MATH 245: QUIZ 3 SOLUTIONS
1)
a)
980 = (2)(2)(5)(7)(7) =
2
2
⋅
5
⋅
7
2
b)
616 = (2)(2)(2)(7)(11) =
2
3
⋅
7
⋅
11
c)
The prime factors of interest are 2, 5, 7, and 11.
For the lcm, we take the larger exponent on each prime factor.
980 =
2
2
⋅
5
1
⋅
7
2
⋅
11
0
616 =
2
3
⋅
5
0
⋅
7
1
⋅
11
1
lcm =
2
3
⋅
5
1
⋅
7
2
⋅
11
1
=
21,560
d)
For the gcd, we take the smaller exponent on each prime factor.
980 =
2
2
⋅
5
1
⋅
7
2
⋅
11
0
616 =
2
3
⋅
5
0
⋅
7
1
⋅
11
1
gcd =
2
2
⋅
5
0
⋅
7
1
⋅
11
0
=
2
2
⋅
7
=
28
2)
Let
a
= 4743 and
b
= 867.
ab
=
gcd
a
,
b
( )
⋅
lcm
a
,
b
( )
4743
( )
867
( ) =
51
( )
⋅
a
,
b
( )
a
,
b
( ) =
4743
( )
867
( )
51
=
80,631
3)
Let
n
be a composite integer.
Then,
n
=
rs
for some integers
r
and
s
strictly between 1 and
n
.
If both
r
and
s
are greater than
, then
rs
>
n
n
=
n
, which contradicts "
n
=
rs
."
So, one or both of
r
or
s
must be less than or equal to
.
4)
Think of 100! as (1)(2)(3)
(100). We want to count the number of "3factors"
in the integers from 1 to 100.
There are
100
3
⎢
⎣
⎢
⎥
⎦
⎥
= 33 multiples of 3 between 1 and 100. (3, 6, 9, …, 99)
We can pick up one 3factor from each of these 33 multiples.
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.
 Spring '11
 staff
 Math, Factors, Prime number, Euclidean algorithm, Prime factor

Click to edit the document details