Factorial Factoids

Copyright ©2003 Sophisticated: The Magazine. All rights reserved.
The cipher 9 does not appear in any of the first 90 factorials.
Will that hold true for all factorials?

s already mentioned, factorials get huge fast.  Here are the first couple of dozen...
 
  0! = 1 (by definition)
  1! = 1
  2! = 2
  3! = 6
  4! = 24
  5! = 120
  6! = 720
  7! = 5,040
  8! = 40,320
  9! = 362,880
10! = 3,628,800
11! = 39,916,800		 12! =  479,001,600
13! =  6,227,020,800
14! = 87,178,291,200
15! = 1,307,674,368,000
16! = 20,922,789,888,000
17! = 355,687,428,096,000
18! = 6,402,373,705,728,000
19! = 121,645,100,408,832,000
20! = 2,432,902,008,176,640,000
21! = 51,090,942,171,709,440,000
22! = 1,124,000,727,777,607,680,000
23! = 25,852,016,738,884,976,640,000

...and we already find ourselves in the sextillions.  The most significant digits are distributed as follows (digit | quantity):

1 | 6,  2 | 5,  3 | 4,  4 | 2,  5 | 2,  6 | 3,  7 | 1,  8 | 1,  9 | 0
...and that bar-chart in the puzzle statement summarizes these distributions as percentages for the first 90 factorials.  What that means is, in formulating Factorial Factoids, somebody had to look at the left-most ciphers in some mighty big integers -- integers capable of representing the number of sub-atomic particles in the known universe.  Actually 90! can cope with gazillions of unknowable universes each corresponding to a sub-atomic particle in the known universe (huh?).

ophisticated solvers will settle the matter concerning all factorials by proving that there is at least one value of n! which can indeed have a 9 as its most significant digit.  If so, it must emerge from some previous factorial having a most significant digit that is not 9.

Let (n - 1)! have a most significant digit m less than 9 such that, with s arbitrary decimal digits to the right of m; we can write…

(n - 1)! = 10s m + x, where  x can have any value between 0 and 10s - 1 (a string of s 9s).
Multiplying the expression for (n – 1)! by n forms n!
n! = n (n – 1)!
n! = 10s m n + n x
As n increases, the number of decimal digits in n! will increase, such that there are s + t digits to the right of the most significant digit in n!, where t > 0.

Now, to test the hypothesis, let the most significant digit of n! be 9, such that the following inequality holds:

10s m n + n x > 9 * 10s+t  + n x  ~~~ for a most significant digit of 9.
10s m n  > 9 * 10s+t    ~~~~~~~~~~~ for x = 0, its smallest permissible value and 
                                                         least helpful in getting to 9.
m n > 9 * 10t ~~~~~~~~~~~~~~~~~ after dividing both sides by 10s
Thus, it would seem that, contrary to the results of extensive observations, there must exist some values for m, n, and t that will satisfy that inequality.  Well then, why did the person who created this puzzle not find any instances even after foozling factorials that reached sizes exceeding anything meaningful to people on earth?  Hmm.
Must have quit too soon.
Guided by the formula above, we will start by letting t = 1, such that...
m n > 90
...then for m = 1, there must exist a value for n! that will indeed have a cipher 9 as its most significant digit -- it lives somewhere out there among the n! integers produced by n > 90.  Now, the creator of this puzzle has already tested n! for  n  = 0, 1, 2,... 89.  Thus it merely remains for us to test larger values of n, specifically n = 90, 91, 92,...  Hey, and let's not quit too soon this time.

Sure enough, for n = 96, we see that...

96! = 991,677,934,870,949,689,209,571,401,541,893,801,158,183,
648,651,267,795,444,376,054,838,492,222,809,091,499,987,689,
476,037,000,748,982,075,094,738,965,754,305,639,874,560,000,
000,000,000,000,000,000

96! = 9.9168*10149

...about 10 quadecoctillion, one might say. 

Oh right, and the answer to the question in the puzzle therefore is…
 

No


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