4719 - 1974 - 7491 - 7419 - 9147 - 9174 - 7914 - 7941 - 4917- 4971 - 4179- 4197
7194 - 7149 - 9417 - 9471 - 1479 -1497 - 4791 - 1947 - 1749 - 1794 -
9741 -9714 - (4791)
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.
what is the greets possible 9 digit number that uses each of the digits 1-3 times
12689 14689 12489
You gave me 5 digits. -- The first digit can be any one of the 5 digits. For each one, -- The second digit can be any one of the remaining 4 digits. For each one, -- The third digit can be any one of the remaining 3 digits. For each one, -- The fourth digit can be any one of the remaining 2 digits. For each one, -- The fifth digit can only be the remaining 1 digit. So there are (5 x 4 x 3 x 2 x 1) = 120 possible different arrangements.
To find how many two-digit numbers have digits whose sum is a perfect square, we first note that the two-digit numbers range from 10 to 99. The possible sums of the digits (tens digit (a) and units digit (b)) can range from 1 (1+0) to 18 (9+9). The perfect squares within this range are 1, 4, 9, and 16. Analyzing each case, we find the valid combinations for each perfect square, leading to a total of 36 two-digit numbers whose digits sum to a perfect square.
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.
Assuming that 2356 is a different number to 2365, then: 1st digit can be one of four digits (2356) For each of these 4 first digits, there are 3 of those digits, plus the zero, meaning 4 possible digits for the 2nd digit For each of those first two digits, there is a choice of 3 digits for the 3rd digit For each of those first 3 digits, there is a choice of 2 digits for the 4tj digit. Thus there are 4 x 4 x 3 x 2 = 96 different possible 4 digit numbers that do not stat with 0 FM the digits 02356.
There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.
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.
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
what is the greets possible 9 digit number that uses each of the digits 1-3 times
12689 14689 12489
You gave me 5 digits. -- The first digit can be any one of the 5 digits. For each one, -- The second digit can be any one of the remaining 4 digits. For each one, -- The third digit can be any one of the remaining 3 digits. For each one, -- The fourth digit can be any one of the remaining 2 digits. For each one, -- The fifth digit can only be the remaining 1 digit. So there are (5 x 4 x 3 x 2 x 1) = 120 possible different arrangements.
There are 10000 such codes. Each of the numbers 0-9 can be in the first position. With each such first digit, each of the numbers 0-9 can be in the second position. With each such pair of the first two digits, each of the numbers 0-9 can be in the third position. etc.
To find how many two-digit numbers have digits whose sum is a perfect square, we first note that the two-digit numbers range from 10 to 99. The possible sums of the digits (tens digit (a) and units digit (b)) can range from 1 (1+0) to 18 (9+9). The perfect squares within this range are 1, 4, 9, and 16. Analyzing each case, we find the valid combinations for each perfect square, leading to a total of 36 two-digit numbers whose digits sum to a perfect square.
In the numbers 1-9 each number has 1 digit and there are 9 of them, so that's 9.In 10-99 each number has 2 digits, and there are 90 of them: 2x90 = 180There are 900 three digit numbers [100 through 999]: 2700 digits.There are 9000 four digit numbers: 36000 digits.90,000 numbers with five digits: 450,000 digits.900,000 numbers with six digits: 5,400,000 digits.Then 1 number with seven digits: 7 digits.Add them up and you have 5,888,896 digits.
It is not possible to answer the question since in standard form, numbers do not have points (full stops) that are five digits apart.