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Mth211
IDEAS & ANSWERS
[DC26-35]
[DC26]
Let the Black Chips be Positive, Red Chips be Negative. Here’s some steps:
a]
3–(4)
Step1: We have a collection of 3 Blacks:
----------------------------------------------------------------------------------------------Step2: We’d like to take away 4 Reds,
So we need to introduce some
“Zero Pairs” (Red & Blacks)…Now
we have the following collection:
-----------------------------------------------------------------------------------------------Step3: Now the 4 Reds can be taken away,
leaving us with 7 Black. And we see
that 3 – (  4 ) = + 7
b]

2(7)
Step1: We have a collection of 2 Reds
------------------------------------------------------------------------------------------------Step2: We’d like to take away 7 Reds, so
let’s introduce some “Zero Pairs”
(Red & Blacks). …Now we have
the following collection:
--------------------------------------------------------------------------------------------------Step3: Now the 7 Reds can be taken away,
leaving us with 5 Black. And we see
that  2  (  7 ) = +5
[DC27]
a) 7560
The actual tree will vary (depending on how you started to break
up 7560…For example, some may have started with 10 x 756, etc.)
But in the end, no matter how you started, you’d end up with
(23)(33)(5)(7)
b) 2700
The actual tree will vary (depending on how you started to break
up 2700…For example, some may have started with 27 x 100, etc.)
But in the end, no matter how you started, you’d end up with
(22) (33)(52)
[DC28]
For these, assume we already did Step1, which is to generate the Factor Trees
(as we did in [DC27] )…So, we assume we already have the Prime Factorization
of 7560 as (23)(33)(5)(7) and 2700 as (22) (33)(52)
a) LCM(7560, 2700)
Step1: As described above.
Step2: Well, we need the Prime Factor of 2 --- (Three of these)
& we need the Prime Factor of 3 --- (Three of these)
& we need the Prime Factor of 5 --- (Two of these)
& we need the Prime Factor of 7 --- (One of these)
Step3: So, we have the LCM(7560, 2700) = (23)(33)(52)(7) = 37800
b) GCF (7560, 2700)
Step1: As described above
Step2: Well, we need the Prime Factor of 2 --- (Two of these)
& we need the Prime Factor of 3 --- (Three of these)
& we need the Prime Factor of 5 --- (One of these)
Step3: So, we have the GCF(7560, 2700) = (22)(33)(5) = 540
[DC29]
Show your work in determining how many factors are in
a) 7560
Knowing the Prime Factorization of 7560 = (23)(33)(51)(71), we
have the Total Number of Factors = (3 + 1)(3 + 1)(1 + 1)(1 + 1) = 64
b) 2700
Knowing the Prime Factorization of 2700 = (22)(33)(52), we
have the Total Number of Factors = (2 + 1)(3 + 1)(2 + 1) = 36
[DC30]
Consider the number 120.
a) Show the use of a Factor Tree (as done in class) in determining the Prime
Factorization of 120
The actual tree will vary (depending on how you started to break
up 120…For example, some may have started with 10 x 12, etc.)
But in the end, no matter how you started, you’d end up with
(23)(3)(5)
b) Clearly illustrate your method (as done in class) of coming up with a list of all
the factors of 120
This would involve combinations as follows:
Make a Choice from List A : { 20, 21 , 22 , 23 }
Make a Choice from List B : { 30, 31 }
Make a Choice from List C : { 50, 51 }
We’ll multiply our choices together to come up with our list of All Factors,
(a List which of course has 4 x 2 x 2 =16 Total Elements. Here’s all the
combinations:
(20)(30)(50) = 1
(20)(31)(50) = 3
(20)(30)(51) = 5
(20)(31)(51) = 15
(21)(30)(50) = 2
(21)(31)(50) = 6
(21)(30)(51) = 10
(21)(31)(51) = 30
(22)(30)(50) = 4
(22)(31)(50) = 12
(22)(30)(51) = 20
(22)(31)(51) = 60
(23)(30)(50) = 8
(23)(31)(50) = 24
(23)(30)(51) = 40
(23)(31)(51) = 120
So All Factors of 120 =
{ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120}
[DC31]
a) This is the GCF(130, 182), which we assume we could show ALL the work
as described in class and in [DC28], for example. Note the GCF(130, 182) =
26, so the answer is: 26 Inches.
b) For Copper Wire, we’d have 130/26 = 5 Pieces of Copper
For Steel Wire, we’d have 182/26 = 7 Pieces of Steel.
Note that we’d have 12 Total Pieces of Wire , as described above.
[DC32]
This is the LCM(120, 220), which we assume we could show ALL the work
as described in class and in [DC28], for example. Note the LCM(120, 220) =
1320, so the answer is: 1320 Miles.
[DC33]
i) GCF(315,180) = 45 Cookies
ii) 7 Piles ChocoChip
iii) 4 Piles Lemon
[DC34]
Note that our Prime Factorization Method works just as well for finding LCM of
three numbers as it did for two numbers. In other words, the exact technique
we used for LCM(x, y) is the same for LCM(x, y, z). In this case:
Prime Factorization of 10 = 2 x 5
of 35 = 5 x 7
of 45 = 3 x 3 x 5
So LCM(10,35,45) means …we need Factor of 2….(One Time)
…we need Factor of 3….(Two Times)
…we need Factor of 5….(One Time)
…we need Factor of 7….(One Time)
So LCM(10,35,45) = 2 x 3 x 3 x 5 x 7 = 630 Minutes, which = 10 Hours & 30
Minutes. Add this to 5:00pm to get our Final Answer of : 3:30am.
[DC35]
Show the steps in the Prime Number Test (as done in class) in determining
whether or not 319 is prime.
Step1: We find the smallest whole number N so that (N – 1)2  319  N2
In this case, N = 18, since [289 = 172 ]  319  [ 182 = 324 ]
Step2: We need a list of the Prime #s that are less than N, and these are
our candidates to test. In this case, we have {2, 3, 5, 7, 11, 13, 17}
(Note: The point of the Prime # Test is that these are the ONLY
candidates that need to be tested.)
Step3: We test those Prime #s:
2…Is Not a Factor of 319, (since 2 doesn’t divide 9)
3…Is Not a Factor of 319, (since 3 doesn’t divide [3+1+9] )
5…Is Not a Factor of 319, (since 5 doesn’t divide 9)
7…Is Not a Factor of 319, (I used a calculator to check)
11..Is Actually a Factor of 319, since 11 x 29 = 319
We can stop there, since we see that 319 is a composite number.
[DC36]
[DC37]
[DC38]
130 People did not show up.
a) 1/6 of Collection b) 360 Pens
17
/18 of a Box at Home
c) 144 Gold Pens