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Proofathon: This Weekend

Hello fellow mathematicians,

This upcoming weekend marks the inaugural Proofathon contest. Essentially, Proofathon is a USAMO-like, online math contest administered over the course of two days, eight problems. Show your support and compete in the inaugural Proofathon! Also, note that if we get over 200 likes on our Facebook page, we will have a special surprise.

Until then, I'll leave you with one of our shortlisted problems:

Show that none of the elements in the set \(\{49,409,4009,\dots\}\) is a perfect cube.

Note by Cody Johnson
3 years, 10 months ago

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Solution: Cubes are always \(-1,0,1\pmod 9\), and since the given numbers are all \(4\pmod 9\), there are no cubes in the set. \(\blacksquare\) Daniel Chiu · 3 years, 10 months ago

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@Daniel Chiu How many perfect fifth powers can be written in the form of \(9\cdot2^k-4\)? Cody Johnson · 3 years, 10 months ago

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@Cody Johnson Let \(m^5=9\cdot 2^k+6\). From this, we see that \(3|m^5\implies 3|m\), and so \(9|m^5\). However, \(9\cdot 2^k+6\equiv 6\pmod 9\), a contradiction. Therefore, the answer is \(\boxed{0}\).

This can also be found by writing out the quintic residues modulo 9. Daniel Chiu · 3 years, 10 months ago

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@Daniel Chiu I messed up (addition error): \(9\cdot2^k-4\) Sorry! Cody Johnson · 3 years, 10 months ago

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@Cody Johnson Let \(m^5=9\cdot 2^k-4\).

If \(k<5\), then we can check manually: \[\begin{align} &k=0\implies m^5=5\text{, no solution} \\ &k=1\implies m^5=14\text{, no solution} \\ &k=2\implies m^5=32,\ m=2 \\ &k=3\implies m^5=68\text{, no solution} \\ &k=4\implies m^5=140\text{, no solution} \end{align}\]

Otherwise, if \(k>4\), then \(2|9\cdot 2^k-4\implies 2|m\). Then, \(32|9\cdot 2^k-4\). Hence, \(32|9\cdot 32\cdot 2^{k-5}-4\), which is not possible.

Therefore, the answer is \(\boxed{1}\). \(\blacksquare\) Daniel Chiu · 3 years, 10 months ago

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@Daniel Chiu I did modulo 8 with three cases, but that works as well! Good job! Cody Johnson · 3 years, 10 months ago

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@Daniel Chiu Is there some easy way to figure out that the numbers are all congruent to \(4 \pmod 9\)? This is how I figured it:

Considered modulo 9, the terms are \(4*10^k\) for all \(k \geq 1\). It can be shown that the numbers \(36, 396, 3996, 39996, 399....96\) are all multiples of \(9\), namely they are \(9 * 44...4\), so thus the numbers \(40, 400, 4000, 40000...\) are congruent to \(4 \pmod 9\). Michael Tong · 3 years, 10 months ago

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@Michael Tong More simply, \(4\cdot10^k+9\equiv4\cdot1^k+9\equiv\boxed{4}\pmod{9}\) Cody Johnson · 3 years, 10 months ago

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@Michael Tong Let \(S(n)\) denote the sum of the digits of \(n\). Then, \(n\equiv S(n)\pmod 9\) (this is not hard to prove). Daniel Chiu · 3 years, 10 months ago

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@Daniel Chiu You can always search for Proofathon in facebook :) P.S.-The link is working fine for me. Soham Chanda · 3 years, 10 months ago

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If anything, look at our logo! Cody Johnson · 3 years, 10 months ago

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Note that the date has changed, and Proofathon is starting this Friday at midnight (which is in 4 hours). Cody Johnson · 3 years, 10 months ago

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how do I submit the problems??is there any registration?? Jawwad Siddique · 3 years, 10 months ago

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@Jawwad Siddique Hey Jawwad,

If you want to submit original problems for consideration for use on future contests, you can email us at Proofathon@gmail.com. If you are asking where to submit solutions to the problems for the contests, visit our contest page http://proofathon.org/Pages/contests.php on the date of the contest. The first contest will be held this weekend from October 26 at 12:00 AM EDT to October 27 at 11:59 EDT.

To register, simply create an account at http://proofathon.org/Pages/account.php and log on with that account on the contest date. Logan Dymond · 3 years, 10 months ago

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The contest is ongoing. One can find the problems here. Cody Johnson · 3 years, 10 months ago

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