Determine the minimum value of a positive base ten integer N, such that each of the last seven digits of N3 is 7 but the eighth digit from the right in N3 is not 7.
This program takes a couple of minutes to run and doesn't produce any results:
10 for N1=1 to 1000000000
20 if N@10<>7 then
30 :N=N1*10000000+7777777
40 :Cr=int(N^(1/3)+0.5)
50 :if Cr*Cr*Cr=N then print N
60 next
The more efficient program below produces results in a couple of seconds, as treating only cubes goes faster than looking at many numbers in search of a cube:
10 for N=999 to 999999999
20 C=N*N*N
30 if C@10000000=7777777 then
40 :if C@100000000<>77777777 then print N,C
50 next
It produces
9660753 901639512372747777777
19660753 7599769868183017777777
29660753 26094352023993287777777
39660753 62385385979803557777777
49660753 122472871735613827777777
59660753 212356809291424097777777
69660753 338037198647234367777777
79660753 505514039803044637777777
89660753 720787332758854907777777
109660753 1318723274070475447777777
119660753 1713385922426285717777777
129660753 2179845022582095987777777
before being stopped manually.
It shows N=9660753, whose cube is 901639512372747777777, as the minimum value.
A run of the below program:
10 for N=999 to 999999999
20 C=N*N*N
30 if C@10000000=7777777 then
40 :print N,C
50 next
shows that the caveat of the eighth digit from the right not being 7 was unnecessary, as such a cube with 8 7's at the end comes much higher:
9660753 901639512372747777777
19660753 7599769868183017777777
29660753 26094352023993287777777
39660753 62385385979803557777777
49660753 122472871735613827777777
59660753 212356809291424097777777
69660753 338037198647234367777777
79660753 505514039803044637777777
89660753 720787332758854907777777
99660753 989857077514665177777777
109660753 1318723274070475447777777
119660753 1713385922426285717777777
129660753 2179845022582095987777777
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Posted by Charlie
on 2009-06-26 13:54:27 |
computer solution
