Nepal TST 2025 Day 1 Problem 2

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Nepal TST 2025 Day 1 Problem 2

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Source: Nepal TST 2025, problem 1

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#1 h Apr 11, 2025, 1:26 PM • 1 Y

Y by khan.academy

Find all integers $n$ such that if
$1 = d_1 < d_2 < \cdots < d_{k-1} < d_k = n$ are the divisors of $n$ , then the sequence
$d_2 - d_1,\, d_3 - d_2,\, \ldots,\, d_k - d_{k-1}$ forms a permutation of an arithmetic progression.

(Kritesh Dhakal, Nepal)

This post has been edited 4 times. Last edited by Bata325, Apr 13, 2025, 2:55 AM
Reason: italics

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#2 h Apr 12, 2025, 12:20 PM

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Assuming integers means naturals and neglecting $n=1$ since then the sequence has no elements.

Claim : $k \geq 5$ is not possible.
Pf : Assume not, note that sum of all of these elements is $n-1$ and $d_k - d_{k-1} = n - \frac{n}{d_2} \geq n - \frac{n}{2} = \frac{n}{2}$ . This must be the last ( $k-1$ th) term of the AP (arranged in increasing order) or the sum of it and the next term will be at least $n > n-1$ which is impossible. Suppose the $n$ th term of the AP is $a_n$ and common difference is $d$ . $n - 1 > a_{k-3} + a_{k-2} + a_{k-1} = 3 a_1 + (3k-9)d > 2(a_1 + (k-2)d) \geq n$ which is not possible.

For $k \leq 4$ the only cases are $n = pq$ or $n=p^2$ or n= $p$ for primes $p$ and $q$ .

$n=p$ always works since a single number is always an AP.

$n=p^2$ always works since 2 numbers are also always in an AP.

$n=pq$ with $q > p$ is the only case that needs to be checked. The numbers we need to check in an AP are $p-1, q-p, pq - q$ . The argument above that showed $d_k - d_{k-1}$ is max element did not use $k \geq 5$ and still holds so $pq - q$ is max element here. So this works $n$ iff either $2(q-p) - (p-1) = pq - q$ or $2(p-1) - (q-p) = pq - q$ .

In the first case SFFT gives $(q+3)(p-3) = -8$ which isn't possible for primes bigger than $2$ so $p=2, q=5$ is only soln and hence only $n=10$ works.

In the second case SFFT gives $p(q-3) = -2$ which be similar argument as above gives $p=q=2$ but $q>p$ so there are no solns in this case.

To conclude the only solutions are $n$ is prime, square of a prime or $10$ (or 1 ig but that really breaks the statement so sush) .

If you really want to consider integers we still have no issue since we are only looking at positive divisors so who cares.

This post has been edited 1 time. Last edited by ThatApollo777, Apr 12, 2025, 12:24 PM
Reason: moral reasons

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