Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
2839
9
avatar

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
avatar+118692 
0

So 2 minutes past 12 does not count because the hand has gone fractionally past the 12?

 Feb 7, 2018
 #2
avatar+130168 
+1

Correct, Melody...!!!

 

cool cool cool

 Feb 7, 2018
 #3
avatar+118692 
+1

Twice  4:24  and  7:36

 Feb 7, 2018
 #4
avatar+118692 
+1

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. 

I can give more info if anyone wants it. 

Melody  Feb 7, 2018
 #5
avatar+130168 
+1

Yeah...post your solution, Melody   [ at your leisure, of course ]

 

I've done somethingl like this, but I don't exactly remember what I did now  !!!!

 

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

 

 

cool cool cool

 Feb 7, 2018
 #6
avatar+118692 
+2

 

.
 Feb 7, 2018
 #7
avatar+26397 
+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.

 

 angular velocity minute hand: ωm=3601 h  angular velocity hour hand: ωh=36012 h angle = angular velocity × time angle minute hand: αm=ωm×thth time in hours  angle hour hand: αh=ωh×thth time in hours  angle difference minute hand-hour hand: αmαh=Δα Δα=αmαhΔα=ωm×thωh×thΔα=(ωmωh)×thΔα=(3601 h36012 h)×thΔα=(3601112)×thΔα=330×th(mod360)

 

Δα=330×th(mod360)

 

Example:

th=6 hΔα=330×6(mod360)Δα=1980(mod360)Δα=19805360Δα=180

 

1.

The hour hand of a clock points exactly at a full minute:36060 min.=x1 min.x=1 min.60 min.360x=6

 

1 minute on the clock conforms 6 degrees.

 

The hour hand of a clock points exactly at a full minute, so αh=6n|n is an integer

 

2. th= ?

αh=ωh×thth=αhωh|αh=6nωh=36012 hth=6n36012 hth=0.2n

 

3.

The hour hand is exactly two minutes away: Δα=±12 Δα=330×th(mod360)|th=0.2nΔα=±12±12=330×0.2n(mod360)±12=66n(mod360)±1266n=360mnN, mN±12=66n+360m|:6±2=11n+60m1. Diophantine equation: 11n+60m=22. Diophantine equation: 11n+60m=2

 

4. Solution diophantine equation 11n + 60m = 2:

The variable with the smallest coefficient is n. The equation is transformed after n11n+60m=2n=260m11=255m5m11=55m+25m11=55m11+25m11n=5m+25m11we set:a=25m1111a=25mThe variable with the smallest coefficient is m. The equation is transformed after m5m=211am=211a5=210aa5=10a+2a5=10a5+2a5m=2a+2a5we set:b=2a55b=2aThe variable with the smallest coefficient is a. The equation is transformed after ano fraction there:a=25b

 

Elemination of the unknowns:m=211a5|a=25b=211(25b)5m=4+11bn=260m11|m=4+11b=260(4+11b)11n=2260b

 

1. Diophantine equation: n=2260bm=4+11bbZb=0:n=22th=0.2nth=0.222=4.4(4:24)

 

5. Solution diophantine equation 11n + 60m = -2:

The variable with the smallest coefficient is n. The equation is transformed after n11n+60m=2n=260m11=255m5m11=55m25m11=55m11+25m11n=5m+25m11we set:a=25m1111a=25mThe variable with the smallest coefficient is m. The equation is transformed after m5m=211am=211a5=210aa5=10a2a5=10a5+2a5m=2a+2a5we set:b=2a55b=2aThe variable with the smallest coefficient is a. The equation is transformed after ano fraction there:a=25b

 

Elemination of the unknowns:m=211a5|a=25b=211(25b)5m=4+11bn=260m11|m=4+11b=260(4+11b)11n=2260b

 

2. Diophantine equation: n=2260bm=4+11bbZb=1:n=22+60=38th=0.2nth=0.238=7.6(7:36)

 

 

The solutions are: 4:24 and 7:36

 

laugh

 Feb 7, 2018
 #8
avatar+118692 
+2

Nice answer Heureka :)

Melody  Feb 7, 2018
 #9
avatar+26397 
+2

Thank you, Melody

 

laugh

heureka  Feb 7, 2018

6 Online Users

avatar
avatar
avatar
avatar
avatar