Prove that no number in the sequence 949, 9449, 94449... is prime.

Interesting.

If the number of 4's is a multiple of 3, then it is clearly divisible by 3, because the sum of the digits is divisible by 3 (that's the test).

If the number of 4's is even, it is clearly divisible by 11, because the sum of the odd digits less the sum of the even digits is 0, which is a multiple of 11 (that's the test).

So that only leaves the case where the number of 4's (mod 6) are 1 or 5.

Because of the mod 6 pattern, let's try looking at 111111. It factors into 3*7*11*13*37.

Trying 7,13, and 37 as factors of 949, we find that 949 = 13*73.

If the number of 4's (mod 6) = 1, then it is therefore a multiple of 13. (We could prove this by calculating remainders mod 13 of 10^n, which necessarily repeat with period 6.)

Trying 7 and 37 as factors of 9444449, we find that 9444449 = 7*1349207. If the number of 4's (mod 6) = 5, then it is therefore a multiple of 7. (We could prove this by calculating remainders mod 7 of 10^n, which necessarily repeat with period 6.)

So, we are done. To summarize, divisors as a function of the number of 4's (mod 6) include the following:

4's mod 6 divisors

----------- --------

0 3 and 11

1 13

2 11

3 3

4 11

5 7

*Edited on ***January 8, 2011, 6:13 pm**