Cool Problems (Part 3)
Hey Brilliantinians! Here I have some amazing problems.
Problem 1 Let be natural numbers. Find all triples such that:
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 .
Problem 3 Natural numbers "a" and "b" satisfy that:
is an integer. Proof that the greatest common divisor of
Note by
Jordi Bosch
6 years, 7 months ago
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Easy Math Editor
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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.
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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 10n−1.
Now by Fermat's Little Theorem, 10p−1−1≡0(modp) 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→106−1=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 1012−1=999999999999≡0(mod13) but in the example we saw 999999=13⋅76923
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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 10modp. Nice problem.
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 9s can be found that the number of nines that other string contains equals to 2p−1 apart from p−1.
Just something interesting I found.
Good observation.
This is another variant of the problem.
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
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That is a very nice solution.