find the sum of all numbers >10000 formed by using digits 0,2,4,6,8 ,no digit being repeated in any number?

Answer 1

What you need to do is write out all possible permutations of those 5 digits, not including the ones that start with 0, and, treating them as five digit numbers, add them all up.

There are 4x4! = 96 such numbers, so you'll be at it for some time, unless you have a computer program to do it for you.

Answer 2

There are, as MHW said, 4 * 4! total such numbers.

Consider the digits of these numbers.
If the first digit is a 2, then there are 4! possibilities for the other digits.
Likewise for 4, for 6, and for 8.

Therefore, the sums of just the first digits of all the numbers can be written as:
(2 * 4! + 4 * 4! + 6 * 4! + 8 * 4!) * 10000, since these are all 10000ths place digits.

This can be rewritten as
(2 + 4 + 6 + 8) * 4! * 10000.

Similarly, for the 1000ths place digit, the total sum of these digits can be written as
(2 + 4 + 6 + 8) * 3 * 3! * 1000
(The reason it's 3 * 3! rather than 4! is because the first digit cannot be 0).

Working through it, the total can be written as
(2 + 4 + 6 + 8)(4! * 10000 + 3 * 3! (1000 + 100 + 10 + 1)) =
20(240000 + 18(1111)) =
5199960.

I wrote a java program to check this. It's not the most elegant in the world, but it works:

public static void main(String [] args)
{
Set set = new HashSet();
for (int i = 0; i < 5; i++)
set.add(2 * i);
int total = 0;
for (int i : set)
{
if (i == 0)
continue;
for (int j : newSet(set, i))
for (int k : newSet(set, i, j))
for (int l : newSet(set, i, j, k))
for (int m : newSet(set, i, j, k, l))
total += 10000 * i + 1000 * j + 100 * k + 10 * l + m;
}
System.out.println(total);
}

private static Set newSet(Set set, T ... trash)
{
Set ret = new HashSet();
ret.addAll(set);
for (T t : trash)
ret.remove(t);
return ret;
}

Answer 3

if we take a series of numbers that are multiples of three say 12 24 369 1239 we locate sum of their digits=> for 12==> [ a million+ 2] == 3 that's divisible by ability of three sum of digits also divisible by ability of three for twenty-four===>[ 2+4 ] == 6 that's divisible by3 sum of digits also divisible by ability of three for 369==>[3+6+9] == 18 that's divisible by3 sum of digits also divisible by ability of three for 1239 ==>[a million+2+3+9] ==15 that's divisible by ability of three sum of digits also divisible by ability of three. as a outcome,the fact is verified by ability of those demonstrtions.