



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.
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Problem A2  
Since 1/2 = 5/10, 1/2^{n} = 5^{n}/10^{n}. Here, 10^{n} dictates the number of decimal places, and 5^{n} dictates the actual digits in the decimal. Since 5^{4} = 625, 1/2^{4} = 0.0625 (four decimal places).
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Problem A3  
Here is the completed table:


Fractions 

Denominator 

Prime Factorization 

Number of Decimal Places 

Decimal Representa tion 


1/2 

2 

2^{1} 

1 

0.5 
1/4 
4 
2^{2} 
2 
0.25 
1/8 
8 
2^{3} 
3 
0.125 
1/16 
16 
2^{4} 
4 
0.0625 
1/32 
32 
2^{5} 
5 
0.03125 
1/64 
64 
2^{6} 
6 
0.015625 
1/1,024 
1,024 
2^{10} 
10 
0.0009765625 
1/2^{n} 
2^{n} 
2^{n} 
n 
0.5^{n} (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, 5^{5} is 3,125, so "3125" shows up in the decimal, with enough leading zeros for it to comprise five digits: .03125. Similarly, 5^{6} is 15,625, so the decimal is .015625 (six digits).
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Problem A4  
We know that 1/2^{n} = 5^{n}/10^{n}, so for all unit fractions with denominators that are the nth power of 2, the decimal will consist of the digits of 5^{n} with enough leading zeros to give n decimal places.
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Problem A5  
Here is the completed table:


Fractions 

Denominator 

Prime Factorization 

Number of Decimal Places 

Decimal Representa tion 


1/5 

5 

5^{1} 

1 

0.2 
1/25 
25 
5^{2} 
2 
0.04 
1/125 
125 
5^{3} 
3 
0.008 
1/625 
625 
5^{4} 
4 
0.0016 
1/3,125 
3,125 
5^{5} 
5 
0.00032 
1/15,625 
15,625 
5^{6} 
6 
0.000064 
1/5^{n} 
5^{n} 
5^{n} 
n 
0.2^{n} (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, 2^{5} is 32, so the decimal is 0.00032 (five digits).
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Problem A6  


Fractions 

Denominator 

Prime Factorization 

Number of Decimal Places 

Decimal Representa tion 


1/10 

10 

2^{1} • 5^{1} 

1 

0.1 
1/20 
20 
2^{2} • 5^{1} 
2 
0.05 
1/50 
50 
2^{1} • 5^{2} 
2 
0.02 
1/200 
200 
2^{3} • 5^{2} 
3 
0.005 
1/500 
500 
2^{2} • 5^{3} 
3 
0.002 
1/4,000 
4,000 
2^{5} • 5^{3} 
5 
0.00025 
1/5^{n} 
2^{n} • 5^{m} 
2^{n} • 5^{m} 
max(n, m) 
For m > n: (10^{m})(2^{mn}) or for n > m: (10^{n})(5^{nm}) 


"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.
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Problem A7  
a.  Yes, they will all terminate. 
b.  A decimal terminates whenever it can be written as n/10^{k} 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 10^{k} 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.) 
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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... 


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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.
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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.
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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.
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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... 


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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).
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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.
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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). 
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