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Zach has chosen five numbers from the set $\{1,2,3,4,5,6,7\}$. If he told Claudia what the product of the chosen numbers was, that would not be enough information for Claudia to figure out whether the sum of the chosen numbers was even or odd. What is the product of the chosen numbers?.

Consider a set,
$S$=$\{N>1|N≠$p_1•p_2.....p_k$\}$

where $p_1•p_2.....p_k$ are primes.

If I write the statement for above expression then I would write that I considered a set "$S$" such that all the elements in "$S$" are greater than $1$ and can't be factorised into primes.

So, from above statement, $N$ can't be a prime number, because every prime "$p$"can be written as $p$ itself.

Let, $M$=min($S$)

So by Construction, $M$=$a$$b$, where $a$,$b$ are composite numbers.

Here, $1$$\lt$$a$$\leq$$b$$\lt$$M$

The above expression contradicts the fact that $M$ is minimal, because $a$,$b$$\lt$$M$.

$\Rightarrow$ $a$,$b$$\notin$S

So, a,b can be factorised into primes.

Let, $a=p_1•p_2....p_k$, $b=q_1•q_2....q_m$

Hence, $M=p_1•p_2..p_k•q_1•q_2..q_m$

So, $M$$\notin$$S$

This implies that $S$=$\varnothing$

Hence, FTA is true$\forall$ N$\gt$1

For $a$,$b$$\in$$R$ , $a$$>0$, $\exists$$n$$\in$$N$, $n$$a$$>$$b$

$$P=\displaystyle\prod_{k=1}^{n}\left[1–tan^2\left(\frac{2^kπ}{2^n+1}\right)\right]$$

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The product would be $\boxed{420}$

By knowing the product of the five numbers, we would also be able to derive the product of the two numbers that we didn't choose (just divide $7!$ by the product of the five numbers). Then, there are only two numbers that can be achieved by multiplying two of these numbers in more than one way: $12 = 3 \cdot 4 = 2 \cdot 6$ and $6 = 1 \cdot 6 = 2 \cdot 3$.

However, when we have $6$, the sum must be odd.

Thus, the remaining possibility is $12$, and the product of the five numbers is $7!/12 = 420$

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  • $\begingroup$ The question is only about the product, not if there'd be enough information. $\endgroup$ – the_fox Jan 29 at 6:19
  • $\begingroup$ I misunderstood the question. I just edited my answer. @the_fox $\endgroup$ – Ekesh Kumar Jan 29 at 6:21

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