The factorial of a nonnegative integer ,
denoted by
, is the product of all
integers from
to
. For
example,
.
Problem 2‑19
Trailing zeros: How many trailing zeros are there in ?
Solution:
We know that each pair of 2 and 5 will give a trailing zero. If we perform
prime number decomposition on all the numbers in , it
is obvious that the frequency of 2 will far outnumber the frequency of 5. Thus,
the frequency of 5 determines the number of trailing zeros. Among numbers
and
20
numbers are divisible by 5 (
). Among these 20
numbers, 4 are divisible by 52 (
). Therefore, the total
frequency of 5 is 24, and there are 24 trailing zeros.
Problem 2‑20
is an integer less than
or equal to 25 and
is not a multiple of
How
many possible values can
have?
Solution:
Since there is no obvious solution, let’s start by testing different values of
Now the pattern emerges: when is
small
,
is not
a factor of
As
becomes
larger,
is not a factor of
if
and only if
is a prime number. Unless
is a prime number, we
can express it as the product of two numbers that are smaller than
Possible
values of
are prime numbers
(seven
choices). Altogether, there are 10 numbers less than 25 that
are
not a multiple of