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MTH 156 Week 5: CheckpointSection 4.1

24.

3120 3222 3024 3126 3228

3420 3522 3324 3426 35283720 3822 3624 3726 38283924

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, 15 x 0 = 0.f. Incorrect, because there is no whole number n, 15 X n = 65.

g. Correct, because n3 xn2=n5h. Correct, because if n is a whole number, there's a whole number n, n x n= n2

46.

a. Correct. S = 2, 4, 6, .., 96, 98, 100. Every other number is a multiple of 4 and thereforeis 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, 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 thesimplifying of fractions easier 

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Sections 4.2

12.51,051 3 =17017, 17017 7 = 2431, etc. So 51,051= 3 7 11 13 17.

26.1078 = 1,2,7,11,14,22,49,77,154,539,10783315 = 1,3,5,13,15,17,39,51,65,85,195,221,255,663,1105,3215

34.

18=18,36,54,72,9030=30,60,90

Chapter 4 Review

All even numbers

2. 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 5E)  Only when it is divisible by both 2 and 3

6.a. Incorrect. 3 divides 24 but 9 doesn't.  b. Correct. If 10 divides n, in that case n = 10(a) = 5(2)(a). Therefore 5 is a factor of nand 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 has 10 factors.

z z v v v v

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12.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 XGCF=180, and

GCF = 3.

16.My assumption is that 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 4is 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.The first set of numbers have only four factors«

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)while the second set does not.

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)Two more numbers with the feature of having 4 factors are: 21(1, 3, 7, 21) and 55(1, 5,

11, 55).

The utilization of prime factors, multiples, prime factorization and prime numbers to

solve problems all are relative to NCTM standards used in this weeks material. Relating,composing and decomposing numbers are all used to solve problems.