7

From STEMS (by CMI) of year 2025:

Find the smallest natural number $k$ such that there exist (possibly negative) integers $a_1, a_2, \dots, a_k$ whose eleventh powers add up to $2057$.

My thoughts:

So I got this question correct that $k = 10$ where $a_1 = 2$ and the rest $a_2 = \dots = a_9 = 1$, the eleventh powers of which add up to $2057$, but I have tried to conclude this in a generalized form because I have no explanation that why $3$ or $4$ or any other integer cannot come in this equation.

I have thought of including $3^{11}$ and then realised that to balance it to around $2000$ I have to subtract $86$ times $2^{11}$. See how much difference there is between $3^{11}$ and $2^{11}$ but after checking, I also understood that there is not much difference between $9^{11}$ and $8^{11}$.

For whatever reason have the $1$ to $20$ numbers each raised to the power 11 pdf file: 1 to 20 each raised to power 11

My question is that how can we generalise this solution to eliminate all the other possibilities?

Regarding to the similar problem:
The similar problem provides two solutions, both focusing on proving that a certain Diophantine equation is completely unsolvable using modular arithmetic. However, my question is specifically about determining the minimum number of terms required to sum to 2057.

Additionally, the generalized form of the answer was added after I posted my question. At the time of posting, my question was not a duplicate, as the generalized solution was later extended to address both types of problems: proving an equation unsolvable and finding the minimum number of terms to sum to 2057.

Blue
  • 83,939
Name not Allowed
  • 447
  • 1
    Are you interested in a quite simple proof that $k=10$ is the correct answer, one which doesn't involve providing an "... explanation that why 3 or 4 or .. any other integer cannot come in this equation" (since I'm not completely sure if that's even true and, if it is, how to prove it, i.e., that there are not another $10$ values for which the sums or differences to the eleventh powers add up to $2057$)? – John Omielan Jan 02 '25 at 08:45
  • k = 10 is the correct that I know, but I was not sure that 3 or 4 or something else can change the answer or not, But because of luck k = 10 was correct, now I am trying and searching for a full answer, also there may be another 10 values for which the sums or differences to the eleventh powers may be 2057 – Name not Allowed Jan 02 '25 at 08:50
  • It's not because of "luck" that $k=10$ is correct. As I stated, if you just want a full proof to the question itself, I can offer that. However, note it doesn't involve showing that are no other $10$ values with sums or differences to the eleventh powers that come up to $2057$, in particular that doesn't involve any other specific values, such as $3$, $4$, etc., cannot come into the equation. If you already know how to prove that $10$ is the actual minimal # of values, I suggest possibly adding that to your question text so that, for example, nobody wastes their time writing it in an answer. – John Omielan Jan 02 '25 at 08:52
  • Please share that kind of answer too, as I am a learner I want to learn every aspect of it. Please share the proof that how 10 is the minimum value for 'k' – Name not Allowed Jan 02 '25 at 09:04
  • 1
    Duplicate of this and many others, which was wrongly reopened due to meta question by OP, which doubled the number of votes on the question and dupe answer (likely most of which are in protest to EoQS). As I mentioned a few days ago in a comment on the answer, the well-known method used in the answer already occurs in many other places. – Bill Dubuque Jan 05 '25 at 20:26
  • 1
    Re: your edit: you accepted an answer which is an instance of the general method made clear in said dupe (and many other answers - as I linked). As such you question is a duplicate per site policy - see abstract duplicates and EoQS – Bill Dubuque Jan 05 '25 at 21:55
  • @BillDubuque Ok I understood the reason. – Name not Allowed Jan 06 '25 at 09:22

1 Answers1

13

By Fermat's little theorem, since $23$ is prime, then for all $a$ which are not a multiple of $23$,

$$a^{22}\equiv 1\pmod{23} \;\to\; (a^{11}-1)(a^{11}+1)\equiv 0\pmod{23} \;\to\; a^{11}\equiv \pm 1\pmod{23}$$

Of course, if $a$ is a multiple of $23$, then $a^{11}\equiv 0\pmod{23}$. Since $2057\equiv 10\pmod{23}$, the smallest possible number of values, each to the eleventh power, which may be added or subtracted to get $2057$ is $10$. In particular, no values are integral multiples of $23$, while each value congruent to $1$ modulo $23$ is added, and each value congruent to $-1$ modulo $23$ is subtracted.

To confirm that $k = 10$ is the actual smallest number of values, we need to just find one set (although there may be more than one) of $10$ values which sum to $2057$, which you've already done with $2^{11} + 9\times 1^{11} = 2057$.

John Omielan
  • 52,653
  • 2
    Duplicate Of course this method already exists in many prior answers, e.g. when handling sums of $\rm\color{#c00}{cubes}$ by working $!\bmod 7 = 1+2\cdot\color{#c00}3,,$ e.g. here, and here and here and surely many others. Please strive to search and organize dupes. – Bill Dubuque Jan 02 '25 at 10:09
  • 11
    I don’t believe this is a duplicate, as I have thoroughly searched for this question and didn't found the same question anywhere else. While it’s possible that a similar method exists somewhere within this extensive library, it’s not feasible for me to review every answer. I request you to please understand my situation – Name not Allowed Jan 02 '25 at 10:35
  • 1
    I believe that the comment from @Bill was directed toward John, not toward you, Name. No need to take it personally. – Gerry Myerson Jan 05 '25 at 14:51