Can you find a 76-digit multiple of 2⁷⁶, written exclusively with 7s and 6s?

6667776766677767667676666776766667777767666677766776777777777777666766667776 = 88247290723637794367680048552939376696792027490086866 × 2⁷⁶.

In general, there exists a k-digit number, written exclusively with 7s and 6s, divisible by 2^k.  The proof is by induction.

For k = 1, 6 is divisible by 2¹.

Let n_k be a k-digit number, written exclusively with 7s and 6s, divisible by 2^k.

If n_k = 0 (mod 2^k), then n_k = 0 or 2^k (mod 2^k+1).

Note that:
6×10^k = 0 (mod 2^k+1)
7×10^k = 2^k (mod 2^k+1)

So we can choose, respectively, n_k+1 = 6×10^k + n_k or 7×10^k + n_k = 0 (mod 2^k+1).

Hence result.

The induction gives an explicit means of constructing n_k.

Edited on March 14, 2004, 2:09 pm

Posted by Nick Hobson on 2004-03-14 13:48:27