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Question Number 67208 by mr W last updated on 24/Aug/19
Find the times in a day when  the hour′s, minute′s and second′s  hand of a clock occupy the same  angular position.  [old question reposted]
$${Find}\:{the}\:{times}\:{in}\:{a}\:{day}\:{when} \\ $$$${the}\:{hour}'{s},\:{minute}'{s}\:{and}\:{second}'{s} \\ $$$${hand}\:{of}\:{a}\:{clock}\:{occupy}\:{the}\:{same} \\ $$$${angular}\:{position}. \\ $$$$\left[{old}\:{question}\:{reposted}\right] \\ $$
Commented by Kunal12588 last updated on 24/Aug/19
at 12 am and 12 pm?
Commented by mr W last updated on 24/Aug/19
12 am and 12 pm are certainly  correct. but there are other times.  observe how a (analog, mechanical)  clock works.
$$\mathrm{12}\:{am}\:{and}\:\mathrm{12}\:{pm}\:{are}\:{certainly} \\ $$$${correct}.\:{but}\:{there}\:{are}\:{other}\:{times}. \\ $$$${observe}\:{how}\:{a}\:\left({analog},\:{mechanical}\right) \\ $$$${clock}\:{works}. \\ $$
Commented by Tinkutara@ last updated on 24/Aug/19
22 times like 1:05,2:10,3:15,4:20,....  ?
$$\mathrm{22}\:{times}\:{like}\:\mathrm{1}:\mathrm{05},\mathrm{2}:\mathrm{10},\mathrm{3}:\mathrm{15},\mathrm{4}:\mathrm{20},…. \\ $$$$? \\ $$
Commented by Kunal12588 last updated on 24/Aug/19
at those times hour′s hand and minute′s hands  have same angular position but second′s hand  has differnt angular position. isn′t it.
$${at}\:{those}\:{times}\:{hour}'{s}\:{hand}\:{and}\:{minute}'{s}\:{hands} \\ $$$${have}\:{same}\:{angular}\:{position}\:{but}\:{second}'{s}\:{hand} \\ $$$${has}\:{differnt}\:{angular}\:{position}.\:{isn}'{t}\:{it}. \\ $$
Commented by mr W last updated on 24/Aug/19
but the question means that all three  hands must be at the same position.
$${but}\:{the}\:{question}\:{means}\:{that}\:{all}\:{three} \\ $$$${hands}\:{must}\:{be}\:{at}\:{the}\:{same}\:{position}. \\ $$
Commented by behi83417@gmail.com last updated on 24/Aug/19
θ=∣((11)/2)m−30h∣  θ=angle between clock and minute hands.  now:θ should be zero.so:  ((11)/2)m−30h=0⇒(m/h)=((60)/(11))   .
$$\theta=\mid\frac{\mathrm{11}}{\mathrm{2}}\mathrm{m}−\mathrm{30h}\mid \\ $$$$\theta=\mathrm{angle}\:\mathrm{between}\:\mathrm{clock}\:\mathrm{and}\:\mathrm{minute}\:\mathrm{hands}. \\ $$$$\mathrm{now}:\theta\:\mathrm{should}\:\mathrm{be}\:\mathrm{zero}.\mathrm{so}: \\ $$$$\frac{\mathrm{11}}{\mathrm{2}}\mathrm{m}−\mathrm{30h}=\mathrm{0}\Rightarrow\frac{\mathrm{m}}{\mathrm{h}}=\frac{\mathrm{60}}{\mathrm{11}}\:\:\:. \\ $$
Answered by mr W last updated on 25/Aug/19
Commented by mr W last updated on 25/Aug/19
this picture shows an analog clock.  its hands don′t move continuously,  but can only occupy 60 possible  angular positions:  0,1,2,...,58,59    the hands move in following way:  the second′s hand moves a step  every second.  the minute′s hand  doesn′t move with each move of   the second′s hand. only after 60  steps of the second′s hand (i.e.  after 60 seconds) the  minute′s hand moves a step, and  so on. similarly the hour′s hand  doesn′t move with each move of  the minute′s hand.  only after  12 steps of the minute′s hand (i.e.  after 12 minutes or 720 seconds)  the hour′s hand moves a step, and  so on.    with this knowledge about how  the hands of a clock work, we  can determine the moments when  the three hands of a clock are at  the same angular position.  let′s say the time is h o′clock exactly.  (0≤h≤11). at this moment the  hour′s hand is at the position 5h,  and the minute′s and second′s  hands are at position 0. after x  seconds, with 0<x<3600,  the position of the  second′s hand is mod(x,60). the  position of the minute′s hand is  ⌊(x/(60))⌋, the position of the hour′s  is 5h+⌊(x/(720))⌋. such that the three  hands are at the same position:  5h+⌊(x/(720))⌋=⌊(x/(60))⌋=mod(x,60)    we get following results:  for h=0: x=0 ⇒time 0h0m0s  for h=1: x=305 ⇒time 1h5m5s  for h=2: x=610 ⇒time 2h10m10s  for h=3: x=976 ⇒time 3h16m16s  for h=4: x=1281 ⇒time 4h21m21s  for h=5: x=1647 ⇒time 5h27m27s  for h=6: x=1952 ⇒time 6h32m32s  for h=7: x=2318 ⇒time 7h38m38s  for h=8: x=2623 ⇒time 8h43m43s  for h=9: x=2989 ⇒time 9h49m49s  for h=10: x=3294 ⇒time 10h54m54s  for h=11: x=3599 ⇒time 11h59m59s  that means in a day there are 24  times when the three hands of  a clock are at the same angular  position.  the picture above shows e.g. the  moment 4h21m21s (am or pm)  when the hands are at the same  position.
