7x(2-1)=7 7x(5-3)=14 7x(9-6)=21 7x(8-4)=28 7x(7+0)=49
7, 14, 28, 56. 903
That's not going to work. Multiples of 5 end in 5 or 0. With three numbers, one of those would have to repeat.
There are a few options for listing five prime numbers using the digits zero to nine only once:2, 3, 5, 7, 8649012, 5, 13, 647, 809
Well, The answer would have to be 40 because 8x5 equals 40. If you can't get the least common multiples using your personal strategy multiply the numbers you are using.
place the digits 1 through 9 into three 3-digit numbers of an addition problem only using 1 through 9 once each
πππ
There is only one combination. In a combination the order of the numbers does not matter so the only combination is 0123456789. This is the same as 1326458097
That's not going to work. Multiples of 5 end in 5 or 0. With three numbers, one of those would have to repeat.
Any number using each of the digits once will be a multiple of 3: eg 1597864302
7, 28, 49, 63, 105 is the series of multiples of 7 using 0-9 only once.
Five multiples of three that use each digit from 0 to 9 only once are: 12, 30, 45, 69, and 87.
7, 42, 63, 98, 105
3! = 6.
5*4+3+2
only using the numbers once
It depends on what is meant by "using" - in a string or with operators; and are leading zeros allowed? The smallest positive string, with no leading zeros, is 1023456789. If a leading zero is permitted, then it is 0123456789. Once you allow operators, it depends on which operators, and the question becomes extremely complex.
2,5,13,647,809
21 and three left over