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57 arithmetic/digits/arabian.p




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This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

57 arithmetic/digits/arabian.p


What is the Arabian Nights factorial, the number x such that x! has 1001
digits? How about the prime x such that x! has exactly 1001 zeroes on
the tail end. (Bonus question, what is the 'rightmost' non-zero digit in x!?)

arithmetic/digits/arabian.s

The first answer is 450!.

Determining the number of zeroes at the end of x! is relatively easy once
you realize that each such zero comes from a factor of 10 in the product

1 * 2 * 3 * ... * x

Each factor of 10, in turn, comes from a factor of 5 and a factor of 2.
Since there are many more factors of 2 than factors of 5, the number of 5s
determines the number of zeroes at the end of the factorial.

The number of 5s in the set of numbers 1 .. x (and therefore the number
of zeroes at the end of x!) is:

z(x) = int(x/5) + int(x/25) + int(x/125) + int(x/625) + ...

This series terminates when the powers of 5 exceed x.

I know of no simple way to invert the above formula (i.e., to find x for
a given z(x)), but I can approximate it by noting that, except for the "int"
function,

5*z(x) - x = z(x)

which gives:

x = 4*z(x) (approximately).

The given problem asked, "For what prime x is z(x)=1001". By the above forumla,
this is approximately 4*1001=4004. However, 4004! has only

800 + 160 + 32 + 6 + 1 = 999 zeroes at the end of it.

The numbers 4005! through 4009! all have 1000 zeroes at their end and
the numbers 4010! through 4014! all have 1001 zeroes at their end.

Since the problem asked for a prime x, and 4011 = 3*7*191, the only solution
is x=4013.

The problem of determining the rightmost nonzero digit in x! is somewhat more
difficult. If we took the numbers 1, 2, ... , x and removed all factors of 5
(and an equal number of factors of 2), the remaining numbers multiplied
together modulo 10 would be the answer. Note that since there are still many
factors of 2 left, the rightmost nonzero digit must be 2, 4, 6, or 8 (x > 1).

Letting r(x) be the rightmost nonzero digit in x!, an expression for r(x) is:

r(x) = (r(int(x/5)) * w * r(x mod 10)) mod 10, x >= 10.

where w is 4 if int(x/10) is odd and 6 if it is even.

The values of r(x) for 0 <= x <= 9 are 1, 1, 2, 6, 4, 2, 2, 4, 2, and 8.

The way to see this is true is to take the numbers 1, 2, ..., x in groups
of 10. In each group, remove 2 factors of 10. For example, from the
set 1, 2, ..., 10, choose a factor of 2 from 2 and 6 and a factor of 5 from
5 and 10. This leaves 1, 1, 3, 4, 1, 3, 7, 8, 9, 2. Next, separate all the
factors that came from multiples of 5. The rightmost nonzero digit of x!
can now (hopefully) be seen to be:

r(x) = (r(int(x/5)) * w * r(x mod 10)) mod 10

where w is the rightmost digit in the number formed by multiplying the numbers
1, 2, 3, ..., 10*int(x/10) after the factors of 10 and the factors left over
by the multiples of 5 have been removed. In the example with x = 10, this
would be (1 * 1 * 3 * 4 * 3 * 7 * 8 * 9) mod 10 = 4. The "r(x mod 10)" term
takes care of the numbers from 10*int(x/10)+1 up to x.

The "w" term can be seen to be 4 or 6 depending on whether int(x/10) is odd or
even since, after removing 10*n+5 and 10*n+10 and a factor of 2 each from
10*n+2 and 10*n+6 from the group of numbers 10*n+1 through 10*n+10, the
remaining factors (mod 10) always equals 4 and 4^t mod 10 = 4 if t is odd and
6 when t is even (t != 0).

So, finally, the rightmost nonzero digit in 4013! is found as follows:

  r(4013) = (r(802) * 4 * 6) mod 10
  r(802)  = (r(160) * 6 * 2) mod 10
  r(160)  = (r(32)  * 6 * 1) mod 10
  r(32)   = (r(6)   * 4 * 2) mod 10

Using a table of r(x) for 0 <= x <= 9, r(6) = 2. Then,

  r(32)   = (2 * 4 * 2) mod 10 = 6
  r(160)  = (6 * 6 * 1) mod 10 = 6
  r(802)  = (6 * 6 * 2) mod 10 = 2
  r(4013) = (2 * 4 * 6) mod 10 = 8

Thus, the rightmost nonzero digit in 4013! is 8.

 

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