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Source: 2013 BAMO-8 #4

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#1 h Feb 27, 2016, 8:47 PM • 5 Y

Y by GoJensenOrGoHome, itslumi, Adventure10, cubres, NicoN9

For a positive integer $n>2$ , consider the $n-1$ fractions $\dfrac21, \dfrac32, \cdots, \dfrac{n}{n-1}$ The product of these fractions equals $n$ , but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of $n$ for which this is possible and prove that you have found them all.

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math90

#2 h Feb 27, 2016, 10:18 PM • 7 Y

Y by MSTang, mathcrazymj, Adventure10, Mango247, ItsBesi, cubres, kiyoras_2001

The answer is all perfect squares greater than $1$ .
To make the product $1$ , the product of the numerators should be equal to the product of the denominators. I.E, $\prod_{k=2}^{n}k(k-1)=(n-1)!^2n$ should be a perfect square. Thus $n$ should be a perfect square.
For perfect squares construct the following example:
$\prod_{i=2}^{\sqrt{n}}\frac{i-1}{i}\cdot\prod_{i=\sqrt{n}+1}^{n}\frac{i}{i-1}=\frac{1}{\sqrt{n}}\cdot\sqrt{n}=1$ .

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#3 h Aug 1, 2019, 3:19 PM • 2 Y

Y by Adventure10, cubres

Maybe just my opinion, but this is a nice ez problem, with a nice idea.
I claim that all square numbers greater than $1$ satisfy.
To see that this is necessary, notice that to have that the multiply of all the fractions equal to 1, then we must have the multiply of all the fraction we have having the same denominator and numerator, which means that the product of all numerators and denominators must be a perfect square. But we know that the product is $(n-1)!^2 n$ . So, $n$ must be a perfect square.

To see that this is sufficient. Notice that we have
$\prod_{i = 2}^{n} \frac{i}{i - 1} = n$ initially, which is a perfect square. We wanted to reverse several fractions such that the final product is $1$ .
This means that we need to find several fractions which product to $\sqrt{n}$ .
But this is obvious, as
$\prod_{i = 2}^{\sqrt{n}} \frac{i}{i - 1} = \sqrt{n}$

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pad

#4 h Sep 5, 2019, 1:29 AM • 3 Y

Y by Adventure10, Mango247, cubres

We claim that $n$ must be a square. A construction for $n=k^2$ is
$\left(\frac12\cdot \frac23\cdot \frac34\cdots \cdot \frac{k-1}{k} \right) \left( \frac{k+1}{k}\cdot \frac{k+2}{k+1}\cdots \cdot \frac{n}{n-1} \right) = \frac{n}{k^2}=1.$ Suppose that $n$ works. In the final list of fractions which have product 1, the product of all the numerators must equal the product of all the denominators. This means that the product of all the numbers overall must be a square. The product of all the numbers is
$(2\cdot 1)(3\cdot 2)\cdots (n(n-1)) = (n-1)!^2n,$ which means $n$ must be a square.

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Eyed

#5 h May 6, 2020, 4:20 AM • 2 Y

Y by pad, cubres

Define a fraction as "good" if it's square root is rational, and "bad" otherwise. Observe that two good fractions multiplied together results in a good fraction and a bad and a good fraction multiplied together results in a bad fraction.

If $n = k^{2}$ , we can flip the first $k-1$ fractions. If $n \neq k^{2}$ , then $\frac{2}{1} \cdot\frac{3}{2}\cdot\ldots \frac{n}{n-1}$ is bad. Whenever we flip a fraction $\frac{p}{p-1} \Rightarrow \frac{p-1}{p}$ , we multiply the original product by $\frac{(p-1)^{2}}{p^{2}}$ , a good fraction. Then, since $1$ is a good fraction, and we start with a bad fraction, it can not be made into $1$ . The only possible values of $n$ is $n = k^{2}$ .

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#6 h Jul 22, 2020, 11:43 PM • 1 Y

Y by cubres

I claim that the problem is true iff $n$ is a perfect square.

Suppose that $n = a^2$ for some positive integer $a \geq 2$ . Now we can turn this expression into $1$ by simply flipping the first $a-1$ fractions.

Now, let a fraction be called a ssquare if, when written in simplest terms, the numerator and denominator are perfect squares. Note that if the $n$ is by definition, not a ssquare. Furthermore, note that with each operation it is impossible to turn a non-ssquare into a ssquare, so we are done.

