Question

investigation how does the prime factorization of a number compare to the prime factorization of its multiples?
a. write the prime factorizations of the first ten multiples of 90 (dont include zero).

multiples of 90	prime factorizations
2 × 90	2 × 2 × 3 × 3 × 5
3 × 90	3 × 2 × 3 × 3 × 5
4 × 90	2 × 2 × 2 × 3 × 3 × 5

|5 × 90|
|6 × 90|
|7 × 90|
|8 × 90|
|9 × 90|
|10 × 90|
b. how many 2s are in the prime factorization of 90? ______
does any multiple of 90 have fewer 2s in its prime factorization than 90 does? ______
c. how many 3s are in the prime factorization of 90? ______
does any multiple of 90 have fewer 3s in its prime factorization than 90 does? ______
d. how many 5s are in the prime factorization of 90? ______
does any multiple of 90 have fewer 5s in its prime factorization than 90 does? ______
e. finish the sentences below by writing at least or at most.
any multiple of 90 must have ______ as many 2s in its prime factorization as 90 does.
any multiple of 90 must have ______ as many 3s in its prime factorization as 90 does.
any multiple of 90 must have ______ as many 5s in its prime factorization as 90 does.
f. does 90 have a 7 in its prime factorization? ______
does any multiple of 90 have a 7 in its prime factorization? ______
the prime factorization of 60 is 2 × 2 × 3 × 5.
without doing any calculations, circle the products that show multiples of 60:
2 × 3 × 5 × 7 × 7 2 × 2 × 5 × 5 × 5 2 × 2 × 2 × 2 × 3 × 5 × 11
how do you decide which products it is true?

Explanation:

Step1: Prime - factorize multiples of 90

To find the prime - factorization of a multiple of 90, we multiply the prime - factorization of 90 ($2\times3\times3\times5$) by the prime - factorization of the multiplier.

Step2: Analyze prime factors of 90

The prime - factorization of 90 is $2\times3\times3\times5$, so there is 1 two, 2 threes, and 1 five.

Step3: Analyze multiples of 90

Since a multiple of 90 is of the form $n\times90$ where $n$ is a positive integer, and prime - factorizations are built by multiplying prime factors, no multiple of 90 can have fewer of the prime factors that are in 90. Also, if $n$ has prime factors not in 90, the multiple will have those additional prime factors.

Step4: Complete sentences

Any multiple of 90 must have at least as many 2s, 3s, and 5s in its prime - factorization as 90 does. Since 7 is not a prime factor of 90, 90 does not have a 7 in its prime - factorization. A multiple of 90 may or may not have a 7 in its prime - factorization depending on the multiplier.

Multiples of 90 and their prime factorizations:

Multiples of 90	Prime factorizations
$2\times90$	$2\times2\times3\times3\times5$
$3\times90$	$3\times2\times3\times3\times5$
$4\times90$	$2\times2\times2\times3\times3\times5$
$5\times90$	$5\times2\times3\times3\times5$
$6\times90$	$2\times3\times2\times3\times3\times5$
$7\times90$	$7\times2\times3\times3\times5$
$8\times90$	$2\times2\times2\times2\times3\times3\times5$
$9\times90$	$3\times3\times2\times3\times3\times5$
$10\times90$	$2\times5\times2\times3\times3\times5$

Answer:

s to sub - questions:
B. How many 2s are in the prime factorization of 90? 1
Does any multiple of 90 have fewer 2s in its prime factorization than 90 does? No
C. How many 3s are in the prime factorization of 90? 2
Does any multiple of 90 have fewer 3s in its prime factorization than 90 does? No
D. How many 5s are in the prime factorization of 90? 1
Does any multiple of 90 have fewer 5s in its prime factorization than 90 does? No
E. Any multiple of 90 must have at least as many 2s in its prime factorization as 90 does.
Any multiple of 90 must have at least as many 3s in its prime factorization as 90 does.
Any multiple of 90 must have at least as many 5s in its prime factorization as 90 does.
F. Does 90 have a 7 in its prime factorization? No
Does any multiple of 90 have a 7 in its prime factorization? Maybe (depending on the multiplier)