well you would probably have to mutiply 9 times ten times ten times ten, because you have four spaces for any number up to nine and zero can't be used in the first digit. My answer would have to be 9000, although I'm only in eighth grade. Anyone who has time to write out every possibility, though, has no life.
There are 22680 numbers, excluding those with leading 0s.
There are 5460 five digit numbers with a digit sum of 22.
2
There are 5760 such numbers.
If you mean, "What is the largest number of digits possible in the product of two 2-digit numbers" then 99 * 99 = 9801, or 4 digits. Anything down to 59 * 17 = 1003 will have 4 digits.
9876543120
9999999999 100000000
Any pair of digits (not including 0), can be used to generate 14 four-digit numbers. If one of the digits is 0, only seven will start with a non-zero digit.
Exactly 2: 243 of them.
1
There are 22680 numbers, excluding those with leading 0s.
that question cant be answered unless it was 3 odd-digits or 4 odd-digits with 1 even digit i think.
Zero; all six digit numbers have six digits. Dude.. i m asking 6 digit numbers, containing only 4 different digits..!! e.g.:123412.. with only 2 digits repeated..!! My guess is: 10*9*8*7*4*3 but I'm also guessing that is wrong- sorry!
With 123 digits you can make 123 one-digit numbers.
There are 5460 five digit numbers with a digit sum of 22.
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.