Show that there are infinitely many integers n such that:

1) All digits of n in base 10 are strictly greater than 1.
2) If you take the product of any 4 digits of n, then it divides n.

Define 9[x] as being an integer composed of x 9s in a row (9[1]=9, 9[2]=99, etc). For any integer i, then, 9[2i]=9[i]*(9[i]+2).
For example:
99=9*11
9999=99*101
999999=999*1001
If i is a multiple of 4, then the left-hand of that equation fits the parameters of this problem, or, n(i)=9[8i] for any integer i.

Put more simply, any string of 9s where the number of digits is a multiple of 8 works. Since 8 has infinitely many multiples (or, since the set of integers is infinite), there are infinitely many integers that fit these parameters.

Posted by DJ on 2003-05-13 14:13:22