It's not difficult to verify that the power of $3^{m}$, where $m=3^{2}$, starts with the digit sequence 1968. - On the other hand, someone claims that in the sequence of powers of $2^{n}$ - where $n$ is a natural number - the first one that starts with the digit sequence 1968 has more than $2^{k}$ digits, where $k=2^{3}$. Let's find the exponent of this power.
I. We need to find an exponent $n$ such that

$$
1,968 \cdot 10^{m}2^{k}=2^{8}=256$, hence

$$
\begin{gathered}
n \cdot 0.30103 > m > 256 \\
n > \frac{256}{0.30103} = 850.4
\end{gathered}
$$

thus $n \geq 851$. For $n=851$, $n \cdot 0.30103 = 256.17653$, so the mantissa of $\log 2^n$ is $0.17653$, which is smaller than the lower bound in (2). We need to search for the solution near $n=851$, increasing $n$ until the mantissa satisfies (2). For $n=852$, $0.17653 + 0.30103 = 0.47756$, which is too much. The mantissa decreases if we increase $n$ by 3, since the logarithm of the power

$$
3 \cdot 0.30103 = 0.90309 = 1 - 0.09691
$$

increases by this value, and the mantissa decreases by $0.09691$ (if it was larger than this). Thus, increasing $n$ to 855, the mantissa becomes $0.38065$, and for $n=858$, it becomes $0.28374$, which is almost correct but still too small. A small increase in $n$ can be achieved by increasing $n$ by 10, since we saw that the mantissa of $2^{10}$ is $0.0103$. Thus, we get a mantissa that satisfies (2):

$$
868 \cdot 0.30103 - 261 = 0.29404
$$

II. However, this does not yet ensure that the exponent $n=868$ also satisfies (1). Indeed, the accuracy of the middle term in (2) significantly deteriorates during the multiplication by 868. To verify, we can use a seven-digit logarithm table, or check our result by calculating the first few digits of $2^{868}$. The latter can be done, for example, in the following steps:

$$
\begin{gathered}
2^{13} = 8 \cdot 1024 = 8192, \quad 2^{14} = 2 \cdot 2^{13} = 16384, \\
2^{27} = 2^{13} \cdot 2^{14} = 134217728 \\
18014396 \ldots 256$ if we had communicated this, we could have shown by performing the procedure with the sequence $1, 2, \ldots$ instead of the exponents $851, 852, \ldots$ that the exponent $n=868$ is the first one for which the first digits of $2^n$ are 1968...

2. The fact that the inequality (2) led to a good result despite the inaccuracies in its formation is due to the fortunate circumstance that the value of $\log 2$ is accurate to seven decimal places: 0.3010300 (to ten decimal places, it is 0.3010299957). However, the verification was complicated by the fact that in $2^{868}$, the required digits $1, 9, 6, 8$ are followed by 0.