When the minute hand of a clock points exactly at a full minute, the hour hand is exactly two minutes away.
Give all possible times in a 12-hour period that satisfy condition.
\(\text{ angular velocity minute hand: $ \omega_m^{\circ} =\dfrac{360^{\circ}}{1~ h} $ }\\ \text{ angular velocity hour hand: $ \omega_h^{\circ} =\dfrac{360^{\circ}}{12~ h} $ }\\ \boxed{\text{angle = angular velocity }\times \text{ time}}\\ \text{ angle minute hand: $ \alpha_m^{\circ} = \omega_m^{\circ} \times t^h \qquad t^h$ time in hours } \\ \text{ angle hour hand: $ \alpha_h^{\circ} = \omega_h^{\circ} \times t^h \qquad t^h$ time in hours } \\ \text{ angle difference minute hand-hour hand: $ \alpha_m^{\circ}-\alpha_h^{\circ} = \Delta\alpha^{\circ}$ } \\ \Delta\alpha^{\circ} = \alpha_m^{\circ}-\alpha_h^{\circ} \\ \Delta\alpha^{\circ} = \omega_m^{\circ} \times t^h - \omega_h^{\circ} \times t^h \\ \Delta\alpha^{\circ} = \left(\omega_m^{\circ} - \omega_h^{\circ} \right) \times t^h \\ \Delta\alpha^{\circ} = \left(\dfrac{360^{\circ}}{1~ h} - \dfrac{360^{\circ}}{12~ h} \right) \times t^h \\ \Delta\alpha^{\circ} = \left(360\cdot \dfrac{11}{12} \right) \times t^h \\ \Delta\alpha^{\circ} = 330 \times t^h \pmod{360^{\circ}} \\\)
\(\mathbf{\boxed{\Delta\alpha^{\circ} = 330 \times t^h \pmod{360^{\circ}} }}\)
Example:
\(t^h = 6\ h \\ \Delta\alpha^{\circ} = 330 \times 6 \pmod{360^{\circ}} \\ \Delta\alpha^{\circ} = 1980^{\circ} \pmod{ 360^{\circ} } \\ \Delta\alpha^{\circ} = 1980^{\circ} - 5\cdot 360^{\circ} \\ \Delta\alpha^{\circ} = 180^{\circ} \)
1.
\(\text{The hour hand of a clock points exactly at a full minute:} \\ \begin{array}{rcll} \dfrac{360^{\circ}}{60~\min.} &=& \dfrac{x}{1~\min.} \\\\ x&=&\dfrac{1~ \min.}{60~ \min.}\cdot 360^{\circ} \\\\ \mathbf{x}&\mathbf{=}&\mathbf{6^{\circ}} \\ \end{array} \)
1 minute on the clock conforms 6 degrees.
\(\text{The hour hand of a clock points exactly at a full minute, so} \\ \text{ $\alpha_h^{\circ} = 6^{\circ} \cdot n \qquad | \qquad n$ is an integer} \)
2. \(\mathbf{t^h =\ ?}\)
\(\begin{array}{rcll} \alpha_h^{\circ} &=& \omega_h^{\circ} \times t^h \\\\ t^h &=& \dfrac{ \alpha_h^{\circ} } {\omega_h^{\circ}} \quad & | \quad \alpha_h^{\circ} = 6^{\circ} \cdot n \qquad \omega_h^{\circ} = \dfrac{360^{\circ}}{12~ h} \\\\ t^h &=& \dfrac{ 6^{\circ} \cdot n } { \dfrac{360^{\circ}}{12~ h}} \\\\ \mathbf{t^h} &\mathbf{=}& \mathbf{0.2n} \\ \end{array}\)
3.