$${this}\:{picture}\:{shows}\:{an}\:{analog}\:{clock}. \\ $$$${its}\:{hands}\:{don}'{t}\:{move}\:{continuously}, \\ $$$${but}\:{can}\:{only}\:{occupy}\:\mathrm{60}\:{possible} \\ $$$${angular}\:{positions}: \\ $$$$\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{58},\mathrm{59} \\ $$$$ \\ $$$${the}\:{hands}\:{move}\:{in}\:{following}\:{way}: \\ $$$${the}\:{second}'{s}\:{hand}\:{moves}\:{a}\:{step} \\ $$$${every}\:{second}.\:\:{the}\:{minute}'{s}\:{hand} \\ $$$${doesn}'{t}\:{move}\:{with}\:{each}\:{move}\:{of}\: \\ $$$${the}\:{second}'{s}\:{hand}.\:{only}\:{after}\:\mathrm{60} \\ $$$${steps}\:{of}\:{the}\:{second}'{s}\:{hand}\:\left({i}.{e}.\right. \\ $$$$\left.{after}\:\mathrm{60}\:{seconds}\right)\:{the} \\ $$$${minute}'{s}\:{hand}\:{moves}\:{a}\:{step},\:{and} \\ $$$${so}\:{on}.\:{similarly}\:{the}\:{hour}'{s}\:{hand} \\ $$$${doesn}'{t}\:{move}\:{with}\:{each}\:{move}\:{of} \\ $$$${the}\:{minute}'{s}\:{hand}.\:\:{only}\:{after} \\ $$$$\mathrm{12}\:{steps}\:{of}\:{the}\:{minute}'{s}\:{hand}\:\left({i}.{e}.\right. \\ $$$$\left.{after}\:\mathrm{12}\:{minutes}\:{or}\:\mathrm{720}\:{seconds}\right) \\ $$$${the}\:{hour}'{s}\:{hand}\:{moves}\:{a}\:{step},\:{and} \\ $$$${so}\:{on}. \\ $$$$ \\ $$$${with}\:{this}\:{knowledge}\:{about}\:{how} \\ $$$${the}\:{hands}\:{of}\:{a}\:{clock}\:{work},\:{we} \\ $$$${can}\:{determine}\:{the}\:{moments}\:{when} \\ $$$${the}\:{three}\:{hands}\:{of}\:{a}\:{clock}\:{are}\:{at} \\ $$$${the}\:{same}\:{angular}\:{position}. \\ $$$${let}'{s}\:{say}\:{the}\:{time}\:{is}\:{h}\:{o}'{clock}\:{exactly}. \\ $$$$\left(\mathrm{0}\leqslant{h}\leqslant\mathrm{11}\right).\:{at}\:{this}\:{moment}\:{the} \\ $$$${hour}'{s}\:{hand}\:{is}\:{at}\:{the}\:{position}\:\mathrm{5}{h}, \\ $$$${and}\:{the}\:{minute}'{s}\:{and}\:{second}'{s} \\ $$$${hands}\:{are}\:{at}\:{position}\:\mathrm{0}.\:{after}\:{x} \\ $$$${seconds},\:{with}\:\mathrm{0}<{x}<\mathrm{3600}, \\ $$$${the}\:{position}\:{of}\:{the} \\ $$$${second}'{s}\:{hand}\:{is}\:{mod}\left({x},\mathrm{60}\right).\:{the} \\ $$$${position}\:{of}\:{the}\:{minute}'{s}\:{hand}\:{is} \\ $$$$\lfloor\frac{{x}}{\mathrm{60}}\rfloor,\:{the}\:{position}\:{of}\:{the}\:{hour}'{s} \\ $$$${is}\:\mathrm{5}{h}+\lfloor\frac{{x}}{\mathrm{720}}\rfloor.\:{such}\:{that}\:{the}\:{three} \\ $$$${hands}\:{are}\:{at}\:{the}\:{same}\:{position}: \\ $$$$\mathrm{5}{h}+\lfloor\frac{{x}}{\mathrm{720}}\rfloor=\lfloor\frac{{x}}{\mathrm{60}}\rfloor={mod}\left({x},\mathrm{60}\right) \\ $$$$ \\ $$$${we}\:{get}\:{following}\:{results}: \\ $$$${for}\:{h}=\mathrm{0}:\:{x}=\mathrm{0}\:\Rightarrow{time}\:\mathrm{0}{h}\mathrm{0}{m}\mathrm{0}{s} \\ $$$${for}\:{h}=\mathrm{1}:\:{x}=\mathrm{305}\:\Rightarrow{time}\:\mathrm{1}{h}\mathrm{5}{m}\mathrm{5}{s} \\ $$$${for}\:{h}=\mathrm{2}:\:{x}=\mathrm{610}\:\Rightarrow{time}\:\mathrm{2}{h}\mathrm{10}{m}\mathrm{10}{s} \\ $$$${for}\:{h}=\mathrm{3}:\:{x}=\mathrm{976}\:\Rightarrow{time}\:\mathrm{3}{h}\mathrm{16}{m}\mathrm{16}{s} \\ $$$${for}\:{h}=\mathrm{4}:\:{x}=\mathrm{1281}\:\Rightarrow{time}\:\mathrm{4}{h}\mathrm{21}{m}\mathrm{21}{s} \\ $$$${for}\:{h}=\mathrm{5}:\:{x}=\mathrm{1647}\:\Rightarrow{time}\:\mathrm{5}{h}\mathrm{27}{m}\mathrm{27}{s} \\ $$$${for}\:{h}=\mathrm{6}:\:{x}=\mathrm{1952}\:\Rightarrow{time}\:\mathrm{6}{h}\mathrm{32}{m}\mathrm{32}{s} \\ $$$${for}\:{h}=\mathrm{7}:\:{x}=\mathrm{2318}\:\Rightarrow{time}\:\mathrm{7}{h}\mathrm{38}{m}\mathrm{38}{s} \\ $$$${for}\:{h}=\mathrm{8}:\:{x}=\mathrm{2623}\:\Rightarrow{time}\:\mathrm{8}{h}\mathrm{43}{m}\mathrm{43}{s} \\ $$$${for}\:{h}=\mathrm{9}:\:{x}=\mathrm{2989}\:\Rightarrow{time}\:\mathrm{9}{h}\mathrm{49}{m}\mathrm{49}{s} \\ $$$${for}\:{h}=\mathrm{10}:\:{x}=\mathrm{3294}\:\Rightarrow{time}\:\mathrm{10}{h}\mathrm{54}{m}\mathrm{54}{s} \\ $$$${for}\:{h}=\mathrm{11}:\:{x}=\mathrm{3599}\:\Rightarrow{time}\:\mathrm{11}{h}\mathrm{59}{m}\mathrm{59}{s} \\ $$$${that}\:{means}\:{in}\:{a}\:{day}\:{there}\:{are}\:\mathrm{24} \\ $$$${times}\:{when}\:{the}\:{three}\:{hands}\:{of} \\ $$$${a}\:{clock}\:{are}\:{at}\:{the}\:{same}\:{angular} \\ $$$${position}. \\ $$$${the}\:{picture}\:{above}\:{shows}\:{e}.{g}.\:{the} \\ $$$${moment}\:\mathrm{4}{h}\mathrm{21}{m}\mathrm{21}{s}\:\left({am}\:{or}\:{pm}\right) \\ $$$${when}\:{the}\:{hands}\:{are}\:{at}\:{the}\:{same} \\ $$$${position}. \\ $$
Commented by JDamian last updated on 25/Aug/19
  But many other analog clocks are not so  discrete − second′s hand moves quickly than  one-second step (fractions of one second)
$$ \\ $$$${But}\:{many}\:{other}\:{analog}\:{clocks}\:{are}\:{not}\:{so} \\ $$$${discrete}\:−\:{second}'{s}\:{hand}\:{moves}\:{quickly}\:{than} \\ $$$${one}-{second}\:{step}\:\left({fractions}\:{of}\:{one}\:{second}\right) \\ $$
Commented by mr W last updated on 25/Aug/19
the solution depents very much on  how the hands move.  if it is assumed that the hands  move continuously, then there is  only one moment when the three  hands are at the same position,  i.e. at 12:00:00.
$${the}\:{solution}\:{depents}\:{very}\:{much}\:{on} \\ $$$${how}\:{the}\:{hands}\:{move}. \\ $$$${if}\:{it}\:{is}\:{assumed}\:{that}\:{the}\:{hands} \\ $$$${move}\:{continuously},\:{then}\:{there}\:{is} \\ $$$${only}\:{one}\:{moment}\:{when}\:{the}\:{three} \\ $$$${hands}\:{are}\:{at}\:{the}\:{same}\:{position}, \\ $$$${i}.{e}.\:{at}\:\mathrm{12}:\mathrm{00}:\mathrm{00}. \\ $$

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