How many zeroes at the end of $3^34^45^56^6 - 3^64^55^46^3$? The quantity $3^34^45^56^6 - 3^64^55^46^3$ will end in how many zeros?
This is a GRE Practice question and I have to do it without using a calculator. Can anyone help me on this? 
 A: Using $a^m\cdot a^n=a^{m+n}; a^{mn}=(a^m)^n$ and $(a\cdot b)^m=a^m\cdot b^m$ where $a,b,m$ are real numbers
$$3^34^45^56^6 - 3^64^55^46^3$$
$$=3^3(2^2)^45^5(2\cdot3)^6 - 3^6(2^2)^55^4(2\cdot3)^3$$
$$= 2^{8+6}5^53^{6+3} - 3^{3+6}2^{10+3}5^4$$
$$=2^{13}5^43^9(2\cdot5-1)$$
As $10=2\cdot5,$ the highest power of $10$ will be min $(13,4)$
A: Step 1: Factor and simplify.
Step 2: Only powers of 2, 5 are relevant in answering the question.
A: S = $3^34^45^56^6 - 3^64^55^46^3$
=$\underbrace{3^32^36^6(5*2)^5}_{\mathcal A}   - \underbrace{  3^62^66^3(5*2)^4}_{\mathcal B}$
So we see above that  both A and B have ( unequal number of )trailing zeroes. A has 5 Zeroes while B has 4 at its end. In this case for S= A-B...

number of zeroes at the end of  S will be  the number of  zeroes at the end of of the lesser of the two numbers amongst A and B  viz 4 

as the fifth digit of S is going to be non zero ( substraction of a non zero digit of B from zero; the fifth trailing zero of A)
EDIT: In response to the query by OP; the logic remains same in case of addition so A+B again will have 4 zeros
A: Note that $$3^34^45^56^6-3^64^55^46^3=5^4(3^34^45^16^6-3^64^56^3)$$
There are always enough factors of $2$ to match the powers of $5$ to give at least four zeros - but the term in brackets cannot be by $5$. This is because $5$  divides the first product in the brackets, but not the second, so cannot divide their difference.