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#7 h Aug 12, 2020, 5:54 AM • 1 Y

Y by cubres

Rephrase reciprocation as multiplying a value of the form $\left( \tfrac{k}{k+1} \right)^2$ for distinct values of $1 \le k < n$ .

Now it is clear that $n$ must be a square, for the product will always be $n$ times the square of a rational by telescoping. To show that this always works, simply multiply the values
$\left( \frac{1}{2} \right)^2 \cdot \left( \frac{2}{3} \right)^2 \cdot \dots \cdot \left( \frac{\sqrt{n}-1}{\sqrt{n}} \right)^2 = \frac1{n}$ to $n$ to obtain $1$ . $\blacksquare$

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#8 h Oct 3, 2020, 5:59 PM • 1 Y

Y by cubres

Solution

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#9 h Oct 8, 2020, 3:01 AM • 1 Y

Y by cubres

We claim the only numbers that work are perfect squares.

If $n$ is a perfect square, then we can use the construction
$\frac{1}{2}\cdot\frac{2}{3}\dots\frac{\sqrt{n}-1}{\sqrt{n}}\cdot\frac{\sqrt{n}+1}{\sqrt{n}}\cdot\frac{\sqrt{n}+2}{\sqrt{n}+1}\dots\frac{n}{n-1}$ And it is easy to see that this works.

We now want to show that $n$ has to be a perfect square for all possible values of $n$ . Assume that it is possible to reciprocate some of the fractions such that the product is equal to $n$ . We divide
$\frac{2}{1}\cdot\frac{3}{2}\dots\frac{n}{n-1}=n$ by the construction that works to get
$\left(\frac{2}{1}\right)^{1+\epsilon_1}\cdot\left(\frac{3}{2}\right)^{1+\epsilon_2}\dots\left(\frac{n}{n-2}\right)^{1+\epsilon_n}$ Where $\epsilon_i$ is either $1$ or $-1$ . Then, the value of $\epsilon_i+1$ must be an even number, so $n$ must be a perfect square, as desired.

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#10 h Oct 29, 2020, 5:54 PM • 1 Y

Y by cubres

The answer is all $n$ such that $n$ is a perfect square greater than $1$ .To see this works, we can take
$\left[\frac{1}{2} \cdot \frac{2}{3} \dotsm\frac{\sqrt{n}-1}{\sqrt{n}}\right] \times \left[\frac{\sqrt{n}+1}{\sqrt{n}} \dotsm \frac{n}{n-1}\right].$ All of the first bracket simplifies to $\tfrac{1}{n}$ an the second to $n,$ and $\tfrac{1}{n} \times n = 1.$

To show this is the only solution, note the numerator $p$ has product $n!$ and the denominator $q$ has product $(n-1)!,$ so $pq = (n-1)!^2 \cdot n.$ Also note that for the value to be $1$ , we must have $p=q$ . Thus, because $pq= (n-1)!^2 \cdot n,$ $n$ must be a perfect square. $\blacksquare$

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#11 h May 15, 2021, 11:23 PM • 2 Y

Y by Mango247, cubres

Posting for storage.
The problem falls after noticing the construction.

If $n$ is perfect square, then following construction works:
$\dfrac21, \dfrac32, \cdots, \dfrac{\sqrt{n}}{\sqrt{n}-1}, \dfrac{\sqrt{n}}{\sqrt{n}+1}, \cdots, \dfrac{n-1}{n}.$
Now we claim that $n$ must be a perfect square.
Notice that for all $p\in\mathbb{P}$ , we must have $2\mid \nu_p(n!)+\nu_p((n-1)!),$ since otherwise for some prime, we have a number of those primes odd and thus we cannot group them into two groups having the same cardinality.
Rewriting we obtain, $2\mid \nu_p(n)+2\nu_p((n-1)!)\implies \nu_p(n)=2k,$ hence $n$ must be a perfect square in order to form such product.

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#12 h Aug 25, 2021, 3:29 PM • 2 Y

Y by centslordm, cubres

Let $a$ be the product of the fractions not flipped and $b$ the product of the fractions which are flipped. Then we have $ab=n$ and $\tfrac{a}{b}=1$ , implying $a^2=n$ . Since $a$ is rational it follows that $n$ must be a perfect square. On the other hand, if $n=m^2$ , we can flip $\tfrac{2}{1},\tfrac{3}{2},\ldots,\tfrac{m}{m-1}$ .

