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...}
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
place the digits 1 through 9 into three 3-digit numbers of an addition problem only using 1 through 9 once each
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.
Five multiples of three that use each digit from 0 to 9 only once are: 12, 30, 45, 69, and 87.
21 and three left over
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
4*3*2 = 24 of them.
six
There are four of them: 739, 793, 937, and 973 .
7, 42, 63, 98, 105
There are 5*4*3 = 60 such numbers.
6 - 123, 132, 213, 231, 312, and 321.
234, 243, 324, 342, 423, and 432.
-8