# Explanation for the answer to how many three digit prime numbers can be formed using only odd prime digits and digits may be used more than once?

The odd prime digits are 3, 5, and 7. but the final digit of your three-digit prime number can't be 5. So there are only 18 combinations to check. You can list these numbers a…nd start going through them to see which are prime. The easiest test is divisibility by 3 (sum the digits) and that quickly eliminates 7 of them. You are left with: 337 353 373 533 553x 557 577 733 737x 757 773 Of these, only two are not prime. I marked them with an "x". I used a table of primes to find these and don't know an easier way.

# How many 5 digit numbers that are divisible by 4 can be formed using the digits from 0 to 6 if no digit is to occur more than once in each number?

There are 7 digits from 0-6 To determine if a number is divisible by 4, we only need look at the last 2 digits. Possibilities are: [135][26] => 12, 16, 32, 36, 52, 56 a…nd [0246][04] => 04, 20, 24, 40, 60, 64 (ignoring 00 and 44) There are 8 combinations without a 0, and 4 with a 0. Of the 8 combinations without a 0, We can only pick 4 of the remaining digits for the ten thousand digit. This leaves us with 4 digits for the thousands digit, 3 for the hundreds digit Of the 4 combinations with a 0, similarly we have 5 X 4 X 3 possibilities for the rest of the digits. 8 * 4 * 4* 3 + 4 * 5 * 4 * 3 = 32 * 12 + 20 * 12 = 52 * 12 = 624 There are 624 possible numbers.

# How many ways are there to write a 3-digit positive integer using digits 1 3 5 7 and 9 if no digit is used more than once?

There are 5 digits, and if none can be repeated then the first digit can be chosen in 5 ways, the second can be chosen in 4 ways and the third in 3. 5 x 4 x 3 = 60.

# How many possible 4 digit combinations are possible without using the same number more than once?

5,040 Assuming that the combination uses ALL single digits from 0 to 9, then for the first digit you will be able to use all 10 numbers, for the second digit you will be able …to use 9, for the third digit 8 and for the last digit 7, giving a total number of combinations of 10 x 9 x 8 x 7 = 5,040 without the same number being used more than once in each combination.

# How many 2-digit numbers can be formed from the digits 4 6 and 8 Assume no number can be used more than once?

Six. 46 48 64 68 84 86

# How many 3 digit numbers can you make using the digits 1 3 5 and 7 by using the digits more than once?

Using the combination C(4,3) I got the answer 24. Its either a combination or a permutation.

# How many even 3-digit numbers can be formed from digits 1 3 5 and 7 if each digit can be used more than once?

If your only option is to assemble the digits into a single number, then the answer is 0, as you need at least one even digit to make an even number, and the digits provided a…re all odd ones. If on the other hand, you can use operators on them (e.g. 1 + 3 + 5 + 7), then you actually have quite a large number of possibilities. For example: 1 + 3 + 5 + 7 - 1 + 3 + 5 + 7 1 - 3 + 5 + 7 (1 + 3) / ( 7 - 5) 1 + √(7 + 5 - 3) 31 + 57 etc. etc. Alternatively, if you're not worried about them being even numbers, then the answer is 24.

# How many 2 digits numbers can you have using 0 through 9 if you can use the numbers more than once?

Assuming that 00, 01, 02, etc. count as two digit numbers, then there are one hundred (1 through 99 and including 0). If you are not allowed to count 00, 01, 02, 03, 04, 05, …06, 07, 08, or 09 because they simplify to one digit numbers, then you have ten less--so a total of ninety two digit numbers.

# How many 4 digit combinations using 0-9 if each number may be used more than once?

10,000 combinations.

Answered

In Algebra

# How many 4 digit numbers can be formed by the digits 1 2 3 4 using the digits more than once in each number example 1112 1122?

256

Answered

In Algebra

# How many different 3 digit identification cards can be made is the digits can be used more than once. if the first digit must be a 5 and repetitions are not permitted?

All the numbers from 500 to 599 EXCEPT 500, 511, 522, 533, 544, 550 - 559, 566, 577, 588 & 599. That is 100 less 19 ie 81. If you regard 505 as being a repetition then there a…re 9 less overall.

Answered

In Technology

# Can You Use digital copy on more than one computer?

It depends on the terms and conditions of the product.

Answered

In Algebra

# How many 2 digit numbers can be formed from the digits 4 6 and 8 Assume no number can be used more than once?

This is a question of permutations. It is equal to 3! / (3 - 2)! = 3! / 1! = 6/1 = 6. If you want to check this, you can actually create the numbers yourself, since there aren…'t that many: 46, 48, 64, 68, 84, 86.