Day 6: A Christmas Math Game

one possible setting is

$4646 \dots 4= 64(1010 \dots 1)+4\times 10^{k-1}=64(1010 \dots 1+\frac{4\times 10^{k-1}}{64})= 64(1010 \dots 1+5^t) \implies 64\times 5^t=4\times 10^{k-1}$

So $k=5$ is the only integer that satisfies the above equality (note that $t=4$ ). I would like to point out the key fact that the contribution of power of $5$ to the sum, should be in a way such that it would not disturb the repetitive behaviour of the sequence.

If we take the sequence $46464646464...$ , we can separate it into the sum: $4 \ \times \ 10^{k-1} \ + \ 6464646464...$ . We can then realize that the sequence $6464646464...$ can be represented as: $64 \ \sum_{n=0}^{k-2} \ 10^{2n}$ . From here, we realize that our sequence is equivalent to: $4 \ \times \ 10^{k-1} \ + \ 64 \ \sum_{n=0}^{k-2} \ 10^{2n}$ . If we find the prime factorization of $10^{k-1}$ , we get $2^{k-1}5^{k-1}$ . Because of this, we can rewrite the previous expression as: $2^2 \ \times \ 2^{k-1}5^{k-1} \ + \ 64 \ \sum_{n=0}^{k-2} \ 10^{2n} \ = \ 2^{k+1}5^{k-1} \ + \ 64 \ \sum_{n=0}^{k-2} \ 10^{2n}$ . In order for this expression to equal a number of the form specified in the question, plus a power of $5$ , we must divide by some number to get rid of the $2^{k+1}$ and the $64$ coefficient on the sum. Setting the two equal to each other, we get: $2^{k+1} \ = \ 64 \ \Rightarrow \ k \ + \ 1 \ = \ 6 \ \Rightarrow \ k \ = \ 5$ .