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.

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**