The measure of an acute angle: $A B^{\circ} C D^{\prime} E F^{\prime \prime}$, where $A, \ldots, F$ (not necessarily distinct) are digits. $A B, C D$, $E F$ are genuine two-digit numbers, and the latter two are less than 60. If we cut the sequence of digits somewhere, move the first part as a whole behind the second part, and again cut the sequence of digits into three two-digit numbers, and consider these as the degrees, minutes, and seconds of an angle, we get the complement of the original angle. Determine all such angles with this property.
Let the angle sought be represented in the prescribed form as $A_{1} B_{1}^{0} C_{1} D_{1}^{\prime} E_{1} F_{1}^{\prime \prime}$, and suppose for now that the two-digit numbers $C_{1} D_{1}$ and $E_{1} F_{1}$ are both less than 60. Thus, the sum of the two angles before combining the seconds and minutes will be $89^{\circ} 59^{\prime} 60^{\prime \prime}$, since we cannot get $120^{\prime \prime}$ or $119^{\prime}$; therefore, the addition remains correct even if we omit the unit symbols and consider the six digits as a single six-digit number:

![](https://cdn.mathpix.com/cropped/2024_05_02_14159d2e8312d4ee42e8g-1.jpg?height=138&width=333&top_left_y=332&top_left_x=885)

Let the first part of the split measurement be $n$-digit, so the second part is $m=6-n$ digits, and the digits of the parts read as single numbers are $u$ and $v$. Thus, the original six-digit number is $10^{n} \cdot u + v$ and the number resulting from the carry-over is $10^{m} \cdot u + v$, and thus

$$
\left(10^{n}+1\right) u + \left(10^{m}+1\right) v = 895960
$$

The roles of the two terms on the left side are symmetric, so it suffices to examine the cases where $n \leq m$, i.e., $n=1,2,3$.

For $n=1$, $m=5$ and the coefficients in (2) are 11 and 100,001, the latter of which is also divisible by 11, but the right side of (2) is not, so there is no solution.

Similarly, for $n=3=m$, both coefficients in (2) are 1001, and the right side is not divisible by it, so there is no solution. For $n=2$, we use our assumption that in (1) there is no carry-over in the tens and hundreds columns, so

$$
\overline{E F} + \overline{A B} = 60, \quad \overline{C D} + \overline{E F} = 59, \quad \overline{A B} + \overline{C D} = 89
$$

Dividing the sum of these three equations by 2,

$$
\overline{A B} + \overline{C D} + \overline{E F} = 104
$$

and subtracting the equations (3) from this, we get $\overline{C D} = 44, \overline{A B} = 45, \overline{E F} = 15$, so a possible value for the angle sought is $45^{\circ} 44^{\prime} 15^{\prime \prime}$, and this indeed fits, as the angle resulting from the split and carry-over, $44^{\circ} 15^{\prime} 45^{\prime \prime}$, is indeed the complement of this. - By swapping the roles of these two angles, we get a solution for the case $n=4$.

II. We will show that we do not get any further solutions even if we allow $\overline{C_{1} D_{1}}$ and $\overline{E_{1} F_{1}}$ to be greater than 60. Let the number of minutes that can be combined from seconds be $p$, and the number of degrees that can be combined from minutes be $f$, so similarly to (3),

$$
\overline{E F} + \overline{E_{1} F_{1}} = p \cdot 60, \quad \overline{C D} + \overline{C_{1} D_{1}} = f \cdot 60 - p, \quad A B + A_{1} B_{1} = 90 - f
$$

Here too, $p, f \leq 2$, since, for example, $C D > 60$ would result, which is excluded. For $n=4$, similarly, $\overline{C D} = 75 > 60$, indeed there is no solution.