Forming three three digit numbers that use the numbers 1-9 without repeating, the highest product possible is 611,721,516. This is formed from the numbers 941, 852, and 763.
81 As there are no limits stated then you can have a number comprising a repeated single digit (such as 2222), two pairs of numbers (e.g. 2244) or three different numbers (such as 2462). The first digit can be one of any of the 3 numbers. The second digit can be one of any of the three numbers, as can the third digit and also the fourth. Then you can have 3 x 3 x 3 x 3 = 81 different 4-digit numbers using the three given numbers.
5 x 10 x 5 = 250 different numbers, assuming there is no limit to each digits' use.
The smallest three-digit number divisible by the first three prime numbers (2, 3, and 5) and the first three composite numbers (4, 6, and 8) is 120.
To make this simpler, I'll assume that numbers with a leading zero like 0179 are "four digit numbers" for the purposes of this question. The number of such numbers that can be formed by those numbers without repeating one of the digits is 4x3x2x1 = 24. We could list them all and add them up, but let's instead just realize that we can choose any digit and put it in any position, and the other digits can then be arranged in 3x2x1 = 6 ways. So the sum for any position is (0+1+7+9)*6 = 102, and the sum of the set of "four digit" numbers is 102x1000+102x100+102x10+102 = 113322. You can figure out for yourself what the sum would be if you exclude the numbers that are really three digit numbers (hint: use the technique above to find the sum of the three digit numbers that can be formed using the digits 1, 7, and 9; then subtract that from the other total).
512
from 100 to 999 it is 900
27 three digit numbers from the digits 3, 5, 7 including repetitions.
Forming three three digit numbers that use the numbers 1-9 without repeating, the highest product possible is 611,721,516. This is formed from the numbers 941, 852, and 763.
81 As there are no limits stated then you can have a number comprising a repeated single digit (such as 2222), two pairs of numbers (e.g. 2244) or three different numbers (such as 2462). The first digit can be one of any of the 3 numbers. The second digit can be one of any of the three numbers, as can the third digit and also the fourth. Then you can have 3 x 3 x 3 x 3 = 81 different 4-digit numbers using the three given numbers.
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
Any 3 out of 4 = 4. Each set can be arranged 6 ways so 24 different numbers.
Choose from 4 for the first number, 3 for the second number, and 2 for the third number. Therefore there are 4*3*2 = 24 three digit numbers can be formed from 3769 if each digit is used only once.
24 of them. So we have a 2 or 6 in the unit position, therefore (2)(4)(3) = 24 even three-digit numbers, can be formed.
For the first digit there would be 4 options For the second digit there would be 3 options for the third digit there would be 2 options Thus 4*3*2 = 24 three digit numbers can be formed if each digit can only be used once in a 3 digit number made.
105
There are only 999 three digit whole numbers.