2) For which values of n does n! have n or fewer digits?

acc=1
for i in range(1,25):
    acc = i * acc
    print(str(i) + "! = " + str(acc) + " len=" + str(len(str(acc))))

xxxxxxxxxx
21! = 51090942171709440000 len=20
22! = 1124000727777607680000 len=22

4) Using only pencil and paper, find the value of 100!/95! .

5!/3! = 5x4x3x2x1 / 3x2x1

100x99x98x97x96

6) There are two 0’s at the end of 10! = 3,628,800. Using only pencil and paper, determine how many 0’s are at the end of the number 100!.

The only way to get 0 at the end of a number is to have a factor of 10. The only way to have a factor of 10 is to have factors of both 2 and 5.

10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 (5x2) x (3x3) x (2x2x2) x (7) x (3x2) x (5) x (2x2) x (3) x (2) x (1) $(7^1)(5^2)(3^4)(2^8)(1^1)$

Looks like 5 is the limiting factor. However many 5 factors exist is the number of 0's in the result.

25, 50, 75, and 100 all have 5x5. 5, 10, 15, 20, 30, 35, 40, 45, 55, 60, 65, 70, 80, 85, 90, 95 each have 5 as a factor. So, 100! has 5²⁴ as a factor and so has 24 zeros at the end. (Wolfram Alpha agrees.)

8) Compute how many 7-digit numbers can be made from the digits 1,2,3,4,5,6,7 if there is no repetition and the odd digits must appear in an unbroken sequence. (Examples: 3571264 or 2413576 or 2467531, etc., but not 7234615.)

Odd numbers starting in pos 1...

1. Specify odd number for pos 1 (4)

2. Specify an unused odd number for pos 2 (3)

3. Specify an unused odd number for pos 3 (2)

4. Specify an unused odd number for pos 4 (1)

5. Specify even number for pos 5 (3)

6. Specify an unused even number for pos 6 (2)

7. Specify an unused even number for pos 7 (1)

   (odds start 1) + (odds start 2) + (odds start 3)

   (4x3x2x1x3x2x1)x3

10) How many permutations of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are there in which the digits alternate even and odd? (For example, 2183470965.)

1. Specify a permutation of 02468 (5!)
2. Specify a permutation of 13579 (5!)
3. Specify whether even or odd begins the list (2)
4. Interleave permutations even-odd-even-odd-etc (1)

5! x 5! x 2