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Afo

#13 h Sep 6, 2021, 9:33 AM • 1 Y

Y by cubres

Solution. The answer is all perfect squares $> 2$ . Note that the product of all the numerators and the denominators must have even powers so $n!(n-1)! = (n-1)!^2 \times n$ is a perfect square and so is $n$ . The construction is to divide into $\sqrt{n}+1$ groups of $\sqrt{n}-1$ fractions and reciprocate every group except the first. Here's an example for $n = 9$ .
$\left( \frac{2}{1} \cdot \frac{3}{2} \right)\cdot \left( \frac{4}{3} \cdot \frac{5}{4} \right)^{-1}\cdot \left( \frac{6}{5} \cdot \frac{7}{6} \right)^{-1}\cdot \left( \frac{8}{7} \cdot \frac{9}{8} \right)^{-1} = 1$

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#14 h Jun 11, 2022, 11:47 PM • 1 Y

Y by cubres

The answer is $n=k^2$ for some positive integer $k>1$ . Upon reciprocating any fraction, the product is multiplied by the perfect square of a rational number. As a result, the original product must also have been a perfect square by invariance.

Furthermore, for $n = k^2$ , simply reciprocate the first $k-1$ fractions (which multiply to $k$ ). Then the product is multiplied by $\frac 1{n^2}$ , and thus it will equal 1.

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#15 h Apr 3, 2023, 1:42 PM • 2 Y

Y by centslordm, cubres

If $n=a^2$ , we flip $\frac{2}{1}$ , $\frac{3}{2}$ , $\dots$ , $\frac{a}{a-1}$ over and our product becomes $\underbrace{\frac{1}{a-1}}_{\text{first } a-2\text{ terms}}\cdot\underbrace{\frac{1}{a+1}}_{\text{terms }a+1\text{ to }a^2-1}\cdot\frac{a-1}{a}\cdot\frac{a+1}{a}\cdot\frac{a^2}{a^2-1}=1$ $\square$

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#16 h Apr 23, 2023, 4:18 PM • 6 Y

Y by Taco12, Jndd, Significant, player01, mulberrykid, cubres

We claim that only positive perfect squares work. Notice that if $n$ is not a perfect square, then we have that the total product of all integers on the top and on the bottom is not a perfect square, meaning that we can never have the top and bottom divide each other out to equal $1$ . Now let $n=k^2$ . We now have
$\frac{1}{2}*\frac{2}{3}*\cdots{}*\frac{k-1}{k}=\frac{1}{k}$ and
$\frac{k+1}{k}*\frac{k+2}{k+1}*\cdots{}*\frac{k^2}{k^2-1}=k$
Multiplying these together gives us our final product of $1$ , as desired, and we are done.

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trk08

#17 h Apr 23, 2023, 5:23 PM • 1 Y

Y by cubres

When multiplying these all out in regular order, we get $n$ as given in the problem. If we take the reciprocal of some of them, let us say that there product is $x$ . This means that the total sum would turn out to be $n\cdot x^2$ which we claim to be $1$ . Simplifying this, we get:
$x=\frac{1}{\sqrt{n}}.$ We also know that $x$ has to be rational meaning that $n$ has to be a perfect square which we can say is $a^2$ . If we substitute this back in, we are basically saying that the original sequence is:
$\frac21, \frac32, \cdots, \frac{a^2}{a^2-1},$ and we want a product of the reciprocal of some of these to be $\frac{1}{a}$ . We know that $a<a^2$ , so we can say that a possible value of $x$ could be:
$\frac{1}{2}\cdot \frac{2}{3}\cdot \dots \frac{a-1}{a}$ $\frac{1}{a},$ which means that $x^2\cdot n=1$ .

Thus, if $n$ is a perfect square, this condition works.

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#18 h Apr 23, 2023, 6:40 PM • 1 Y

Y by cubres

It is quite intuitive that to get a product of $1$ , we need to have a product that cancels in the numerator and denominator. For these to equal the same product, $n$ would have to be the product of these - or $n=k^2$ for the product $k$ , thus $n$ must be a perfect square.

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#19 h May 5, 2023, 6:28 PM • 1 Y

Y by cubres

$\textbf{Claim:}$ $n$ must be a perfect square.

