0+1+2+3+4+5 = 15 so, if a 5-digit number, formed from these digits, is divisible by 3 it must exclude 0 or 3.
If it excludes 0, all 5*4*3*2*1 = 120 permutations are valid.
If it excludes 3, then 4*4*3*2*1 = 96 permutations are valid (the others start with a zero and so are 4-digit numbers).
That makes a total of 120+96 = 216 numbers.
192 ways
All of them. That is, 1,000 of them (including zero).
A number is divisible by 234 if:its last digit is 0, 2, 4, 6 or 8; andthe sum of its digits (the calculation repeated if required), is divisible by 9; andthe result of nine times the last digit subtracted from the number formed by the rest (the calculation repeated if required) is divisible by 13.
94205 if digits are not to be repeated and 99999 if digits can be repeated
500
192 ways
There are 5! = 120 such numbers.
All of them. That is, 1,000 of them (including zero).
Stop trying to cheat in math, it's your grade not mine...
A number is divisible by 234 if:its last digit is 0, 2, 4, 6 or 8; andthe sum of its digits (the calculation repeated if required), is divisible by 9; andthe result of nine times the last digit subtracted from the number formed by the rest (the calculation repeated if required) is divisible by 13.
94205 if digits are not to be repeated and 99999 if digits can be repeated
500
24
Whole numbers are divisible by 4 if the number formed by the last two individual digits is evenly divisible by 4. For example, the number formed by the last two digits of the number 2628 is 28, which is evenly divisible by 4 so the number 2628 is evenly divisible by 4.
1 set
It is 120 if the digits cannot be repeated.
5*4*3*2*1 = 120