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
a d1 answer (+background)