$\emph{Proof:}$ Let $\{t_1, t_2, \dots, t_k\}$ be a subset of $\{2, 3, \dots, n\}$ such that all the numbers of the form $\tfrac{t_i}{t_i-1}$ are reciprocated.

Notice that reciprocating a factor $a$ of $n$ gives the resulting number as $\tfrac{n}{a^2}$ . Therefore, reciprocating all the numbers of the form $\tfrac{t_i}{t_i-1}$ gives our end result as
$\frac{n}{\left(\frac{t_1}{t_1-1}\cdot\frac{t_2}{t_2-1}\cdot\dots\cdot\frac{t_k}{t_k-1}\right)^2}=1$ implying that we require
$\frac{t_1}{t_1-1}\cdot\frac{t_2}{t_2-1}\cdot\dots\cdot\frac{t_k}{t_k-1} =\sqrt{n}.$ Since the LHS product is rational, we require the RHS to be rational as well. But $n$ is an integer, it isn't a fraction. Hence, $n$ must be a perfect square. $\square$

$\textbf{Construction:}$ Let $n=k^2$ , notice that
$\frac{k^2}{k^2-1}\cdot\frac{k+1}{k}\cdot\frac{k-1}{k-2}\cdot\frac{k-2}{k-3}\dots\frac{3}{2}\cdot\frac{2}{1}=k.$ So the answer is $\boxed{\text{all perfect squares greater than 1}}$ .
$\blacksquare$

motivation

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Alcumusgrinder07

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Alcumusgrinder07

#21 h Sep 10, 2023, 5:32 PM • 1 Y

Y by cubres

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#22 h Sep 28, 2023, 4:06 PM • 1 Y

Y by cubres

The product is equal to $n$ . To flip a fraction $\frac{m}{m+1}$ , multiply by $\frac{(m+1)^2}{m^2}$ . Define "perfect square" as the square of a rational number. If $n$ is not a perfect square, then multiplying by a perfect square keeps the product not a perfect square. So the product can never reach $1$ , which is a perfect square. So $n$ has to be a perfect square. If $n$ is a perfect square, flipping the first $\sqrt{n}-1$ fractions results in the product being $1$ .

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#23 h Oct 7, 2023, 5:19 PM • 1 Y

Y by cubres

The answer is $n$ such that $n$ is a perfect square. This can be achieved by
$\left( \frac{2}{1}\cdot \frac{3}{2} \cdot \dots \cdot\frac{\sqrt{n}}{\sqrt{n}-1} \right) \cdot \left( \frac{\sqrt{n}}{\sqrt{n}+1} \cdot \frac{\sqrt{n}+1}{\sqrt{n}+2} \cdot \dots \cdot \frac{n-2}{n-1}\right) \cdot \frac{n-1}{n} = 1$ Notice the product of the reciprocated fractions is $n$ if we flip a fraction $\frac{a}{b}$ we multiply by $\frac{b^{2}}{a^{2}}$ so each prime factor gets changed by a multiple of $2$ hence $n$ must be a square since $1$ is a square.

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#24 h Oct 17, 2023, 9:31 PM • 1 Y

Y by cubres

We claim the answer is all perfect squares. A construction for $n=k^2$ is

$\left(\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdots \cdot \frac{k-1}{k} \right) \left( \frac{k+1}{k}\cdot \frac{k+2}{k+1}\cdots \cdot \frac{n}{n-1} \right).$
Now, we must prove that only perfect squares work. Notice that the numerator and denominator must be equal, or the product of the numerator and denominator must be a perfect square. Now, this value always stays constant throughout, so we have that

$n!(n-1)!=n((n-1)!)^2$
is a perfect square, implying that $n$ must be a perfect square.

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#25 h Dec 8, 2023, 4:48 PM • 1 Y

Y by cubres

The answer is when $n$ is a perfect square greater than one.

If the product can be made into $1$ , the product of the numerator and the denominator must be a square. Thus $(n-1)!\cdot n!$ is a square, so $n$ is a square. When $n$ is a square, we can flip over $\frac21,\frac32,\dots,\frac{\sqrt{n}}{\sqrt{n}-1}$ and win. $\blacksquare$

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#26 h Dec 29, 2023, 3:54 AM • 1 Y

Y by cubres

The answer is all $n$ which are perfect squares. Note that the product of all the numerators must be equal to the product of all of the denominators of the fractions (after some are flipped) for the product to be $1$ . Hence $1 \cdot 2^2 \cdot 3^2 \cdots (n - 1)^2 \cdot n$ must be a perfect square, so $n$ is a perfect square.

