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Source: Indian TST 2019 Practice Test 1 P3

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#1 h Jul 17, 2019, 12:52 PM • 2 Y

Y by harshmishra, Adventure10

Let $n\ge 2$ be an integer. Solve in reals:
$|a_1-a_2|=2|a_2-a_3|=3|a_3-a_4|=\cdots=n|a_n-a_1|.$

This post has been edited 1 time. Last edited by ayan.nmath, Jul 17, 2019, 1:03 PM

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#2 h Jul 17, 2019, 1:28 PM • 8 Y

Y by Pluto1708, meet18, ashrith9sagar_1, aops29, harshmishra, khina, Adventure10, kiyoras_2001

The title is spoiler

ayan.nmath wrote:

Let $n\ge 2$ be an integer. Solve in reals:
$|a_1-a_2|=2|a_2-a_3|=3|a_3-a_4|=\cdots=n|a_n-a_1|.$

Solution. If all are equal, we get an obvious solution. Assume the otherwise. The problem statement is nothing but $\pm\frac{1}{1}+\pm \frac{1}{2} + \ldots + \pm \frac{1}{n} = 0.$ Pick a prime $p$ between $\frac{n}{2}$ and $n$ (Bertrand's). We will have $p\mid 1\cdot 2\cdot \ldots \cdot (p-1) \cdot (p+1)\cdot \ldots n$ . Therefore $n\geq 2p$ which is false. Therefore the only solution is when $a_1 = a_2 = \ldots = k$ for some constant $k$ .

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#3 h Jul 17, 2019, 1:29 PM • 2 Y

Y by ashrith9sagar_1, Adventure10

Answer
solution

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#4 h Jul 17, 2019, 1:30 PM • 3 Y

Y by aops29, Adventure10, Mango247

Seems like you all had great fun at IMOTC

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#5 h Jul 17, 2019, 1:32 PM • 2 Y

Y by Adventure10, Mango247

Yes we did.

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Kayak

#6 h Jul 17, 2019, 3:05 PM • 2 Y

Y by AlastorMoody, Adventure10

Of course we need to show that $\sum_{1 \leq i \leq n} \frac{\varepsilon_i}{i}$ is never zero for $\varepsilon_i \in \{-1, 1 \}$ ; this is easily done by inducting to prove $\nu_2 \left( \sum_{1 \leq i \leq n} \frac{\varepsilon_i}{i} \right) = \lfloor \log_2 n \rfloor$ using $\nu_2(a+b) \leq \max( \nu_2(a), \nu_2(b) )$ with equality if $\nu_2(a) \neq \nu_2(b)$ .

This post has been edited 2 times. Last edited by Kayak, Jul 17, 2019, 3:18 PM
Reason: remark is unfotunately left out inadvertently

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#7 h Mar 21, 2023, 1:46 PM

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We show that the only solution is $a_1=a_2=\cdots=a_n$ , which clearly satisfies.

Suppose $|a_1-a_2|=d\ne0$ , then we have $d\left(\pm\frac{1}{1}\pm\frac{1}{2}\pm\cdots\pm\frac{1}{n}\right)=0$ .
By Bertrand's Postulate, there is always a prime $p$ such that $\frac{n}{2}<p\leq n$ if $n\geq2$ .
Take the inverse of this prime on left side of the equation and the others on the right side. Clearly, the denominator of the RHS is not divisible by $p$ , but that of the LHS is. This gives the required contradiction.

So, the only solution is $a_1=a_2=\cdots=a_n$ , as claimed. $\blacksquare$

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#8 h Oct 11, 2024, 9:45 PM

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The answer is $a_1 = a_2 = \cdots = a_n$ . This clearly works. Now we show it is the only solution. It suffices to show that $|a_1 - a_2| = 0$ . Suppose otherwise. Let $|a_1 - a_2| = k \ne 0$ . We have $|a_i - a_{i + 1}| = \frac ki$ for all $i \in [n]$ , where $a_{n+1} = a_1$ . Now, for each $i \in [n]$ , let $c_i \in \{-1,1\}$ satisfy $a_{i + 1} - a_i = c_i \cdot \frac ki$ . We have $0 = a_{n+1} - a_1 = \sum_{i = 1}^{n} (a_{i + 1} - a_i) = \sum_{i=1}^n c_i \cdot \frac ki$
Dividing both sides by $k$ (since $k$ is nonzero) gives that $\sum_{i=1}^n \frac{c_i}{i} = 0$
Let $p \in \left( \frac n2 ,n \right]$ be a prime, which exists by Bertrand's Postulate since $n > 1$ ).

Note that for each $i \in [n]$ not equal to $p$ , we have $\nu_p \left( \frac{c_i}{i} \right) = 0$ (as there are no other multiples of $p$ in $[1,n]$ ), but $\nu_p \left( \frac{c_p}{p} \right) < 0$ . So the $\nu_p$ of the sum must be less than zero (as we are summing a number with negative $\nu_p$ with numbers of nonnegative $\nu_p$ ). Hence the sum cannot equal to zero, which is a contradiction.

Therefore, the sequence is constant.

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243 posts

pie854

#9 h Oct 12, 2024, 9:58 AM

Y by

Let $a_{n+1}:=a_1$ . We have $(i-1)|a_{i-1}-a_i|=i|a_i-a_{i+1}|\Rightarrow i a_{i+1} \in \left \{a_i+(i-1)a_{i-1}, \ (2i-1)a_i-(i-1)a_{i-1} \right\},$ for all $2\leq i\leq n$ . From this it's easy to prove by induction that for every $2<i\leq n+1$ there is a number $c_i$ such that $a_i=c_i a_2+(1-c_i)a_1 \qquad (1)$ So $a_1=c_{n+1}a_2+(1-c_{n+1})a_1$ , i.e. $a_1=a_2$ . Now from $(1)$ it follows that $a_i=a_1$ for all $i\leq n$ .

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