We see that $$a$$◇$$a=1$$, and think of division. Testing, we see that the first condition $$a$$◇$$(b$$◇$$c)=(a$$◇$$b)\cdot c$$ is satisfied, because $$\frac {a}{\dfrac bc}=\frac ab\cdot c$$. Therefore, division can be the operation ◇. Solving the equation, $$\frac {2016}{\dfrac 6x}=\frac {2016}6\cdot x=336x=100\Rightarrow x=\frac {100}{336}=\frac {25}{84}$$, so the answer is $$25+84=109$$.  We can manipulate the given identities to arrive at a conclusion about the binary operator ◇. Substituting $$b=c$$ into the first identity yields $$(a$$◇$$b)\cdot b=a$$◇$$(b$$◇$$b)=a$$◇$$1=a$$◇$$(a$$◇$$a)=(a$$◇$$a)\cdot a=a$$. Hence, $$(a$$◇$$b)\cdot b=a$$, or, dividing both sides of the equation by $$b$$$, $$(a$$◇$$b)=\frac ab$$.  Hence, the given equation becomes $$\frac {2016}{\dfrac 6x}=100$$. Solving yields $$x=\frac {100}{336}=\frac {25}{84}$$, so the answer is $$25+84=109$$.  One way to eliminate the ◇ in this equation is to make $$a=b$$ so that $$a$$◇$$(b$$◇$$c)=c$$. In this case, we can make $$b=2016$$.  $$2016$$◇$$(6$$◇$$x)=100\Rightarrow (2016$$◇$$6)\cdot x=100$$  By multiplying both sides by $$x$$, we get:  $$(2016$$◇$$6)\cdot 6=\frac {600}x\Rightarrow 2016$$◇$$(6$$◇$$6)=\frac {600}x$$  Because $$6$$◇$$6=2016$$◇$$2016=1$$:  $$2016$$◇$$(2016$$◇$$2016)=\frac {600}x\Rightarrow (2016$$◇$$2016)\cdot 2016=\frac {600}x\Rightarrow 2016=\frac {600}x$$  Therefore, $$x=\frac {600}{2016}=\frac {25}{84}$$, so the answer is $$25+84=109$$. 
某運算◇對所有的非零實數$$a$$、$$b$$、$$c$$滿足：$$a$$◇$$(b$$ ◇$$c)=(a$$◇$$b)\cdot c$$(此處 $$\cdot$$ 是一般的乘號)，且 $$a$$◇$$a=1$$ ．若方程式$$2016$$◇$$(6$$◇$$x)=100$$ 的解爲$$\frac pq$$,其中$$p$$、$$q$$爲互質的正整數，則 $$p+q$$之値爲何？  A binary operation ◇ has the properties that $$a$$◇$$(b$$ ◇$$c)=(a$$◇$$b)\cdot c$$ and that $$a$$◇$$a=1$$ for all nonzero real numbers $$a$$, $$b$$, and $$c$$. (Here represents multiplication). The solution to the equation $$2016$$◇$$(6$$◇$$x)=100$$ can be written as $$\frac pq$$, where $$p$$ and $$q$$ are relatively prime positive integers. What is $$p+q$$?+q?
Options:
[a] $$109$$ 
[b] $$201$$ 
[c] $$301$$ 
[d] $$3049$$ 
[e] 33,601 