Divide divide only divide...

$\begin{aligned} Q & = 1 \div (2 \div (3 \div (4 \div \cdots \div (2015 \div (2016 \div 2017))\cdots ) \\ & = \frac {1\times 3 \times 5 \times \cdots \times 2015 \times 2017}{2 \times 4 \times 6 \times \cdots \times 2016} \\ & = \frac 32 \times \frac 54 \times \frac 76 \times \cdots \times \frac {2k+1}{2k} \times \cdots \times \frac {2017}{2016} > 1 & \small \color{#3D99F6} \text{where }k \in \mathbb N \end{aligned}$

Note that the nominator of each term is 1 larger than denominator. Therefore $Q > 1$ and the expression is true .

If the divisor is greater than the dividend, the quotient will be less than $1$ .

If the divisor is less than $1$ , the quotient will be greater than the dividend.

$2016\div2017 < 1$

Therefore, $2015\div(2016\div2017) > 2015$

Therefore, $2014\div(2015\div(2016\div2017)) < 1$

Therefore, $2013\div(2014\div(2015\div(2016\div2017))) > 2013$

And so on.

When the dividend is odd, the quotient will be greater than $1$ . Therefore, the inequality is $\boxed{true}.$

look carefully,

starting from the middle-- $\frac{2016}{2017}<1$

and then,if we divide the gotten value by $2015$ .it will be greater than $1$ . example: $\frac{2015}{1}=2015$ but $\frac{2015}{.9(less than 1)}=2238.88..$

so,the value will increase day by day and surely the final value will be > $1$ .

$2016\div2017<1$

$2015\div(2016\div2017>1$

$2014\div(2015\div(2016\div2017))<1$

$2013\div(2014\div(2015\div((2016\div2017))))>1$

$\quad \vdots$ $\quad \vdots$ $\quad \vdots$

Here we can see that if this division we start from an even number, then whole result always $< 1$

Again if this division we start from an odd number, then whole result always $> 1$

In the question, this division starts from $1$ which is an odd number.

So logically the whole result must be $> 1$

So it's $\color{#20A900}{\boxed{true}}$

When looking at the two choices I chose one at random with a 50/50 chance.

You can make the equation shorter -----> 1/(2/3) ----> 3/2

LOL Everyone is thinking too hard I think.

I could be wrong but I noticed:

If the highest number (e.g. 2017) = odd number then it's always > 1.

If the highest number (e.g. 2018) = even number then it's always < 1

It alternates 1÷2, 1÷.66, but as 2015-1 is even and alternations on even intervals end with the opposite position as the input, it is larger than the 1.

1/(2/(3/(4/5))))>1

As simply I would tell my answer If 1 is divided by 2 it reduces to a number lesser than one .this goes on till 2017 so the final result would be smaller than 1 Sorry I am a newbie And this what best I can do.

When dividing by a fraction it is the same as the numerator and denominator flipping then multiplying... 1÷ 2÷3 = 1÷2/3 = 1×3/2 > 1. Each additional '÷' added just flips another bigger number up into the denominator.

So I focused on the important part of the question, in my opinion. This was 2015/(2016/2017).

I think you can do this as everything else stems from it. Turn it into algebra as you do ... n/(n+1/n+2). That equals n*(n+2/n+1) which equals (n^2+2n)/(n^2+n). This produces a top heavy fraction as long as n is greater than 0, which it is.

Now I think this is alright for a solution as the whole inequality can just be represented in the same with the same method repeated all the way across.

if n = a/b and b>a then n <1 and if b<a then n >1. The solution therefore oscillates around 1. i.e n=(>1) / (<1) / (>1) / <1) ... Therefore, if the final element is odd the solution will be >1 and if the final element is even then the solution will be <1

Odd number 2017 :)

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def problem(year):
    result=year
    while year>1:
        result=(year-1)/result
        year-=1
    print (result)

This gives the answer 35.83825372...

I just started programming this week, I almost don't know anything about programming. How do you improve this code? Is it correct and how do I make it more efficient?

We see a pattern here which will help the sum be solved faster. In the first case the divisor is odd ,2017 which divides 2016 where the quotient is smaller than 1 so when it divides the odd dividend the quotient is greater than 1,and this pattern continues. So when an odd dividend is there quotient is greater than 1 , and the last term is 1 which is odd and thus gives quotient >1.