Can you find a 76 digits multiple of 2^76, written exclusively with sixes and sevens (as asked in

"Spirit of 76"), starting with the (given) value of 2^76, and doing nothing but a (relatively) few sums?

(To be more precise: you should be able to do this by hand, with just pencil and paper, in a reasonably short time -- minutes, not hours or days.)

My algorithm:

Start with N=0, and S=2^76 (given).

Do the following 76 times, for I=1 to 76

while the I-th digit of N (from the right)

isn't 6 or 7, keep adding S to N

multiply S by 10

After the 76 times, the last (right) 76 digits of N are the solution

To see why this works, obviously N is always a multiple of 2^76, and taking the right 76 digits is ths same as taking N modulus 10^76 -- but as 10^76 is a multiple of 2^76, the modulus is a multiple of 2^76 too.