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Problem H1
Write the base five numbers 1234_{five} and 1.234_{five} as base ten numbers.
Problem H2
Find the base ten fractions represented by the following:
a. a. 0.1_{five}, 0.2_{five}, 0.3_{five}, and 0.4_{five
}b. b. 0.01_{five}, 0.02_{five}, 0.03_{five}, and 0.04_{five
}c. c. 0.12_{five}, 0.23_{five}, 0.34_{five}, 0.43_{five}
Problem H3
Find the base five representation for these base ten fractions:
a. 9/25
b. 23/125
Problem H4
a. If you were counting in base five, what number would you say just before you said 100?
b. In base five, what number is one more than 344?
c. What is the greatest threedigit number that can be written in base five? What numbers come just before and just after this number?
Problem H5
What number in base five behaves the way 3 does in base four?
Problem H6
a. Count by twos to 30_{five}.
b. In base five, how can you tell if a number is even?
c. Count by threes to 30_{five}.
TAKE IT FURTHER
Problem H7
Find the base five representation for the base ten fraction 1/2.
Problem H8
How might you tell if a number is even or odd in bases two, three, four, five, six, seven, eight, nine, and ten? Can you generalize to base n?
Problem H9
In order to use base sixteen, we need 16 digits. However, we only know 10 digits — 0, 1, 2, … , 8, and 9 — so to represent 10, 11, 12, 13, 14, and 15 in base sixteen, we’ll use A, B, C, D, E, and F, respectively. This gives us the following representation for the base sixteen digits:


Remember that 16 in this base is written as 10 (onezero). So, for example, the number A6_{sixteen} becomes (10 • 16) + 6, or 166, in base ten. The number 123 in base ten is (7 • 16) + 11 , or 7B, in base sixteen.
Now translate these base sixteen numbers into base ten numbers:
a. 6D_{sixteen }b. AE_{sixteen }c. 9C_{sixteen }d. 2B_{sixteen}
Problem H10
Using the same system, translate these base ten numbers into base sixteen numbers:
a. 97
b. 144
c. 203
d. 890
Problem H1
1234_{five} = (1 • 5^{3}) + (2 • 5^{2}) + (3 • 5^{1}) + (4 • 5^{0}) = 125 + 50 + 15 + 4 = 194 in base ten.
1.234_{five} = (1 • 5^{0}) + (2 • 5^{1}) + (3 • 5^{2}) + (4 • 5^{3}) = 1 + 2/5 + 3/25 + 4/125 = 194/125 = 1 69/125 in base ten.
Problem H2
a. The base ten fraction for 0.1_{five} is 1/5. The others are 2/5, 3/5, and 4/5.
b. The base ten fraction for 0.01_{five} is 1/25. The others are 2/25, 3/25, and 4/25.
c. The base ten fraction for 0.12_{five} 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.
Problem H3
a. 9/25 = 5/25 + 4/25 = 1/5 + 4/25 = 0.14_{five
}b. 23/125 = 20/125 + 3/125 = 4/25 + 3/125 = 0.043_{five}
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 threedigit number that can be written in base five is 444. Just before this number is 443, and just after it is 1000.
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.
Problem H6
a. The count goes 2, 4, 11, 13, 20, 22, 24. Note that 30_{five} 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.
Problem H7
We know that in base five, we write a fraction as (A • 5^{0}) + (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 • 5^{0} + …. Since 5^{0} = 1, which is larger than 1/2, we cannot make 5^{0} 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^{1}s 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^{2}s 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^{3}s 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 onefifth 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 onefifth of the whole. 
This line splits the whole into two halves. 
Half of the whole is twofifths 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 onefifth. 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….
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.
Solution H9
a. 6D_{sixteen} = (6 • 16) + 13 = 109_{ten
}b. AE_{sixteen} = (10 • 16) + 14 = 174_{ten
}c. 9C_{sixteen} = (9 • 16) + 12 = 156_{ten
}d. 2B_{sixteen} = (2 • 16) + 11 = 43_{ten}
Solution H10
a. 97 = (6 • 16) + 1 = 61_{sixteen
}b. 144 = (9 • 16) = 90_{sixteen
}c. 203 = (12 • 16) + 11 = CB_{sixteen
}d. 890 = (3 • 256) + 7 • 16 + 10 = 37A_{sixteen}