(In reply to
Generalisation by broll)
"Finding 2^n ending in 8888 is a bit more of a challenge..."
Indeed, as the mod 10000 cycle goes in a repeating cycle of 3,1,9,7,5,... for the digit before 888:
39 3888
139 1888
239 9888
339 7888
439 5888
539 3888
639 1888
739 9888
839 7888
939 5888
1039 3888
1139 1888
1239 9888
1339 7888
1439 5888
1539 3888
1639 1888
1739 9888
1839 7888
1939 5888
2039 3888
2139 1888
2239 9888
2339 7888
2439 5888
2539 3888
2639 1888
2739 9888
2839 7888
2939 5888
3039 3888
3139 1888
3239 9888
3339 7888
3439 5888
3539 3888
3639 1888
3739 9888
3839 7888
3939 5888
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Posted by Charlie
on 2013-01-08 11:33:42 |
re: Generalisation
