_Wolfdieter Lang (wolfdieter.lang(AT)kit.edu), _, Dec 20 2010

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Minute Seconds (rounded down) at which the minute hand overlaps with hour hand overlap problem on analog the analogue clock. Full seconds.

At which a.m. times h:m:s (with fractions of seconds)overlaps does the minute hand overlap with the hour hand on an analog clock? This is problem 43 of the quoted Loyd/Gardner book where also the solution is given (pp. 41-2, solution pp. 180-1 in the German version).

a(n) gives the full second for the (a.m.) hour h=n = 0,1,2,...,10, when the minute hand overlaps the hour hand on an analog clock, provided the minute is

A178181(n), and the fraction of the second is A183033(n)/11.

a(n) gives the full second for the (a.m.) hour h=n = 0,1,2,...,10, when the minute hand overlaps the hour hand on an analog clock, provided the minute is

A178181(n), and the fraction of the second is A183033(n)/11.

a(n)= floor((300/11)*n) (mod 60), n=0..10. See the Mathematica code by R. G. Wilson, and the solution in the Loyd book with (27+3/11)s = 300/11 s.

See the eleven times given in EXAMPLE.

a(n)= floor(300*n/11) (mod 60), n=0..10.

nonn,fini,full,easy,new

At which a.m . times h:m:s (with fractions of seconds)overlaps the minute hand with the hour hand on an analog clock? This is problem 43 of the quoted Loyd/Gardner book where also the solution is given (pp. 41-2, solution pp. 180-1 in the German version).

For the same problem on an analog quartz clock (discrete seconds) the best approximation with rounded seconds is given in A178182A181874.

a(n)= floor((300/11)*n) (mod 60), n=0..10. See the Mathematica code by R. G. Wilson, and the solution in the Loyd book with (27+3/11)s = 300/11 s.

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approved

proposed

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Minute with hour hand overlap problem on analog clock. Full seconds.

Table[ Floor@ Mod[300/11 n, 60], {n, 0, 10}]

approved

proposed

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approved

proposed

reviewed