Math 55: Discrete Mathematics, Fall 2008
Homework 3 Solutions
3.5: 6. There are 24 zeroes at the end of 100!. To see this, let
e
be the exponent of 2
and
f
exponent of 5 in the prime factorization of 100!. Then the largest power of 10 that
divides 100! is 10
min(
e,f
)
. To calculate
f
, observe that 20 of the factors in 100! = 1
·
2
· · ·
100
are multiples of 5, and of those, four (25, 50, 75, and 100) contain an extra factor of 5. Since
none of them contains a factor 5
3
, we get
f
= 24. Since there are 50 even factors, we see
that
e
≥
50, and therefore the minimum of
e
and
f
is 24, without further calculating
e
.
*18. We can assume
n >
1, so
n
is not relatively prime to itself. Therefore
φ
(
n
) is equal
to the number of members of the set
{
1
,
2
, . . . , n

1
}
that are relatively prime to
n
, and
thus
φ
(
n
) =
n

1 if and only if every positive integer less than
n
is relatively prime to
n
. If
n
is prime, then clearly this condition holds. Conversely, if the condition holds, then
n
has
no divisor 1
< d < n
, since then
d
would not be relatively prime to
n
. So
n
is prime.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 STRAIN
 Math, Number Theory, Prime number, congruences

Click to edit the document details