\(\text{The hour hand is exactly two minutes away: $\Delta\alpha^{\circ} = \pm 12^{\circ}$ } \\ \begin{array}{rcll} \Delta\alpha^{\circ} &=& 330 \times t^h \pmod{360^{\circ}} \quad & | \quad \mathbf{t^h=0.2n} \qquad \Delta\alpha^{\circ} = \pm 12^{\circ} \\ \pm 12^{\circ} &=& 330 \times 0.2n \pmod{360^{\circ}} \\ \pm 12^{\circ} &=& 66n \pmod{360^{\circ}} \\ \pm 12^{\circ} -66n &=& 360^{\circ}m \qquad & n \in N,\ m \in N \\ \pm 12^{\circ} &=& 66n + 360^{\circ}m \quad & | \quad : 6 \\ \pm 2^{\circ} &=& 11n + 60^{\circ}m \\ &&\boxed{1. \text{ Diophantine equation: } 11n + 60^{\circ}m = 2} \\ &&\boxed{2. \text{ Diophantine equation: } 11n + 60^{\circ}m = -2} \\ \end{array} \)
4. Solution diophantine equation 11n + 60m = 2:
\(\text{The variable with the smallest coefficient is $n$. The equation is transformed after $n$: }\\ \begin{array}{rcll} 11n + 60m &=& 2 \\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \\ &=& \dfrac{ 2 - 55m - 5m } {11} \\ &=& \dfrac{ - 55m + 2 - 5m } {11} \\ &=& -\dfrac{55m}{11} + \dfrac{ 2 - 5m } {11} \\ n &=&-5m+ \dfrac{ 2 - 5m } {11} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ 2 - 5m } {11} \\ & 11a &=& 2 - 5m \\ \end{array} \\ \text{The variable with the smallest coefficient is $m$. The equation is transformed after $m$: }\\ \begin{array}{rcll} 5m &=& 2 - 11a \\ \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ 2 - 11a } {5}} \\ &=& \dfrac{ 2 - 10a-a } {5} \\ &=& \dfrac{ - 10a + 2 -a } {5} \\ &=& -\dfrac{10a}{5} + \dfrac{ 2 -a } {5} \\ m &=& -2a+ \dfrac{ 2 - a } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} & b &=& \dfrac{ 2 - a } {5} \\ & 5b &=& 2 - a \\ \end{array}\\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ 2 - 5b } \\ \end{array} \)
\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ 2 - 11a } {5}} \quad & | \quad \mathbf{a = 2 - 5b }\\ & = & \dfrac{ 2 - 11 (2 - 5b) } {5} \\ \mathbf{m} & \mathbf{=} & \mathbf{-4 + 11b} \\\\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \quad & | \quad \mathbf{m = -4 + 11b }\\ & = & \dfrac{ 2 - 60(-4 + 11b) } {11} \\ \mathbf{n} & \mathbf{=} & \mathbf{22 - 60b } \\ \end{array} \)
\(\boxed{ \text{1. Diophantine equation: }\\ \mathbf{n = 22 - 60b} \\\mathbf{m=-4 + 11b} \qquad b \in Z } \)\(\begin{array}{|lrcll|} \hline b = 0: & n &=& 22 \\ & t^h &=& 0.2n \\ & t^h &=& 0.2\cdot 22 \\ & &=& 4.4\quad (4:24) \\ \hline \end{array}\)
5. Solution diophantine equation 11n + 60m = -2:
\(\text{The variable with the smallest coefficient is $n$. The equation is transformed after $n$: }\\ \begin{array}{rcll} 11n + 60m &=& -2 \\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ -2 - 60m } {11}} \\ &=& \dfrac{ -2 - 55m - 5m } {11} \\ &=& \dfrac{ - 55m - 2 - 5m } {11} \\ &=& -\dfrac{55m}{11} + \dfrac{ -2 - 5m } {11} \\ n &=&-5m+ \dfrac{ -2 - 5m } {11} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ -2 - 5m } {11} \\ & 11a &=& -2 - 5m \\ \end{array} \\ \text{The variable with the smallest coefficient is $m$. The equation is transformed after $m$: }\\ \begin{array}{rcll} 5m &=& -2 - 11a \\ \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ -2 - 11a } {5}} \\ &=& \dfrac{ -2 - 10a-a } {5} \\ &=& \dfrac{ - 10a - 2 -a } {5} \\ &=& -\dfrac{10a}{5} + \dfrac{ -2 -a } {5} \\ m &=& -2a+ \dfrac{ -2 - a } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} & b &=& \dfrac{ -2 - a } {5} \\ & 5b &=& -2 - a \\ \end{array}\\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ -2 - 5b } \\ \end{array}\)
\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ -2 - 11a } {5}} \quad & | \quad \mathbf{a = -2 - 5b }\\ & = & \dfrac{ -2 - 11 (-2 - 5b) } {5} \\ \mathbf{m} & \mathbf{=} & \mathbf{4 + 11b} \\\\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \quad & | \quad \mathbf{m = 4 + 11b }\\ & = & \dfrac{ -2 - 60(4 + 11b) } {11} \\ \mathbf{n} & \mathbf{=} & \mathbf{-22 - 60b } \\ \end{array}\)
\(\boxed{ \text{2. Diophantine equation: }\\ \mathbf{n = -22 - 60b} \\\mathbf{m=4 + 11b} \qquad b \in Z } \)\(\begin{array}{|lrcll|} \hline b = -1: & n &=& -22+60 \\ & &=& 38 \\ & t^h &=& 0.2n \\ & t^h &=& 0.2\cdot 38 \\ & &=& 7.6\quad (7:36) \\ \hline \end{array} \)
The solutions are: 4:24 and 7:36