For $n = k^2$ , notice that
$\frac{k^2}{k^2 - 1} \cdot \frac{k^2 - 1}{k^2 - 2} \cdots \frac{k + 1}{k} \cdot \frac{k - 1}{k} \cdot \frac{k - 2}{k - 1} \cdots \frac{1}{2} = \frac{k^2}{k \cdot k} = 1.$

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cursed_tangent1434

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cursed_tangent1434

#27 h Dec 29, 2023, 4:07 AM • 2 Y

Y by GeoKing, cubres

We claim that the answer is any square number. Let an integer which satisfies the required condition be called good.

Claim : Any $n$ such that $v_p(n)=k$ where $k$ is odd is not good
Proof : Notice that if the numerator of a certain fraction is divisible by a certain prime, the denominator is not, and vice versa. Thus, if $v_p(n!)+v_p(n-1)!$ is an odd number it is impossible to divide it by two (such that $v_p$ of the numerator is equal to the denominator - so that the final product is 1). Now, notice that when $n=p^km$ for odd $k$ , and $p \nmid m$
$v_p(2^km!) + v_p((2^km-1)!) = 2^k + (2^k - k) = 2^{k+1} - k$ Thus, if $k$ is odd $v_p(n!)+v_p(n-1)!$ , is also clearly odd. Thus, all odd numbers are not good as claimed.

Now, we will show that sqaure numbers are in fact good.

Algorithm : All squares numbers $n$ are clearly good, by the following algorithm.
Simply reciprocate the first $k-1$ fractions. Thus, we obtain,
$\frac{1}{2}\cdot \frac{2}{3} \cdots \frac{k-1}{k}\cdot \frac{k+1}{k} \cdot \frac{k+2}{k+1} \cdots \frac{k^2}{k^2-1}$
Proof : Clearly, this product $P = \frac{k^2}{k\cdot k} = 1$ , thus all squares obviously work.

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#28 h Dec 30, 2023, 10:49 PM • 1 Y

Y by cubres

The boolean "product is the square of a rational" is invariant. Thus $n$ must be a $\boxed{\text{perfect square}}$ . Our construction is to flip the first $\sqrt{n}-1$ fractions. $\blacksquare$

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#29 h Jun 30, 2024, 5:07 AM • 1 Y

Y by cubres

Note that flipping any of the fractions is equivalent to dividing the product by a rational square, so if $n$ is a value such that the product can be $1$ , then $n$ must be a perfect square $>1$ . But note that all such n work; indeed, let $n = k^2$ .Then we can flip the fractions $\frac{2}{1}, \cdots, \frac{k}{k-1}$ . This divides the original product by $k^2 = n$ , so the final product becomes $1$ , which is what we wanted. $\square$

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#30 h Oct 8, 2024, 4:56 PM • 1 Y

Y by cubres

Solution

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eg4334

#31 h Dec 6, 2024, 3:06 AM • 1 Y

Y by cubres

First off, considering any prime $p$ after we flip it the $v_p$ of the numerator is equal to the $v_p$ of the denominator, hence the product of the numerator and denominator is a square, i.e: $n! \cdot (n-1)! \text{ = square} \implies n \text{ = square}$ I claim $\boxed{\text{all squares}}$ work. We use induction. $2^2$ case is obvious. If the result is true for $k^2$ , then to construct $(k+1)^2$ look at the final $2k+1$ terms because the first $k^2$ already work. Out of the last $2k+1$ terms, flip the last $k+1$ . Keep the rest the same. The product of the last $2k+1$ then turns into $\frac{k^2+k}{k^2} \cdot \frac{k^2+k}{(k+1)^2} = 1$ , done.

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#32 h Dec 31, 2024, 4:26 PM • 1 Y

Y by cubres

We claim that all perfect squares $n \ge 4$ work.

Let the product of the fractions we wish to reciprocate be $\frac{a}{b}$ and the product of the remaining fractions be $k$ . Clearly we have that:
$k\cdot\frac{a}{b} = n \:\:\: \text{and} \:\:\: k\cdot\frac{b}{a} = 1.$ Dividing equations yields
$n = \left(\frac{a}{b}\right)^2$ since $n$ is an integer it must be a perfect square.

