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# Minute

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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.

Feb 7, 2018

#1
+102459
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So 2 minutes past 12 does not count because the hand has gone fractionally past the 12?

Feb 7, 2018
#2
+101871
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Correct, Melody...!!!

Feb 7, 2018
#3
+102459
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Twice  4:24  and  7:36

Feb 7, 2018
#4
+102459
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I based this on the fact that the little hand moves 5 minute units in an hour so the little hand is on a minute mark every 12 minutes.

I worked out that ever 12 minutes the hands move apart by a further 11 units.  Of course this is modular so 60units=0units

Then i just did some grunt work and found the pattern.

There would be a better modular way of doing it but that is how I did it.

Melody  Feb 7, 2018
#5
+101871
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I've done somethingl like this, but I don't exactly remember what I did now  !!!!

I want to see what you did......

Feb 7, 2018
#6
+102459
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Feb 7, 2018
#7
+22550
+2

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

Feb 7, 2018
#8
+102459
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