The last non-zero digit of n!, (n+1)!, (n+2)! at intervals of 25 in n

 
 
Let d(n) be the last non-zero digit of n!

In the first triple, the last non-zero digit is 8, that is,   d(9) = d(10) = d(11) = 8

  9! = 362880
10! = 3628800
11! = 39916800
 

The second triple:     d(34)   =   d(35)   =   d(36)   =   2

34! = 295232799039604140847618609643520000000
35! = 10333147966386144929666651337523200000000
36! = 371993326789901217467999448150835200000000
 

The third triple:     d(59)   =   d(60)   =   d(61)   =   6

59! = 1386831185456898357379390197203894063459028767726874325408\\
21294940160000000000000
60! = 8320987112741390144276341183223364380754172606361245952449277696\\
409600000000000000
61! = 507580213877224798800856812176625227226004528988036003099405939\\
480985600000000000000
 

d(84)   =   d(85)   =   d(86)   =   2

d(109)   =   d(110)   =   d(111)   =   2

d(134)   =   d(135)   =   d(136)   =   4

d(159)   =   d(160)   =   d(161)   =   6

d(184)   =   d(185)   =   d(186)   =   8

…………………………………
 

It appears that triples   d(n),   d(n+1),   d(n+2)   occur at intervals of 25 in n.

Explain why this is so.

 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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