When will i get a number that ends in $6 \cdot 10^n$?

Question

I am attempting Brocards problem, and I have it pinned down to one simple question:

When will $n(n-1)(n-2)(n-3)...6$ end in $6\cdot(10^n)$?

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So far I only have when $n$ is $6, 8, 14,$ and $19$. I have checked up to $50$, and then decided to come here. If there is no more, how would I go proving that? (If you have any ideas how to prove there are no more, or found any more, please comment them.)

Answer 1

There should be more than just the ones you listed, assuming I am interpreting your problem correctly. On a (very bad) SAGE program, I've checked up to 100.

prod = 1
for n in range(6,100):
    prod = prod * n
    while(mod(prod, 10)==0):
        prod = prod/10
    if(mod(prod, 10)==6):
        print((prod, n))

Output:

(6, 6)
(336, 8)
(72648576, 14)
(10137091700736, 19)
(2460273325330034507063488413696, 34)
(8610956638655120774722209447936, 35)
(309994438991584347889999540125696, 36)
(435852181222167593133339353416728576, 38)
(67992940270658144528800939133009657856, 40)
(2787710551096983925680838504453395972096, 41)
(5034605255281152969779594339042833125605376, 43)
(2155193679259734838691369626280099833076646936576, 47)
(2534507766809448170301050680505397403698136797413376, 50)
(129259896107281856685353584705775267588604976668082176, 51)
(356240273671668797024834479449116637474195315697234477056, 53)
(26224977716989947938044268629125636740010233997715193493469306873184256, 62)
(3039259243182390440687488258050455368930636276170793716141802375022330203078656, 67)
(2756571201266155343549608568510448184547360431841124759755099390295093630841429712160423936, 74)
(27618667788044610558328229826084427400005552385350412392657050269858826094097640534415714358072639609634816, 84)
(234758676198379189745789953521717632900047195275478505337584927293800021799829944542533572043617436681895936, 85)
(20189246153060610318137936002867716429404058793691151459032303747266801874785375230657887195751099554643050496, 86)
(154568868547832032595664038037955236983517474124499455570351317489074635153356832765916784370670418190347194597376, 88)
(123809663706813458109126894468402144823797496773724063911851405308748782757838823045499344280907004970468102872498176, 90)
(11266679397320024687930547396624595178965572206408889815978477883096139230963332897140440329562537452312597361397334016, 91)
(96397708923470131229933763525520036351229435798034461265511856767770567260122276267933607459737070441986583024115589840896, 93)
(801606330687350998777736882913030822602865115993108134650870644327781213437348962490048993129908938760510701580663984730392231936, 97)

Of course, there are less barbaric ways to solve this. See here for a discussion. And analyzing this argument should show you that, in fact, you should be able to get a last nonzero digit of 6 infinitely often.

That said, if we are referring to the same Brocard's problem ($n! + 1 = m^2$) this approach is unlikely to get you anywhere.

Answer 2

It is "amazing" that the sequence $$\{6,8,14,19,34,35,36,38,40,41,43,47,50,51,53\}$$ given by @Paco Adajar in his answer is sequence $A045547$ in $OEIS$.

It also corresponds "to numbers $n$ such that final nonzero digit of" $n!$ is $2$.