questions regarding x^y?

Question

4^2004 x 5^2005 x 6^2006 is written in full digit expression. How many zeroes are there at the end of this number?

Answer 1

To solve this you have to find the maximal number n, such that 10^n is a factor of the whole big number. It cannot be larger than 2005.
4^2004 X 5^2005 X 6^2006 = 4^2004 X 5^2005 X (2 X 3)^2006 =
= 4^2004 X 5^2005 X 2^2006 X 3^2006 = 4^2004 X 5^2005 X 2^2005 X 2 X 3^2006 = 10^2005 X 4^2004 X 2 X 3^2006.
4^2004 X 3^2006 X 2 is not divisible by 10.
Hence the number of zeros at the end of the number is 2005.

Answer 2

To determine the number of zeros at the end of a number, we just have to determine how many times is it divisible by 10. Now, to be divisible by 10, it has to be divisible by by 5.

4^2004 * 5^2005 * 6^2006 = (2^2)^2004 * 5^2005 * (2*3)^2006
= 2^6014 * 5^2005 * 3^2006

Since it is only divisible by 5 2005 times (although a lot more times divisible by 2), then it would only be divisible by 10 2005 times. 5 limits the number of times divisible it is by 10.

Thus, that should have 2005 zeros at its end.