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Week 5 Chapter 4 CheckPoint
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Text
Page
Section Numbers
Section
4.1
Section
4.2
Chapter 4
Review
4.1
Exercises
Pg. 217-221
24, 32, 46, 84
Pg. 241-244
12 , 26, 34
Pg. 245
1-20 (all even-numbered questions)
#
24
Problem and Final Answer
In Exercises 24, fill the omitted numbers in as many methods as
possible so that the figure will be divisible by 6.
3120 3222 3024 3126 3228
3420 3522 3324 3426 3528
3720 3822 3624 3726 3828
3924
32
a. Incorrect, because there's no whole number n, in a way that 7n = 23
b. Incorrect, 24 is a multiple of 8, not a factor.
c. Correct, 16 does divide 16, since 16 X 1 = 16.
d. Incorrect, because there is no whole number n, in a way that 0 X n = 6.
e. Correct, because there is a whole number, , in a way that 15 x 0 = 0.
f. Incorrect, because there is no whole number n, in a way that 15 X n = 65.
g. Correct, because n  n  n
h. Correct, because if n is a whole number, there's a whole number n, in a way
3
2
that n  n  n
46
2
5
Week 5 Chapter 4 CheckPoint
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name on the File and also to name the NCTM standards that this work is connected to for content.
Name__________________________
a. Correct. Think about the set of non-zero even numbers up to 100: S = 2, 4, 6,
.., 96, 98, 100. Every other number is a multiple of 4 and therefore is divisible by
4. Therefore one half of the set is divisible by 4.
b. Incorrect. Think about 12. Both 2 and 4 are factors of 12 and therefore divide
12 however 8 isn't a factor of 12.
c. Correct. Because 12 divides the number N, and then N = 12 × Y, Y some
natural number. However N = 6 × (2 × Y). Therefore 6 is a factor of N and 6
divides N.
d. Incorrect. Both 4 and 6 divide 12 however 24 isn't a factor of 12.
84
Divisibility and multiples assist with finding the GCF and LCM of fractions
making the simplifying of fractions easier.
4.2
12
51,051  3 =17017, 17017  7 = 2431, etc. So 51,051= 3  7  11  13  17.
26
1078 = 1,2,7,11,14,22,49,77,154,539,1078
3315 = 1,3,5,13,15,17,39,51,65,85,195,221,255,663,1105,3215
34
Week 5 Chapter 4 CheckPoint
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18=18,36,54,72,90
30=30,60,90
2
Because, 841  29 by the Factor Test Theorem we just need to test factors
under 29.
4
A) Only when its units digit is ,2,4,6,or 8
B) Only when the sum of its digit is divisible by 3
C) Only when the figure represented by its final 2 numbers is divisible
by 4
D) Only when its units digit is a or a 5
E) Only when it is divisible by both 2 and 3
6
a. Incorrect. 3 divides 24 however 9 doesn't.
b. Correct. If 10 divides n, in that case n = 10(a) = 5(2)(a). Therefore 5 is a factor
of n and therefore divides n.
c. Incorrect. 2 and 4 both divide 12 however 8 doesn't divide 12.
8
a. 97 is prime. It's not divisible by any prime below 10.
b. 51 isn't prime. It's divisible by 3 and 17.
c. 143 isn't prime. It's divisible by 11 and 13.
d. 223 is prime. It isn't divisible by any prime under 15.
10
48= 2  3 Because the product of the exponents, each enhanced by 1, is 5 X 2,
or 10, 48 has 10 factors.
4
12
Week 5 Chapter 4 CheckPoint
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a. The set of multiples of 24 is {24, 48, 72, 96, 120, 144, 168, 192, ...} The set of
multiples of 32 is {32, 64, 96, 128, 160, 192, .... }. The lowest element of the
junction of these sets is the LCM, 96.
b. The two prime factorizations are 24 = 2 × 2 × 2 × 3 and 32 = 2 × 2 × 2 × 2 × 2.
The LCM is 2 × 2 × 2 × 2 × 2 × 3 = 96.
14
The GCF of the numbers is 3. Because GCF (a,b) X LCM(a,b)=ab , and then 60 X
GCF=180, and
GCF = 3.
16
Reasons will change. A prospect is: we observe that 2 and 3 are relatively prime.
A number divisible by one of them isn't always divisible by the other.
Conversely, any number divisible by 4 is always divisible by 2 since 2 is a part of
4. For example, think about 12. It's divisible by both 2 and 3 and even by 6. It's
divisible by both 2 and 4, however not by 8.
18
8(1, 2, 4, 8), 10(1, 2, 5, 10), 15(1, 3, 5, 15), 26(1, 2, 13, 26), and 33(1, 3, 11, 33)
all have 4 factors. The numbers 5(1, 5), 9(1, 3, 9), 16(1, 4, 16), 18(1, 2, 9, 3, 6,
18), and 24(1, 2, 12, 3, 8, 4, 6, 24) do not. Two more numbers with the feature
of having 4 factors are: 21(1, 3, 7, 21) and 55(1, 5, 11, 55). All numbers which
are the product of 2 primes will have this feature.
20
The utilization of factors, multiples, prime factorization and relatively prime numbers to solve problems.
Relating, composing and decomposing numbers.