A clock has an hour hand of length 3 and a minute hand of length 4. From 1:00 am to 1:00 pm of the same day, find the number of occurrences when the distance between the tips of the two hands is an integer.
(In reply to
solution by Charlie)
The 132 times:
Time third
side
1:00:11.39973 2
1:05:27.27273 1
1:10:43.14572 2
1:14:12.97838 3
1:17:48.82568 4
1:21:49.09091 5
1:26:46.68668 6
1:38:10.90909 7
1:49:35.13150 6
1:54:32.72727 5
1:58:32.99250 4
2:02:08.83980 3
2:05:38.67246 2
2:10:54.54545 1
2:16:10.41845 2
2:19:40.25111 3
2:23:16.09841 4
2:27:16.36364 5
2:32:13.95941 6
2:43:38.18182 7
2:55:02.40422 6
3:00:00.00000 5
3:04:00.26523 4
3:07:36.11253 3
3:11:05.94519 2
3:16:21.81818 1
3:21:37.69117 2
3:25:07.52384 3
3:28:43.37113 4
3:32:43.63636 5
3:37:41.23214 6
3:49:05.45455 7
4:00:29.67695 6
4:05:27.27273 5
4:09:27.53796 4
4:13:03.38525 3
4:16:33.21792 2
4:21:49.09091 1
4:27:04.96390 2
4:30:34.79656 3
4:34:10.64386 4
4:38:10.90909 5
4:43:08.50487 6
4:54:32.72727 7
5:05:56.94968 6
5:10:54.54545 5
5:14:54.81069 4
5:18:30.65798 3
5:22:00.49064 2
5:27:16.36364 1
5:32:32.23663 2
5:36:02.06929 3
5:39:37.91659 4
5:43:38.18182 5
5:48:35.77759 6
6:00:00.00000 7
6:11:24.22241 6
6:16:21.81818 5
6:20:22.08341 4
6:23:57.93071 3
6:27:27.76337 2
6:32:43.63636 1
6:37:59.50936 2
6:41:29.34202 3
6:45:05.18931 4
6:49:05.45455 5
6:54:03.05032 6
7:05:27.27273 7
7:16:51.49513 6
7:21:49.09091 5
7:25:49.35614 4
7:29:25.20344 3
7:32:55.03610 2
7:38:10.90909 1
7:43:26.78208 2
7:46:56.61475 3
7:50:32.46204 4
7:54:32.72727 5
7:59:30.32305 6
8:10:54.54545 7
8:22:18.76786 6
8:27:16.36364 5
8:31:16.62887 4
8:34:52.47616 3
8:38:22.30883 2
8:43:38.18182 1
8:48:54.05481 2
8:52:23.88747 3
8:55:59.73477 4
9:00:00.00000 5
9:04:57.59578 6
9:16:21.81818 7
9:27:46.04059 6
9:32:43.63636 5
9:36:43.90159 4
9:40:19.74889 3
9:43:49.58155 2
9:49:05.45455 1
9:54:21.32754 2
9:57:51.16020 3
10:01:27.00750 4
10:05:27.27273 5
10:10:24.86850 6
10:21:49.09091 7
10:33:13.31332 6
10:38:10.90909 5
10:42:11.17432 4
10:45:47.02162 3
10:49:16.85428 2
10:54:32.72727 1
10:59:48.60027 2
11:03:18.43293 3
11:06:54.28022 4
11:10:54.54545 5
11:15:52.14123 6
11:27:16.36364 7
11:38:40.58604 6
11:43:38.18182 5
11:47:38.44705 4
11:51:14.29434 3
11:54:44.12701 2
12:00:00.00000 1
12:05:15.87299 2
12:08:45.70566 3
12:12:21.55295 4
12:16:21.81818 5
12:21:19.41396 6
12:32:43.63636 7
12:44:07.85877 6
12:49:05.45455 5
12:53:05.71978 4
12:56:41.56707 3
solct=0; t=[0,0;0,0];
for t0=0:12/11:11
for s3=1:7
if s3==1
solct=solct+1;
t(solct,:)=[t0,1];
elseif s3==7
solct=solct+1;
t(solct,:)=[t0+6/11, 7];
else
angle=acosd((16+9-s3^2)/(2*4*3))/360;
solct=solct+1;
t(solct,:)=[t0+angle*12/11, s3];
if t(solct,1)<1 t(solct,1)=12+t(solct,1); end
solct=solct+1;
t(solct,:)=[t0-(angle-1)*12/11,s3];
end
if t(solct,1)<1 t(solct,1)=12+t(solct,1); end
end
end
t=sortrows(t);
for i=1:solct
h=floor(t(i,1)); m=floor(60*(t(i,1)-h)); sc=3600*(t(i,1)-h-m/60);
fprintf('%2d:%02d:%08.5f %d\n',h,m,sc,t(i,2))
end
|
|
Posted by Charlie
on 2020-11-09 12:18:44 |
re: solution -- the times
