Hey Brilliantinians! Here I have some amazing problems.

**Problem 1** Let \(x, y, z\) be natural numbers. Find all triples such that:

\(\sqrt[6x-y^{2}-7]{\sqrt[6y-z^{2}-7]{\sqrt[6z-x^{2}-7]{728x+729y+730z}}}\) is a positive integer.

**Problem 2** Proof that for any prime diferent that 2 and 5, there exists a multiple of p whose digit numbers are all 9's. For example, for \(p = 13 \rightarrow 999999 = 13 \cdot 76923\).

**Problem 3**
Natural numbers "a" and "b" satisfy that:

\(\frac{a+1}{b} + \frac{b+1}{a}\) is an integer. Proof that the greatest common divisor of \((a,b) \le \sqrt{a + b}\)

## Comments

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TopNewestFor Problem 2, we know that strings of n 9's are equal to 10^n - 1. Now by Fermat's Little Theorem, 10^(p-1) - 1 = 0 mod p where p is coprime to 10. As p is not 2 or 5 this statement is therefore true hence the result follows. – Michael Ng · 2 years, 2 months ago

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-We know that strings of n 9's are in the form \(10^n - 1.\)

Now by Fermat's Little Theorem, \(10^{p-1}-1 \equiv 0 \pmod p\) if \(\gcd(p,10) = 1\) As \(p\) is not \(2\) or \(5\) this statement is therefore true hence the result follows.

EXAMPLE: That means if we choose for example \(p = 7 \rightarrow 10^{6} - 1 = 7k \)

NOTE: Fermat's Teorem doesn't give us the minimum 9's string that achieves the condition.

For example for \(p = 13\) using Fermat's Theorem we get \( 10^{12}- 1 = 999999999999 \equiv 0 \pmod{13}\) but in the example we saw \(999999 = 13 \cdot 76923 \) – Jordi Bosch · 2 years, 2 months ago

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Just something interesting I found. – Julian Poon · 2 years, 2 months ago

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– Michael Ng · 2 years, 2 months ago

Yes, thank you for the formatting! In fact I think the lowest power of 10 which works is called the order of \(10\bmod p \). Nice problem.Log in to reply

This is another variant of the problem.

– Calvin Lin Staff · 2 years, 2 months agoLog in to reply

If you don't remove the 0s then it becomes a proof for the problem below by @Calvin Lin – Joel Tan · 2 years, 2 months ago

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– Michael Ng · 2 years, 2 months ago

That is a very nice solution.Log in to reply