The house numbers on Clock Street are the same as the seconds on a digital clock, namely:
00:00:00, 00:00:01, ..., 00:00:59, 00:01:00, ..., 00:59:59, 01:00:00, ..., 23:59:59
(A) The sum of the addresses up to and including Jack's house equals the sum of the addresses past Jack's house up to and including Jake's house. What are their addresses? (There are several solutions. Please give the largest.)
(B) The sum of the even addresses up to Jane's house equals the sum of the odd addresses past Jane's house up to and including June's house. What are their addresses? (There are several solutions. Please give the largest.)
Note: The addresses are added like one adds times. Like, 00:50+00:50=01:40
Writing the number of seconds into the day in clock format is isomorphic to writint them as ordinary arabic numbers:
00:01:40 is 100 seconds into the day
00:03:50 is 230 seconds into the day
-------- ---
5:30 330 are both representations of 330 seconds into the day.
so it doesn't matter the form in which the addresses appear.
Let Jack's house be a and Jake's house be b.
a(a+1) (b-a)(a+1+b)
------ = ------------
2 2
a(a+1) = (b-a)(a+1+b)
a^2 + a = b*a + b + b^2 - a^2 - a - a*b
2*a^2 + 2*a = b^2 + b
for t2=86399:-1:10
syms t
% s=solve(@(t)2*t^2+2*t==t2^2+t2);
s=roots([2 2 -t2^2-t2]);
if abs(s(2)-round(s(2)))<.00001
disp([s(2) t2])
s1=seconds(s(2)); s2=seconds(t2);
s1.Format='hh:mm:ss'; s2.Format='hh:mm:ss';
s1, s2
% break
end
end
Starting with b=23*3600+59*60+59 = 86399, solving for a, and trying decreasing values from there, the first integer value of a we get to is the largest possible:
Jack at 16730 or 04:38:50 Jake at 23660 or 06:34:20
Verification:
>> sum(1:16730),sum(16731:23660)
ans =
139954815
ans =
139954815
Jane and June present a more difficult problem especially as the cases of Jane at an odd position and an even position differ. June is apparently already known to be at an odd house number.
clearvars
evensum=0; oddsum=1; june=1;
for jane=1:86400
if mod(jane,2)==0
evensum=evensum+jane-2;
else
oddsum=oddsum-jane;
end
while oddsum<evensum
june=june+2;
oddsum=oddsum+june;
end
if evensum==oddsum
disp([jane june evensum])
end
end
Showing on each line:
Jane's house, June's house, sum of eves prior to Jane
which equals sum of odds after Jand
up to and including June's
>> houseNumbersOnClockStreet
1 1 0
2 1 0
3 3 0
33 45 240
50 69 600
91 127 1980
1105 1561 304152
1682 2377 706440
3075 4347 2360832
37521 53061 351918840
57122 80781 815702160
>> sum(2:2:57120),sum(57123:2:80781)
ans =
815702160
ans =
815702160
>>
The two sums at the bottom verify the sums obtained through addition and subtraction in the program.
The next step is to convert the house numbers to the time format, and also to limit the addresses to the 24-hour range:
clearvars
evensum=0; oddsum=1; june=1;
for jane=1:86400
if oddsum>86400
break
end
if mod(jane,2)==0
evensum=evensum+jane-2;
else
oddsum=oddsum-jane;
end
while oddsum<evensum
june=june+2;
oddsum=oddsum+june;
end
if evensum==oddsum
janes=seconds(jane);
junes=seconds(june);
evensums=seconds(evensum);
janes.Format='hh:mm:ss';
junes.Format='hh:mm:ss';
evensums.Format='hh:mm:ss';
disp([janes junes evensums])
end
end
>> houseNumbersOnClockStreet
00:00:01 00:00:01 00:00:00
00:00:02 00:00:01 00:00:00
00:00:03 00:00:03 00:00:00
00:00:33 00:00:45 00:04:00
00:00:50 00:01:09 00:10:00
00:01:31 00:02:07 00:33:00
>>
The largest case has Jane at 00:01:31, June at 00:02:07, and each of the two totals as 00:33:00, previously reported numerically as 91, 127, and 1980 respectively.
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Posted by Charlie
on 2023-11-10 13:31:29 |
computer solution
