Number of Trailing Zeros

In the above multiplication, we have $4^{5} = 2^{10}$ and $5^{6}$

So there are 10 of TWO and 6 of FIVE.

We also know that $2 \times 5 = 10$ is what account for a trailing zero. There are 6 of $2 \times 5$ in the above multiplication. Hence, there would be 6 trailing zero.

$3^4*4^5*5^6 \\ 3^4*4^4*5^4*4^1*5^2 \\ 60^4*100 \\ \text{Therefore, there are 6 trailing zero's}$

What is the title thing?

3^4=81, 4^5=1024, 5^6=15625. 81 1024 15625=3^4 4^5 5^6. These are the same. 3^4 4^5 5^6=3^3 3, 4^5=4^4 4, 5^6=5^5 5 So 27 3 256 4 3125 5 This is 1.296e+9

First $10^n=2^n \times 5^n$ so lets find them

We have find $4^5$ which $4=2^2$ and $5^6$

Then we start:

$4^5 \times 5^6=(2^2)^5 \times 5^6=2^{2 \times 5} \times 5^6=2^10 \times 5^6=2^{10-6+6} \times 5^6=2^{10-6} \times 2^6 \times 5^6=2^4 \times (2 \times 5)^6=2^4 \times 10^6$

So the number of zero digits is $\boxed{\large{6}}$

2x5=10

Simple, but important.

You see, 5^6 is 6 fives, and 4^5 is 5 fours.

But one four is two twos.

So 4^5 is 10 twos.

That means there are 6 pairs of fives and twos multiplied together to be a ten

3^4x4^5x5^6

=3^4x4^4x5^4x4x5^2

=60^4x10^2

=6^2x10^2x6^2x10^2x10^2

Which is 10^2x10^2x10^2 and that is 10^6 so it has 6 zeroes.

It so happens that all the above expressions are square numbers as well as higher powers. The square roots of each part of the above equation are 9, 32 and 125, so we are looking at (9 x 32 x 125) x (9 x 32 x 125) = (288 x 125) x (288 x 125) = 36,000 x 36,000. In that final equation we are multiplying a number with three trailing zeros by itself, so we will have six trailing zeros at the end of our final sum. For the record one last (straightforward) calculation gives us 1,296,000,000 as the actual answer.

Relevant wiki: Trailing Number of Zeros

For trailing number of zeros, we need to find number of 10's we can get. For us, $3^{4}$ is useless as it will not help in finding 10. Here, highest power of 2 $\rightarrow$ $\boxed{2^{10}} \\$ . Also, highest power of 5 $\rightarrow \boxed{5^{6}} \\$ . So, we require to have highest number of 2 and 5. As a result we will take $\boxed{2^{6} \times 5^{6}} = \boxed{10^{6}}$ . So, total number of trailing zeros are 6 .