I think the question means to ask for "four-digit" odd numbers, and left out a word. Each candidate is of the form (T H N U), and we're told that (H = N + 2). ' T ' can be: 1, 2, 3, 4, 5, 6, 7, 8, 9 = 9 possibilities. For each of these ... ' H N ' can be : 97, 86, 75, 64, 53, 42, 31, and 2,0 = 8 possibilities. For each of these ... ' U ' can be: 1, 3, 5, 7, 9 = 5 possibilities. So the total number of possibilities is: (9 x 8 x 5) = 360different 4-digit odd numbers with the hundreds digit greater by two than the tens digit.
There are 3 values (1, 2, 3) for each of the 4 digits. Therefore, there are 3*3*3*3 or 81 four digit numbers that can be formed.
5*5*4*4 = 400
10, 24, 48, 80, 82
If numbers can be repeated and zeroes are allowed to lead, this is simply all natural numbers in the set {0000 - 9999}, for a total of 10,000 possible combinations. If numbers cannot be repeated, this becomes a permutation problem; out of 10 possible digits, permute four of them. This evaluates as: nPr = n! / (n-r)! 10P4 = 10! / (10-4)! 10P4 = 1 * 2 * 3 * ... * 6 * 7 * 8 * 9 * 10 / 1 * 2 * 3 * 4 * 5 * 6 10P4 = 7 * 8 * 9 * 10 10P4 = 5040 There are thus 5,040 possible combinations of four of the digits in 0-9 if any digit is not used twice and 0 is allowed to be a leading digit.
The numbers are 215, 216, 217 and 218.
Add the two greatest possible four digit numbers. 9999 + 9999
625 if numbers can have leading 0s, 500 otherwise.
Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.
The number is a 13-digit number.
Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991
There is no English number "924 hundred." The possible numbers are: 924 x 100, which is 92400 (ninety-two thousand four hundred) 924, which is "nine hundred twenty-four."
34.48 million in numbers is 34,480,000 In words it is ' Thirty four million, four hundred and eighty thousand'. NB Note the use of COMMAS at everythird digit.
The number 500,000,000,444,444,444,444 is a 12-digit numeral. In this number, the digit 5 is in the hundred millions place, followed by 11 consecutive 4s. Each place value in a number represents a power of 10, with the rightmost digit representing 10^0 (1), the next representing 10^1 (10), and so on. Therefore, in this number, the digit 4 in the trillion's place is multiplied by 10^12.
2.4 million in numbers would be 2,400,000. In non-digit form it would be written out as two million and four hundred thousand.
9 x 10 x 10 x 5 = 45000
Lowest four-digit number: 1000 Highest four-digit number: 9999
A four digit whole number can be found from 1000 to 9999