consider a number greater than 3499 then rounding off yields 4000. So, 3499 is the required number.
3000
The digit in the hundreds is a 2, so we round the number down to get 3000 as the answer.
If you include the number 3,000, there are exactly 2,001 4-digit numbers between 500 and 3000.
Between 1 and 3000, ie the sum of 2, 3, , ... , 2999 is 4498499
3000-999=2001.
2001 is the correct answer.
There are 900 three-digit numbers, all of which are between 99 and 3,000.
All the numbers between 999 and 3000 are four-digit numbers. You need to subtract 3000 - 999 - 1. The extra -1 is to account for the fact that the numbers 999 and 3000, in this case, are not included.
There are 900 three digit numbers between 99 and 3,000 - ranging from 100 to 999
Well honey, the numbers between 99 and 3000 are inclusive, so you gotta count 'em all. To find out how many three-digit numbers there are, you subtract the first three-digit number (100) from the last three-digit number (999) and add 1, giving you a grand total of 900 three-digit numbers. So, buckle up and start counting, darling!
1000 to 2999 inclusive so 2000 numbers.
There are 2001 numbers.
None. But if you wanted divisible, the answer is 2001 (if both points ends are included).
A palindrome reads the same forwards as backwards. Therefore the first and last digits must be the same, the second and penultimate the same, and so on. As 1000 & 3000 both have 4 digits, there are two pairs of digits that must be the same: the first and fourth, the second and third. The first digit (and hence last digit) can only be 1 or 2 - a choice of 2 The second digit (and hence third digit) can be any of 0, 1, 2, ..., 9 - a choice of 10 for each of the choices of the first/last digit. Thus there are 2 × 10 = 20 palindromes between 1000 and 3000.