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Learning Math Home
Number and Operations Session 7: Solutions
 
Session 7 Part A Part B Part C Homework
 
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A B C 
Homework

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Solutions for Session 7, Part A

See solutions for Problems: A1 | A2 | A3 | A4 | A5 | A6 | A7 | A8 | A9 | A10 | A11 | A12 | A13 | A14 | A15 | A16


Problem A1

It appears that the number of decimal places equals the power of 2. Therefore, 1/16 should have four decimal places. Checking the value by long division or using a calculator confirms this, since 1/16 = 0.0625.

<< back to Problem A1


 

Problem A2

Since 1/2 = 5/10, 1/2n = 5n/10n. Here, 10n dictates the number of decimal places, and 5n dictates the actual digits in the decimal. Since 54 = 625, 1/24 = 0.0625 (four decimal places).

<< back to Problem A2


 

Problem A3

Here is the completed table:

Fractions

Denominator

Prime Factorization

Number of Decimal
Places

Decimal Representa-
tion

1/2

2

21

1

0.5

1/4

4

22

2

0.25

1/8

8

23

3

0.125

1/16

16

24

4

0.0625

1/32

32

25

5

0.03125

1/64

64

26

6

0.015625

1/1,024

1,024

210

10

0.0009765625

1/2n

2n

2n

n

0.5n (with enough leading zeros to give n decimal places)

Notice that the decimal will include the power of 5, with some leading zeros. For example, 55 is 3,125, so "3125" shows up in the decimal, with enough leading zeros for it to comprise five digits: .03125. Similarly, 56 is 15,625, so the decimal is .015625 (six digits).

<< back to Problem A3


 

Problem A4

We know that 1/2n = 5n/10n, so for all unit fractions with denominators that are the nth power of 2, the decimal will consist of the digits of 5n with enough leading zeros to give n decimal places.

<< back to Problem A4


 

Problem A5

Here is the completed table:

Fractions

Denominator

Prime Factorization

Number of Decimal
Places

Decimal Representa-
tion

1/5

5

51

1

0.2

1/25

25

52

2

0.04

1/125

125

53

3

0.008

1/625

625

54

4

0.0016

1/3,125

3,125

55

5

0.00032

1/15,625

15,625

56

6

0.000064

1/5n

5n

5n

n

0.2n (with enough leading zeros to give n decimal places)

Notice that the decimal will include the power of 2, with some leading zeros. For example, 25 is 32, so the decimal is 0.00032 (five digits).

<< back to Problem A5


 

Problem A6

Fractions

Denominator

Prime Factorization

Number of Decimal
Places

Decimal Representa-
tion

1/10

10

21 • 51

1

0.1

1/20

20

22 • 51

2

0.05

1/50

50

21 • 52

2

0.02

1/200

200

23 • 52

3

0.005

1/500

500

22 • 53

3

0.002

1/4,000

4,000

25 • 53

5

0.00025

1/5n

2n • 5m

2n • 5m

max(n, m)

For m > n:
(10-m)(2m-n)
or
for n > m:
(10-n)(5n-m)

"Max(n, m)" means the larger of m or n -- in other words, the greater exponent between the power of 2 and the power of 5 in the third column.

<< back to Problem A6


 

Problem A7

a. 

Yes, they will all terminate.

b. 

A decimal terminates whenever it can be written as n/10k for some integer n and k. Then n will be the decimal, and there will be k decimal places. Since any number whose factors are 2s and 5s must be a factor of 10k for some k, the decimal must terminate. Specifically, k will be the larger number between the powers of 2 and 5 in the denominator. (See the table in Problem A6 for some examples.)

<< back to Problem A7


 

Problem A8

Answers will vary.

<< back to Problem A8


 

Problem A9

Here is the completed table:

Fraction

Denominator

Period

Decimal Representation

1/2

2

terminating

0.5

1/3

3

1

0.333...

1/5

5

terminating

0.2

1/7

7

6

0.142857...

1/11

11

2

0.090909...

1/13

13

6

0.076923076923...

1/17

17

16

0.05882352941176470588...

1/19

19

18

0.05263157894736842105...

<< back to Problem A9


 

Problem A10

Since the decimal cannot terminate (because the denominator contains factors other than powers of two and/or five), the remainder 0 is not possible. That means that there are only six possible remainders when we divide by 7: 1 through 6. When any remainder is repeated, the decimal will repeat from that point. If, after six remainders, you have not already repeated a remainder, the next remainder must repeat one of the previous remainders, because you only had six to choose from. Therefore, there can be no more than six possible remainders before the remainder begins repeating itself.

<< back to Problem A10


 

Problem A11

Yes, the period is one less than the denominator -- it can never be more. For example, when dividing by 19, there are 18 possible remainders.

<< back to Problem A11


 

Problem A12

The period for each of these is not one less than the denominator, but it is a factor of the number that is one less than the denominator. For example, 12 is one less than 13; 1/13 has a period of six, and 6 is a factor of 12.

<< back to Problem A12


 

Problem A13

Here is the completed table:

Fraction

Denomin-
ator

Period

Decimal Representation

1/23

23

22

 0.0434782608695652173913...

1/29

29

28

 0.0344827586206896551724137931...

1/31

31

15

 0.032258064516129...

1/37

37

3

 0.027027...

1/41

41

5

 0. 0243902439...

1/43

43

21

 0.023255813953488372093...

<< back to Problem A13


 

Problem A14

In all cases, the period is a factor of one less than the prime number in the denominator. For example, 1/41 has a period of five, and 5 is a factor of 40 (i.e., 41 - 1).

<< back to Problem A14


 

Problem A15

Judging from the pattern, we might expect the period to be a factor of 46. The possible factors are 1, 2, 23, and 46. The actual period is 46.

<< back to Problem A15


 

Problem A16

a. 

1/7 = , 6/7 = . Sliding the expansion of 1/7 by three digits yields the expansion of 6/7. All of the expansions of 2/7 through 6/7 can be built this way, by sliding the expansion of 1/7 by one through five digits. If the digits are written in a circle, the first digit of 6/7 will be directly opposite the first digit of 1/7. Similarly, the first digit of 5/7 will be opposite the first digit of 2/7, and the first digit of 4/7 will be opposite the first digit of 3/7. In every case, the two fractions add up to 7/7, or 1.

b. 

Thirteen has two rings:

 

1/13 = can be used to generate 10/13, 9/13, 12/13, 3/13, and 4/13.

 

2/13 = can be used to generate the others: 7/13, 5/13, 11/13, 6/13, and 8/13.

c. 

Answers will vary, but if you try this with enough prime numbers, you should find that the size of a ring is the same as the period of the expansion, and this determines the number of rings. For 41, each ring has five numbers, and there are eight rings (since there are 40 possible fractions from 1/41 to 40/41).

<< back to Problem A16 Non-interactive page

 

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