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Learning Math Home
Number and Operations Session 3: Solutions
 
Session 3 Part A Part B Part C Homework
 
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A B C 
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Solutions for Session 3 Homework

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


Problem H1

1234five = (1 • 53) + (2 • 52) + (3 • 51) + (4 • 50) = 125 + 50 + 15 + 4 = 194 in base ten.

1.234five = (1 • 50) + (2 • 5-1) + (3 • 5-2) + (4 • 5-3) = 1 + 2/5 + 3/25 + 4/125 = 194/125 = 1 69/125 in base ten.

<< back to Problem H1


 

Problem H2

a. 

The base ten fraction for 0.1five is 1/5. The others are 2/5, 3/5, and 4/5.

b. 

The base ten fraction for 0.01five is 1/25. The others are 2/25, 3/25, and 4/25.

c. 

The base ten fraction for 0.12five is 7/25 (1/5 + 2/25). The others are 13/25 = 2/5 + 3/25, 19/25 = 3/5 + 4/25, and 23/25 = 4/5 + 3/25.

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Problem H3

a. 

9/25 = 5/25 + 4/25 = 1/5 + 4/25 = 0.14five

b. 

23/125 = 20/125 + 3/125 = 4/25 + 3/125 = 0.043five

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Problem H4

a. 

The number just before 100 is 44. The count goes ...40, 41, 42, 43, 44, 100....

b. 

The number that is one larger than 344 is 400. Carrying a 1 to the next digit is required for both the ones digit and the fives digit.

c. 

The greatest three-digit number that can be written in base five is 444. Just before this number is 443, and just after it is 1000.

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Problem H5

Four. In base four, 3 is the greatest digit; adding 1 to a 3 requires regrouping. In base five, this is true of the number 4. In general, in any base n, the number n - 1 will be the greatest digit and will require regrouping when 1 is added to it.

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Problem H6

a. 

The count goes 2, 4, 11, 13, 20, 22, 24. Note that 30five is not a multiple of 2.

b. 

It is more difficult to decide this in base five. The easiest way is to look for the sum of the digits of a number. If the sum is even, the number is even.

c. 

The count goes 3, 11, 14, 22, 30.

<< back to Problem H6


 

Problem H7

We know that in base five, we write a fraction as (A • 50) + (B • 5-1) + (C • 5-2) + (D • 5-3) + ..., where A, B, C, and D can only be 0, 1, 2, 3, or 4. To tackle this problem, refer to the algorithm we used in Problem A2. There, we found the largest number we could make as a power of 2, then subtracted to find a remainder, then continued to the next power of 2. Here, we'll do this with powers of 5.

First we find A: 1/2 = A • 50 + .... Since 50 = 1, which is larger than 1/2, we cannot make 50 from 1/2. Therefore, A = 0.

Now find B: 1/2 = B • 5-1 + .... Since 5-1 = 1/5 = 0.2 and 1/2 = 0.5, we can make two 5-1s from 1/2. Therefore, B = 2. Now subtract 2/5 from 1/2 to leave 1/10, the remainder.

So far, our base five decimal is 0.2____. Now find C using the remainder from the previous step: 1/10 = C • 5-2 + .... Since 5-2 = 1/25 = 0.04 and 1/10 = 0.1, we can make two 5-2s from 1/10. Therefore, C = 2. Now subtract 2/25 from 1/10 to leave 1/50, the remainder.

At this point, our base five decimal is 0.22___. Now find D using the remainder from the previous step: 1/50 = D • 5-3 + .... Since 5-3 = 1/125 = 0.008 and 1/50 = 0.02, we can make two 5-3s from 1/50. Therefore, D = 2. Now subtract 2/125 from 1/50 to leave 1/250, the remainder.

You might have noticed a repeating pattern. This is caused by the fact that our remainder has been one-fifth of the previous remainder at each step, and that we are using base five. This means that the pattern will continue indefinitely, and 1/2 = 0.2222222..., a repeating decimal, in base five. This is a perfect case to support the saying "A picture is worth a thousand words." Here is this proof in visual form:

Each piece is one-fifth
of the whole.

This line splits the
whole into two halves.

Half of the whole is two-fifths and half of a third fifth.

The third fifth can be further divided into five 25ths, and the drawings will look like the ones above, except that every piece will be 1/25 instead of one-fifth. This process continues.

For each place value, we need two of the five parts (i.e., two fifths, two 25ths, two 125ths, and so on). The decimal is 0.222222....

<< back to Problem H7


 

Problem H8

Here is a count of the first seven multiples of 2, from base two to base ten:

Base two: 10, 100, 110, 1000, 1010, 1100, 1110
Base three: 2, 11, 20, 22, 101, 110, 112
Base four: 2, 10, 12, 20, 22, 30, 32
Base five: 2, 4, 11, 13, 20, 22, 24
Base six: 2, 4, 10, 12, 14, 20, 22
Base seven: 2, 4, 6, 11, 13, 15, 20
Base eight: 2, 4, 6, 10, 12, 14, 16
Base nine: 2, 4, 6, 8, 11, 13, 15
Base ten: 2, 4, 6, 8, 10, 12, 14

An important note is that the numeral 10 is a multiple of 2 only when the base is an even number. When the base is an even number, the units digits of even numbers repeat, so we need only look at the units digit of a number to determine if it is odd or even. If the units digit is even, the number is even.

If the base is an odd number, the units digit is not enough information to determine if a number is even. In odd bases, it is the sum of the digits that determines whether a number is even -- if the sum is even, the number is even.

<< back to Problem H8


 

Problem H9

a. 

6Dsixteen = (6 • 16) + 13 = 109ten

b. 

AEsixteen = (10 • 16) + 14 = 174ten

c. 

9Csixteen = (9 • 16) + 12 = 156ten

d. 

2Bsixteen = (2 • 16) + 11 = 43ten

<< back to Problem H9


 

Problem H10

a. 

97 = (6 • 16) + 1 = 61sixteen

b. 

144 = (9 • 16) = 90sixteen

c. 

203 = (12 • 16) + 11 = CBsixteen

d. 

890 = (3 • 256) + 7 • 16 + 10 = 37Asixteen

<< back to Problem H10


 

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