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
First, factor: 16 = 2*2*2*2 18 = 2*3*3 Multiply the factors (only count overlapping ones once) to find the least common multiple (LCM) LCM = 2*2*2*2*3*3 = 144 The multiples of 16 and 18 are {144, 288, 432, 576...}
To find the second common multiple of two numbers, you first need to identify the least common multiple (LCM) of the two numbers. The LCM is the smallest number that is a multiple of both numbers. Once you have the LCM, you can then find the second common multiple by adding the LCM to the LCM itself. This will give you the second common multiple of the two numbers.
place the digits 1 through 9 into three 3-digit numbers of an addition problem only using 1 through 9 once each
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.
7, 28, 49, 63, 105 is the series of multiples of 7 using 0-9 only once.
Any number using each of the digits once will be a multiple of 3: eg 1597864302
Five multiples of three that use each digit from 0 to 9 only once are: 12, 30, 45, 69, and 87.
9
7, 42, 63, 98, 105
3! = 6.
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.
5*4+3+2
only using the numbers once
If I'm allowed to use the same digit more than once, then there are (10 x 10 x 10) = 1,000 ways. If I'm not allowed to use the same digit more than once then there are (10 x 9 x 8) = 720 ways. Either way, I'm afraid you'll have to generate the list on your own.