To do this problem, we could simply divide the big number by 2, then 3, then 4, then 5, and see which ones leave no remainder. But that approach would be very time consuming. There are some tricks we can use to simplify the problem considerably. Let us consider each of the divisor's, 2, 3, 4 and 5, in turn.

A number is evenly divisible by 2 if it's rightmost digit is even. Obviously, the rightmost digit of 6,846,624 is 4, which is even. So, the number must be evenly divisible by two.

A number is evenly divisible by 4 if all the digits of the number are even and the rightmost digit is divisible by 4. Obviously, all the digits of the big number are even and the rightmost digit is evenly divisible by 4. So, the entire number is evenly divisible by four.

A number is evenly divisible by 3 if all the digits of the number add up to a number that is divisible by 3. Well, if we add up the digits of 6,846,624, we get 36. Is 36 divisible by 3? If you don't know, you can add that number's digits together, which yields 9. It is obvious that 9 is divisible by 3, so the entire big number must be evenly divisible by 3.

A number is evenly divisible by 5 only if the rightmost digit is either a 0 or a 5. Obviously, the number 6,846,624 is not divisible by 5.

So, the big number is evenly divisible by 2, 3 and 4 but not 5.