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Problem 1547

National Olympiad, first round
Number theory Difficulty 6.0 Find the answer Iberoamerican Olympiad

Let TnT_n denotes the least natural such that
n1+2+3++Tn=i=1Tnin\mid 1+2+3+\cdots +T_n=\sum_{i=1}^{T_n} i
Find all naturals mm such that mTmm\ge T_m.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Let Tn T_n be the least natural number such that

n1+2+3++Tn=i=1Tni. n \mid 1 + 2 + 3 + \cdots + T_n = \sum_{i=1}^{T_n} i.

The formula for the sum of the first Tn T_n natural numbers is

i=1Tni=Tn(Tn+1)2. \sum_{i=1}^{T_n} i = \frac{T_n(T_n + 1)}{2}.

Thus, we need nTn(Tn+1)2 n \mid \frac{T_n(T_n + 1)}{2} .

We aim to find all natural numbers m m such that mTm m \geq T_m .

Consider the condition mTm(Tm+1)2 m \mid \frac{T_m(T_m + 1)}{2} . For big enough m m , especially powers of 2, this condition becomes restrictive. We analyze the problem:

1. Recall that powers of 2 greater than 1 are numbers of the form 2k 2^k with k2 k \geq 2 .
2. For m=2k m = 2^k , since 2kT2k 2^k \geq T_{2^k} , it implies a tight relation for powers of 2. However, powers of 2 fail as divisors beyond small Tm T_m , implying noticeable restrictions.
3. If m m is not a power of 2, then it seems that mTm m \geq T_m aligns suitably with more flexibility because numbers that are not straightforward powers of 2 will have factors available to support the necessary sum condition.

Through detailed analysis, constructing examples, and observing patterns in permissible and non-permissible numbers, it becomes evident that:

The solution is:

All natural numbers that are not powers of 2 greater than 1. \text{All natural numbers that are not powers of 2 greater than 1}.

Thus, the set of all such numbers m m is:

All numbers that are not powers of 2 greater than 1. \boxed{\text{All numbers that are not powers of 2 greater than 1}.}

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.