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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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 niners | Comment 9 of 15 |
every number from the set {9,99,999,9999,...} could be represented as 10^k-1.
we have that (2,n)=1,(5,n)=1.as we use the function of Oyler we obtain that:
10^pfi(n)-1 is divisible by n.
also we know that phi(n) is less than n.
this is enough now to say that for every number n relatively prime with 2 and 5 we have at least one number from the set that is divisible by n.
in the end i will say that the function phi(n) is explained with the number of the positive integers less than n and relatively prime with n.
that's all.
 Posted by martin on 2002-12-07 14:36:36

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