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Clock a doodle do (Posted on 2020-11-09) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution re: solution -- the times | Comment 2 of 3 |
(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
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