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Learning Math Home
Number and Operations Session 6: Solutions
 
Session 6 Part A Part B Homework
 
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Solutions for Session 6 Homework

See solutions for Problems: H1 | H2 | H3 | H4 | H5


Problem H1

The number must be a square; otherwise, it would have an even number of factors. Try some square numbers:

1: 1 (one factor)
4: 1, 2, 4 (three factors)
9: 1, 3, 9 (three factors)
16: 1, 2, 4, 8, 16 (five factors)
25: 1, 5, 25 (three factors)
36: 1, 2, 3, 4, 6, 9, 12, 18, 36 (nine factors)
49: 1, 7, 49 (three factors)

The first four numbers with this property are 4, 9, 25, and 49. The next three after that are 121, 169, and 289. In general, the way to categorize these numbers is that they are squares of the prime numbers.

<< back to Problem H1


 

Problem H2

Here is the completed table:

Integer

Prime Factori-
zation

Number of
Factors

2

21

2

4

22

3

8

23

4

16

24

5

2n

2n

(n + 1)

3

31

2

9

32

3

27

33

4

81

34

5

3m

3m

m + 1

6

21 • 31

4

12

22 • 31

6 (i.e., 3 • 2)

18

21 • 32

6 (i.e., 2 • 3)

36

22 • 32

9 (i.e., 3 • 3)

2n • 3m

2n • 3m

(n + 1) • (m + 1)

To find the number of factors for any number, write the prime factorization of your number. Then record one more than each exponent. The number of factors will be the product of the augmented exponents.

So, for 72 = 23 • 32, the exponents are 3 and 2. One more than each exponent gives the numbers 4 and 3. The number of factors is 4 • 3, or 12.

<< back to Problem H2


 

Problem H3

For this problem and the next, we need the following list of numbers and the sum of their factors:

Number

Factors

Sum

1

 1

1

2

 1, 2

3

3

 1, 3

4

4

 1, 2, 4

7

5

 1, 5

6

6

 1, 2, 3, 6

12

7

 1, 7

8

8

 1, 2, 4, 8

15

9

 1, 3, 9

13

10

 1, 2, 5, 10

18

11

 1, 11

12

12

 1, 2, 3, 4, 6, 12

28

13

 1, 13

14

14

 1, 2, 7, 14

24

15

 1, 3, 5, 15

24

16

 1, 2, 4, 8, 16

31

17

 1, 17

18

18

 1, 2, 3, 6, 9, 18

39

19

 1, 19

20

20

 1, 2, 4, 5, 10, 20

42

21

 1, 3, 7, 21

32

22

 1, 2, 11, 22

36

23

 1, 23

24

24

 1, 2, 3, 4, 6, 8, 12, 24

60

25

 1, 5, 25

31

26

 1, 2, 13, 26

42

27

 1, 3, 9, 27

40

28

 1, 2, 4, 7, 14, 28

56

The first two perfect numbers are 6 and 28, since their factors add to exactly twice the value of the number.

<< back to Problem H3


 

Problem H4

Refer to the table for Problem H3.

Abundant numbers less than 25 are 12, 18, 20, and 24. All others (besides 6, which is perfect) are deficient.

<< back to Problem H4


 

Problem H5

You could use a sieve-like table similar to the one used in this session:

You can see that the first and the third columns get crossed out immediately. Thus, the prime numbers will be located in the second or fourth columns.

Alternatively, you could argue that every number is either zero, one, two, or three more than a multiple of 4. If a number is zero or two more, then it can't be prime (unless it's 2), for such numbers are divisible by 2. This leaves "one more" and "three more" as the only choices. Three more than a multiple of 4 is the same as one less than the next multiple of 4. So again, the prime numbers will be located in the columns that are one more or one less than a multiple of 4.

<< back to Problem H5


 

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