It is also easy to see that the first $\sqrt{n}-1$ terms multiply to $\sqrt{n}$ , and as a result the last $n-\sqrt{n}$ terms also multiply to $\sqrt{n}.$ Reciprocate either the first $\sqrt{n}-1$ terms or last $n-\sqrt{n}$ terms and we are done. $\square$

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#33 h Dec 31, 2024, 4:33 PM • 1 Y

Y by cubres

Suppose that we reciprocate the fractions $a_1, a_2, \cdots, a_k,$ which should obviously be distinct. Then we have that $\frac{n}{(a_1a_2 \cdots a_k)^2} = 1 \implies n = (a_1a_2 \cdots a_k)^2.$ Thus $n$ is a perfect square.

Now, we show that all perfect squares greater than $1$ work. For the construction, consider $a_1=\frac{2}{1}, a_2 = \frac{3}{2}, \cdots a_k = \frac{\sqrt{n}}{\sqrt{n}-1}.$

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sansgankrsngupta

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sansgankrsngupta

#34 h Feb 12, 2025, 12:40 PM • 1 Y

Y by cubres

OG! The answer is all perfect squares $n$
Construction:
Let $n=k^2$ , reciprocate the fractions $\frac{2}{1}, \frac{3}{2} \ldots \frac{k}{k-1}$ Proof of necessity:
$P= \frac{2}{1}* \frac{3}{2} \cdots * \frac{n}{n-1}=n$ Assume you reciprocate some of the fractions among the fractions to get 1 .
We view reciprocation of a fraction $f$ as $\frac{1}{f}=f.\frac{1}{f^2}$ . Hence we are essentially multiplying $P$ by the square of some rational number.
Hence we get $Pq^2=nq^2=1 \iff n= (\frac{1}{q})^2$ , where $q$ is a rational number. so $n$ is also the square of some rational number.
$n= \frac{p^2}{q^2}$ where $(p,q)=1$ and $p$ and $q$ are integers since $n$ is an integer we have that $p^2|q^2$ but we know that $gcd(p^2,q^2)=1$ . Thus, $p^2=1$ . Hence $n=p^2$ as desired.

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#35 h Apr 14, 2025, 1:33 AM

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The answer is all $n=k^2$ .

Let $c=p_1^{e_1}p_2^{e_2}\cdots$ be the ``prime factorization" of the product of (possibly reciprocated) fractions, where each $e_i$ is a not necessarily positive integer. Clearly, the parity of each $e_i$ is invariant no matter how we reciprocate the fractions. Because $1=p_1^0p_2^0\cdots$ , we know that $n$ must be a perfect square (as the product of the fractions can equal $n$ ).

When $n=k^2$ , it's easy to see that reciprocating the first $k-1$ fractions yields a product of $1$ , as desired. $\blacksquare$

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#36 h Apr 21, 2025, 3:53 PM

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We claim that we can do this for all $n=k^2>2$ . Note that each time we flip a fraction $\frac{a}{b}$ it is equivalent to multiplying it by $\frac{b^2}{a^2}$ . So this is saying that for some $S^2$ where $S \in \mathbb Q$ , representing the product of all the inverses of fractions we are flipping, we have $nS^2=1 \implies n=(\frac{1}{S})^2$ . Since $n \in \mathbb Z$ we must have $\frac{1}{S} \in \mathbb Z$ and thus $n$ is a perfect square.

Now we will flip every fraction up until $\frac{\sqrt{n}}{\sqrt{n}-1}$ and keep the rest the same. Then we have $\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{\sqrt{n}-2}{\sqrt{n}-1}\frac{\sqrt{n}-1}{\sqrt{n}}\frac{\sqrt{n}+1}{\sqrt{n}}\cdots\frac{n}{n-1}=\frac{n}{\sqrt{n}\sqrt{n}}=1$

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#37 h Apr 25, 2025, 3:18 PM

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I claim that all perfect squares $n > 1$ are valid. Let $n = k^2.$ Thus, $1$ can be made from: $\frac12 \cdot \frac23 \cdot \ldots \cdot \frac{k - 1}{k} \cdot \frac{k + 1}{k} \cdot \ldots \cdot \frac{k^2}{k^2-1} = 1$ Now, I claim that if $n$ isn't a perfect square, then it won't work. Notice that flipping a fraction $\frac c{c-1}$ will just multiply the current value by the square number $\frac{(c-1)^2}{c^2}.$ Thus, since $n$ isn't a perfect square, you can never achieve another perfect square, $1.$

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