A B C 
Homework

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).

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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 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).

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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.

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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 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).

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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.

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Problem A7

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Problem A8

Answers will vary.

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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

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