-- If 'zero' is always the first digit, then we're pretty darn sure that no other digit can be repeated in the same 2-digit number, since there's only one digit left to be filled in. -- If 'zero' is always the first digit, then there are exactly six possible 2-digit numbers: 00 01 02 03 04 and 05.
0
108
The question is a bit ambiguous: "the sum of the product ... " does not seem to make sense. The first 5 digit number is 10000 the first 5 digit multiple of 6 is 10002 Their product is 100020000. I am not sure what to sum, though.
7
132 (11x12)
-- If 'zero' is always the first digit, then we're pretty darn sure that no other digit can be repeated in the same 2-digit number, since there's only one digit left to be filled in. -- If 'zero' is always the first digit, then there are exactly six possible 2-digit numbers: 00 01 02 03 04 and 05.
0
The first digit would have to be 1, the remaining digits, zero.
110
108
If you mean digit, the answer is NO.
I suspect you mean "without repeated digits", and I'll answer it that way.Here's how I would construct all the 5-digit numbers without repeated digits:The first digit can be any one of 9 (1 thru 9 but not zero). For each of these . . .The second digit can be any one of 9 (zero thru 9 but not the same as the first one). For each of these . . .The third digit can be any one of the remaining 8. For each of these . . .The fourth digit can be any one of the remaining 7. For each of these . . .The fifth digit can be any one of the remaining 6.Total number of possibilities = (9 x 9 x 8 x 7 x 6) = 27,216
The question is a bit ambiguous: "the sum of the product ... " does not seem to make sense. The first 5 digit number is 10000 the first 5 digit multiple of 6 is 10002 Their product is 100020000. I am not sure what to sum, though.
7
Nine. The sum of the digits must be a multiple of 9; because of the repeated digits, this is only possible if the first two digits add up to 9.
Assuming no repeated digits, lowest first, 20; in any order 120; Allowing repeated digits: 216