# ps1-p10-daquila

```Problem Set 1
Question 10
How many zeros are at the end of the number 100!?
Solution 1
We can ask Mathematica to compute the number 100! and count the number of zeros
at the end.
After entering the command
100!
Mathematica returned the value
9332621544394415268169923885626670049071596826438162146859296389521
7599993229915608941463976156518286253697920827223758251185210916864
000000000000000000000000
Now we just count the zeros at the end.
We see there are 24 zeros at the end of 100!
Solution 2
To get a zero at the end of a number we need to count the number of pairs of 2 and 5
that are in the prime factorization of the number.
In any factorial there will always be more factors of 2 than factors of 5 it suffices to
count the factors of 5 since there will be plenty of 2's to pair with them.
We break 100! into factors and count factors of 5.
1
2
3
4
5-1
6
7
8
9
10 - 1
2
11
12
13
14
15 - 1
16
17
18
19
20 - 1
21
22
23
24
25 - 2
26
27
28
29
30 - 1
31
32
33
34
35 - 1
36
37
38
39
40 - 1
41
42
43
44
45 - 1
46
47
48
49
50 - 2
51
52
53
54
55 - 1
56
57
58
59
60 - 1
61
62
63
64
65 - 1
66
67
68
69
70 - 1
71
72
73
74
75 - 2
76
77
78
79
80 - 1
81
82
83
84
85 - 1
86
87
88
89
90 - 1
91
92
93
94
95 - 1
96
97
98
99
100 - 2
2
3
2
3
2
2
3
2
3
Add up the number of factors of 5: 2 + 2 + 3 + 2 + 3 + 2 + 2 + 3 + 2 + 3 = 24
We see there are 24 factors of 5 so in conclusion the number of zeros at the end of
100! is going to be 24.
```