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
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
NOT AN ANSWER:More acceptable combinations:1231 3213 2321More unacceptable combinations:1112 2222 3113Thanks.
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.
Five multiples of three that use each digit from 0 to 9 only once are: 12, 30, 45, 69, and 87.
Any number using each of the digits once will be a multiple of 3: eg 1597864302
21 and three left over
7, 28, 49, 63, 105 is the series of multiples of 7 using 0-9 only once.
six
4*3*2 = 24 of them.
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