Let N = 111 ... 1222 ... 2, i.e. 1999 digits of digit 1 followed by 1999 digits of 2.
Express N as a product of four integers, none of them equal to 1.
111 ... 1222 ... 2 can be written as 111...1 * (10^1999) + 111...1 + 111...1
= 111...1 * (10^1999 + 2)
= 111...1 * 2 * (5*10^1998 + 1)
Now, 10 == 1 mod 3, so 5*10^1998 == 5 mod 3 and the expression (5*10^1998 + 1) is therefore divisible by 3. Specifically, it's 1666...7 (1997 6's). So there's one more factoring available to yield:
= 111...1 * 2 * 3 * 1666...7
and the four integers are:
2
3
166...7 (1997 sixes)
111...1 (1999 ones)

Posted by Paul
on 20151106 09:48:00 