# This quantity (see question) will end with how many zeros?

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.

• To get a zero you need a 5 and a 2; there are 4 such pairs because of the $5^4$ term. – dbx Feb 23 '18 at 13:33
• The first term ends with 5 zeroes and the second term ends with 4. – Yanko Feb 23 '18 at 13:33
• See how many products of 2 and 5 you can make. The power of 5 is 4 is a hint. – Samuel Feb 23 '18 at 13:33
• @Samuel Got it! Thanks – OGC Feb 23 '18 at 13:38

$$3^92^{13}5^4(10-1)=5^42^42^93^99=10^42^93^99$$
Since $2^93^99$ does not end in zero, the number ends in four zeros.
$$3^34^45^56^6 - 3^64^55^46^3$$ $$=3^34^45^46^3 ( 5.6^3 - 3^3.4)$$ $$= 10^4. 2^4.3^3.6^3 ( 6^3.5 - 3^3. 4)$$