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 Niners (Posted on 2002-10-23)
The set of numbers {9, 99, 999, 9999, ...} has some interesting properties. One of these has to do with factorization. Take any number n that isn't divisible by 2 or by 5. You will be able to find at least one number in the set that is divisible by n. Furthermore, you won't need to look beyond the first n numbers in the set.

Prove it.

 See The Solution Submitted by levik Rating: 4.2500 (8 votes)

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 Thanks to Euler | Comment 10 of 15 |
Solution follows straight from
1. Euler's Totient Function 'phi(x)': number of numbers less than 'x' and relatively prime to 'x'.
2. Euler's Theorem: If gcd(a,m)=1 then a^phi(m) = 1 (mod m)

Explaination:
since gcd(10,n)=1 due to 'n' not being divisible by 2 or 5, 10^phi(n) = 1 (mod n). That is, (10^phi(n) - 1) is divisible by 'n'. Also, phi(n)<=n by the very definition of it.

That was a very good question, it got me thinking at the first glance! i am new to this group... any tips?
I believe:
Maths is the king of all sciences.
Number Theory is the queen of Maths.
Problem Solving is the life of life.

 Posted by Vikram on 2002-12-31 07:43:12

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