The number of books is *not* constrained, they are infinite. The number of unique books is constrained by the number a characters in a book and the possible values of the characters. No different than the possible numbers constructed of a finite number of digits with a finite number of possible values for each digit. Clearly not infinite and in fact countable.
-- Rick
It looks that way, but I have to wonder if this mechanism isn't equivalent to a Turing tape; all the books taken together form one Turing tape. That's probably a dead end and the permutation constraint you mentioned is the relevant one.
"Countable" also means "infinite in an aleph-sub-0 way" as a way of saying "it's not on a one-to-one and onto map of the reals."
-- Les Cargill
What are the last two digits of 3^1234 ?
69?
-- Rick
My calculator says it's
5856367529932071269049608726415028439757143623443063785468784843536946634255003858796842453459905186846046851774292445753149532138648409140561593823222428559200283816367593571556942770711203894947318922509986128995034233177140247221695941679146171215675158154358685224309073135481486540427499644363484815218991488230440687359021965115067678545212365725678242878485920885436927030372580147256775788882374301866845441065076726499613479868210410548366056401445912760529623658838510267592199677099676046090140514526228797525926977514932394954796457727416398081482661262807288229389463819882569.so, yes.
--sp
-- Best regards, Spehro Pefhany Amazon link for AoE 3rd Edition: http://tinyurl.com/ntrpwu8 Microchip link for 2015 Masters in Phoenix: http://tinyurl.com/l7g2k48
Wow. Is that an HP-3^1234 calculator?
-- Rick
Ooops
On Mon, 07 Sep 2015 13:58:17 -0400, Spehro Pefhany Gave us:
42
Cuz your brain has failed you, it has to be an odd number because 3=2+1.
On Mon, 7 Sep 2015 12:40:21 -0700 (PDT), snipped-for-privacy@gmail.com Gave us:
You missed it. "42" is "the answer to everything". Being a brit, one would think you would have gotten it right away.
[All arithmetic is modulo 100]
81^n = (80+1)^n = n*80+1 (80^n = 0 for n >= 2) = n*(100-20)+1 = 1-20*n
81^n = [1, 81, 61, 41, 21, 1, ...]i.e. 81^n forms a cycle of length 5, 81 = 9^2 = 3^4 => 3^n forms a cycle of length 20.
1234 = 1220 + 14 => 3^1234 = 3^14 = 9^7 = 9^(2*3+1) = (9^2)^3 * 9 = 81^3 * 9 = 41 * 9 = 69
Excellent- a nice systematic solution. I think they're looking for the same binomial expansion trick for the well known sum of digits divisibility by 3 result. so...or C(n,m)=combinatorial m out of n,
3^1234=9^617=(10-1)^617= sigma C(617,n) x 10^(617-n) x (-1)^n, n=0...617, so modulo 100 is C(617,616) x 10^1 x(-1)^616 + C(617,617) x 10^0 x(-1)^617= =6170-1=6169= 69 mod 100
Try this one:
Don't really care for Euler's Totient function approach since it's generally a table lookup. The strictly numerical "answers" are just hacks that happen to work out.
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