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.
So 2 minutes past 12 does not count because the hand has gone fractionally past the 12?
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.
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......
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: ω∘m=360∘1 h angular velocity hour hand: ω∘h=360∘12 h angle = angular velocity × time angle minute hand: α∘m=ω∘m×thth time in hours angle hour hand: α∘h=ω∘h×thth time in hours angle difference minute hand-hour hand: α∘m−α∘h=Δα∘ Δα∘=α∘m−α∘hΔα∘=ω∘m×th−ω∘h×thΔα∘=(ω∘m−ω∘h)×thΔα∘=(360∘1 h−360∘12 h)×thΔα∘=(360⋅1112)×thΔα∘=330×th(mod360∘)
Δα∘=330×th(mod360∘)
Example:
th=6 hΔα∘=330×6(mod360∘)Δα∘=1980∘(mod360∘)Δα∘=1980∘−5⋅360∘Δα∘=180∘
1.
The hour hand of a clock points exactly at a full minute:360∘60 min.=x1 min.x=1 min.60 min.⋅360∘x=6∘
1 minute on the clock conforms 6 degrees.
The hour hand of a clock points exactly at a full minute, so α∘h=6∘⋅n|n is an integer
2. th= ?
α∘h=ω∘h×thth=α∘hω∘h|α∘h=6∘⋅nω∘h=360∘12 hth=6∘⋅n360∘12 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∘)±12∘−66n=360∘mn∈N, m∈N±12∘=66n+360∘m|:6±2∘=11n+60∘m1. Diophantine equation: 11n+60∘m=22. Diophantine equation: 11n+60∘m=−2
4. Solution diophantine equation 11n + 60m = 2:
The variable with the smallest coefficient is n. The equation is transformed after n: 11n+60m=2n=2−60m11=2−55m−5m11=−55m+2−5m11=−55m11+2−5m11n=−5m+2−5m11we set:a=2−5m1111a=2−5mThe variable with the smallest coefficient is m. The equation is transformed after m: 5m=2−11am=2−11a5=2−10a−a5=−10a+2−a5=−10a5+2−a5m=−2a+2−a5we set:b=2−a55b=2−aThe variable with the smallest coefficient is a. The equation is transformed after a: no fraction there:a=2−5b
Elemination of the unknowns:m=2−11a5|a=2−5b=2−11(2−5b)5m=−4+11bn=2−60m11|m=−4+11b=2−60(−4+11b)11n=22−60b
1. Diophantine equation: n=22−60bm=−4+11bb∈Zb=0:n=22th=0.2nth=0.2⋅22=4.4(4:24)
5. Solution diophantine equation 11n + 60m = -2:
The variable with the smallest coefficient is n. The equation is transformed after n: 11n+60m=−2n=−2−60m11=−2−55m−5m11=−55m−2−5m11=−55m11+−2−5m11n=−5m+−2−5m11we set:a=−2−5m1111a=−2−5mThe variable with the smallest coefficient is m. The equation is transformed after m: 5m=−2−11am=−2−11a5=−2−10a−a5=−10a−2−a5=−10a5+−2−a5m=−2a+−2−a5we set:b=−2−a55b=−2−aThe variable with the smallest coefficient is a. The equation is transformed after a: no fraction there:a=−2−5b
Elemination of the unknowns:m=−2−11a5|a=−2−5b=−2−11(−2−5b)5m=4+11bn=2−60m11|m=4+11b=−2−60(4+11b)11n=−22−60b
2. Diophantine equation: n=−22−60bm=4+11bb∈Zb=−1:n=−22+60=38th=0.2nth=0.2⋅38=7.6(7:36)
The solutions are: 4:24 and 7:36