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.
(from http://www.ocf.berkeley.edu/~wwu/riddles/)
def. N(n) = 10^n - 1 (set)
note the property of differences:
N(n) - N(n') = 10^x * N(n-n') for some x.
example:
N(3) - N(1) = 999-9 = 990 = 10 * N(2) = N(3-1)
consider a certain number p (for the purpose
of divisibility by p) note the remainder to
be 0 for divisibility.
for the elements of the set N(1) .. N(p)
assign for each a remainder r, arising from
N()/p. note that the possible values for
r are only r = 0 to p-1, p different values
in total. the key i think is:
because the property of differences, each N()
must have a different r. since there are p
values of N and p remainders, at least one
of them must be r = 0.
so, there is one N(n) for n ≤ p such
that N(n) is divisible by p
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