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 Strings of Nines (Posted on 2013-05-29)
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.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 a d1 answer (+background) | Comment 1 of 2

BACKGROUND

1.      Take a  rational number n that has neither 2 nor 5 in its factorization.

2.      Calculate the quotient q = 1/n by continuous division.

3.      Consider   the numbers to the right of the decimal point. Those numbers will contain a repetend - a repeating series of digits.

4.      Call it p.

5.      The number of p's  digits is  always less than n,   that is at most n-1.

6.      Converting  the periodic decimal fraction into regular fraction is done by the following formula :
1/n= p/99..999 ,  where the number of nines equals to the number of repetend's digits.(WHY? -see an example below)

A D1  Solution:

BEGIN

1/n= p/9..9999 ( les than n  nines ).

9…9999=n*p

Since  n is an integer  9…9999 is divisible by n.

END

EXAMPLE

n=7   1/7=per,frac. =,142857142857142857…=F

1000000*f-f= 999999f=p

1/7= 142857/999999

S0 999999 must be divisible by 7.

Edited on May 29, 2013, 11:48 am
 Posted by Ady TZIDON on 2013-05-29 11:41:22

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