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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 
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Solutions for Session 7 Homework

See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6 | H7 | H8 | H9 | H10 | H11


Problem H1

a. 

0.142857 = 142,857/1,000,000.

b. 

F = 0.142857142857. . . . There are six digits repeated, so multiply by 1,000,000 (106) to get 1,000,000F = 142,857.142857142857. . . . Subtracting F from each side gives you 999,999F = 142,857, so F = 142,857/999,999, which can be reduced to F = 1/7.

<< back to Problem H1


 

Problem H2

Adding the two groups of nine numbers gives you 99,999,999.

Note that if the period length is even, you can always break the number into two rows where columns of digits will add up to 9.

<< back to Problem H2


 

Problem H3

Adding the three groups of six numbers gives you 999,999.

If the period length is a multiple of three, you can always break the number into three rows where each column will either add up to 9 or will be a two-digit number ending in 9. Note that this works only if we break a number into two or three rows. No other break-up of the number will work all the time.

<< back to Problem H3


 

Problem H4

Since the sum should be all 9s, the pattern would continue:

0.021

276

595

744

680

851

063

82

  978

723

404

255

319

148

936

17

Note that the repetition starts after 46 decimal places, so the period is 46.

<< back to Problem H4


 

Problem H5

Yes, it makes sense. As with other prime numbers, the period is a factor of one less than the number itself. In this case, with the prime number 47, the period would need to be either 46, 23, 2 or 1. We already have 23 digits and they haven't repeated, so the period must be 46.

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Problem H6

Consuela's method is not helpful because 3 is not a factor of the period, 46. Her method relies on being able to arrange the non-repeating digits into three rows of identical lengths.

<< back to Problem H6


 

Problem H7

We can primarily refer to Consuela's numbers, starting from the right side. If the sum is going to be all 9s, the right-most covered digits must be 775 51 (don't forget to carry). We can then use Shigeto's numbers 408 163 (uncovered) to find the rest of Consuela's covered bottom row: 346 938 (the 34 is visible on Shigeto's page). Finally, we can complete Consuela's covered top row: 020. The completed number is

.

As a check, you might verify that these numbers work by using Shigeto's rule; i.e., that they add (in pairs) to 9s.

<< back to Problem H7


 

Problem H8

Going on the assumption that the number is in the form 1/n, n must be less than 50 (since 1/50 = 0.02) but greater than 47 (1/47 = 0.021276. . ., as seen in Problem H3). So n is either 48 or 49; 1/48 = 0.208333. . ., which is too large, and 1/49 is just right.

If we could not make this assumption, we could use the multiply-and-subtract method from Part B, but we'd need a really accurate calculator!

<< back to Problem H8


 

Problem H9

Yes, 1 can be represented as 0.99999.... The technique of Part B can be applied to this decimal; if F = 0.999..., then 10F = 9.999..., and, subtracting F from each side, 9F must equal 9. This means that F = 1!

Another way to convince yourself that this is true is that 0.333... represents 1/3. Then 2/3 is represented as 0.666..., and 3/3 is represented as 0.999. ...Since 3/3 = 1, 1 and 0.999... are the same number.

All terminating decimals have an alternate representation in this form. For example, 0.25 can also be represented as 0.24999999....

<< back to Problem H9


 

Problem H10

Yes. Because 2 is a factor of 10, it will have no effect on the period; it will only delay the repetition by one decimal place. So 1/14 = 0.0714285714285. . ., which, like 1/7, has a period of six. Similarly, 1/28 will also have a period of six, delayed by two decimal places.

<< back to Problem H10


 

Problem H11

One convincing argument is that when dividing by n, there are only n possible remainders: the numbers (0, 1, 2, 3, 4, ..., n - 1). If we divide and get a remainder of 0, then the division is complete, and the resulting decimal is terminating. If we divide and get a remainder that we have seen earlier in our division, this means that the decimal is about to repeat. So if we want the longest possible period for our decimal, we want to avoid 0 but run through every possible number before repeating. To do this, we would hit all the non-zero remainders (1, 2, 3, ..., n - 1) while missing 0 -- a total of n - 1 possible remainders. Since each remainder corresponds to continuing the decimal by one place, there is a maximum of n - 1 decimal places before repetition begins.

<< back to Problem H11

 

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