Selected Solutions to Homework # 1
•
Please post questions to the discussion forum or stop by during office
hours if you have questions about the first or third problem (truth
tables and the problem about quotients and remainders using your
calculator).
•
# 2:
Here is a solution to part (b) ii: Divide
a
by
b
where
a
= 2011
and
b
= 13 represent numbers written in base 7.
This means that
a
= (2011)
7
and
b
= (13)
7
.
There are two approaches. The first approach is to convert to base 10,
divide as usual, and then convert the answers back to base 7.
Convert to base 10:
a
= (2011)
7
= 2
*
7
3
+ 0
*
7
2
+ 1
*
7
1
+ 1
*
7
0
= 2
*
343 + 0 + 7 + 1 = 694
b
= (13)
7
= 7 + 3 = 10
Divide:
694 = 69
*
10 + 4
Convert back to base 7:
(2011)
7
= (126)
7
*
(13)
7
+ (4)
7
(To obtain the base 7 representation of 69, you can successively divide
by 7:
69 = 9
*
7 + 6
,
9 = 1
*
7 + 2
,
1 = 0
*
7 + 1
The remainders are 6 =
r
0
, 2 =
r
1
, and 1 =
r
2
, where the notation
r
i
follows the fourth problem on the homework.)
The second approach is to perform long division in base 7. (For sim
plicity of notation, I will supress the base seven subscript.)
We see
that 13 goes into 20 once...
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 Spring '10
 R.Bell
 Algebra, Division, Remainder, #, Natural number

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