How many zeros will $ 3^{3}4^{4}5^{5}6^{6} - 3^{6}4^{5}5^{4}6^{3}$ ends with?

Question

I have this GRE practice question, asking me to find the quantity $$ 3^{3}4^{4}5^{5}6^{6} - 3^{6}4^{5}5^{4}6^{3}$$ will end in how many zeros?

The answer given is $4$, but I don't quite understand how to get it.

I know that for a zero to occur there has to be a multiplication of $5$ and $2$.

So for the first term, I have: $$3^{3}4^{4}5^{5}6^{6} = 3^{9}2^{14}5^{5}$$ and for the second term, I have : $$3^{6}4^{5}5^{4}6^{3} = 3^{9}2^{13}5^{4}.$$

I factor out the common term to get $$3^{9}2^{13}5^{4}(10-1).$$

Not sure where to go from here.

Answer 1

You almost have it.

$$3^92^{13}5^4(10-1)=5^42^42^93^99=10^42^93^99$$

Since $2^93^99$ does not end in zero, the number ends in four zeros.

Answer 2

Maybe directly take whatever you can take common at the beginning.

$$3^34^45^56^6 - 3^64^55^46^3$$ $$=3^34^45^46^3 ( 5.6^3 - 3^3.4)$$ $$ = 10^4. 2^4.3^3.6^3 ( 6^3.5 - 3^3. 4)$$

Now it is clear that it would contain 4 zeroes at the end.