because there are 32890431 in the pie scenario it concludes the
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.
Multiplying by multi-digit numbers is similar to multiplying by two-digit numbers in that both processes involve breaking down the numbers into place values and multiplying each digit by each digit in the other number. The key similarity lies in the application of the distributive property, where each digit in one number is multiplied by each digit in the other number, and then the products are added together to get the final result. This process is consistent whether you are multiplying by a two-digit number or a multi-digit number.
To multiply two digit numbers, multiply each place value of a factor by each place value digit and add the results.
It would help to know which digit. 0 appears in 9 numbers and each of the others in 18 numbers.
I suspect you mean "without repeated digits", and I'll answer it that way.Here's how I would construct all the 5-digit numbers without repeated digits:The first digit can be any one of 9 (1 thru 9 but not zero). For each of these . . .The second digit can be any one of 9 (zero thru 9 but not the same as the first one). For each of these . . .The third digit can be any one of the remaining 8. For each of these . . .The fourth digit can be any one of the remaining 7. For each of these . . .The fifth digit can be any one of the remaining 6.Total number of possibilities = (9 x 9 x 8 x 7 x 6) = 27,216
There are four of each.
All decimal numbers are simply a way of representing numbers in such a way that the place value of each digit is ten times that of the digit to its right.
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
The final digit must be a 2 to form an even number. The first digit may be any of three remaining digits (1, 3, 5) while the second digit may be any of the two remaining digits. All together, that makes 3*2 = 6 distinct even numbers.
93 and 42
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.
These numbers are selections from the numbers from 100 to 999. That's 9 choices for the first digit. Each time, the second digit has 9 choices (0 to 9 excluding the hundreds digit), and then the last digit has 8 choices. Total is then 9x9x8 = 648