Cool Problems (Part 3)

Hey Brilliantinians! Here I have some amazing problems.

Problem 1 Let x,y,zx, y, z be natural numbers. Find all triples such that:

728x+729y+730z6zx276yz276xy27\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=13999999=1376923p = 13 \rightarrow 999999 = 13 \cdot 76923.

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

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

Note by Jordi Bosch
5 years ago

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For 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 - 5 years ago

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Another way (generalised for any integer n not coprime to 2 or 5) is pigeonhole principle: out of the infinitely many powers of 10: 1, 10, 100, ... there exists two with the same remainder when divided by n. Subtracting gives a number 999...999000...000, but n is coprime to 2 or 5 so it divides 999...999.

If you don't remove the 0s then it becomes a proof for the problem below by @Calvin Lin

Joel Tan - 5 years ago

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That is a very nice solution.

Michael Ng - 5 years ago

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Good observation.

This is another variant of the problem.

Prove that for any integer, there exists a multiple whose digits are all 0's or 9's. For example, 6×15=90 6 \times 1 5= 90 .

Calvin Lin Staff - 5 years ago

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That's exactly how it is done! With your permission I'll format your solution, just to make it more eye-pleasure for other people.

-We know that strings of n 9's are in the form 10n1.10^n - 1.

Now by Fermat's Little Theorem, 10p110(modp)10^{p-1}-1 \equiv 0 \pmod p if gcd(p,10)=1\gcd(p,10) = 1 As pp is not 22 or 55 this statement is therefore true hence the result follows.

EXAMPLE: That means if we choose for example p=71061=7kp = 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=13p = 13 using Fermat's Theorem we get 10121=9999999999990(mod13) 10^{12}- 1 = 999999999999 \equiv 0 \pmod{13} but in the example we saw 999999=1376923999999 = 13 \cdot 76923

Jordi Bosch - 5 years ago

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Just adding on, 3, 13, 31, 37, 41, 43, 53, 67, 71, 79, 83, 89, 107, 151, 157, 163, 173, 191, 197, 199, 227, 239, 241, 271, 277, 281, 283, 293, 307, 311, 317, 347, 359, 373, 397, 401, 409, 431, 439, 443, 449, 467, 479, 521, 523, 547, 557, 563, 569, 587, 599, 601, 613, 631, 641, ... are some of the numbers such that another string of 99s can be found that the number of nines that other string contains equals to p12\frac{p-1}{2} apart from p1p-1.

Just something interesting I found.

Julian Poon - 5 years ago

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Yes, thank you for the formatting! In fact I think the lowest power of 10 which works is called the order of 10modp10\bmod p . Nice problem.

Michael Ng - 5 years ago

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