Solutions
- 1.
- Define the function f (i, j) = ij. From the Matrix submenu, choose Fill Matrix with 10 rows and 10 columns, and use
the function f to generate a matrix. Then, reduce the matrix
mod11 to get the following: BITMAPSETProbSolvHint0.1807in0.2197in0inq1
Select the matrix, from the Edit menu choose Insert
Column(s), and add 1 column at position 1. You have now added a column on
the left. Repeat this procedure using Insert Row(s), adding a row
at position 1. Fill in the empty boxes with × and the integers 1
through 10 to generate the final multiplication table,
From the table,
2⋅6 = 1 implies 2-1 = 6, and
3⋅4 = 1 implies
-1 = 4. As a check,
2-1
mod11 = 6 and
3-1
mod11 = 4.
- 2.
- The solution is given by
x = (8 - 4)/5
mod13 = 6. As a check,
6⋅5 + 4
mod13 = 8.BITMAPSETProbSolvHint0.1807in0.2197in0inq2
- 3.
- The problem requires the solution to the system
of congruences. The system is equivalent to the equation
x = 3 + 5a = 5 + 7b, or
+ 5a≡5 
mod7
, which has a solution
a = (5 - 3)/5
mod7 = 6, which means x = 3 + 5a = 33 jelly beans. Other possible
solutions are x = 33 + 35n, where n is any positive integer.BITMAPSETProbSolvHint0.1807in0.2197in0inq3
- 4.
- Define the function
nextp as indicated in
this chapter. Then
nextp(1099) produces a number with lots of
zeroes that ends in 289. The prime p can be written as
p = 1099 + 289.BITMAPSETProbSolvHint0.1807in0.2197in0inq4
- 5.
- Note that
2p-1
modp = 1 and
2(p-1)/2
modp = 1, whereas
2(p-1)/4
modp produces another
number with lots of zeroes that ends in 288. More precisely,
(p-1)/4≡ -1
modp. This congruence illustrates the fact
that, if p is a prime, then
x2≡1 
modp
has only two solutions,
x≡1 
modp
and
x≡ -1 
modp
.BITMAPSETProbSolvHint0.1807in0.2197in0inq5
- 6.
- We have


![$\left.\vphantom{
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & 4 \\
1 & 4 & 9
\end{array}
}\right]^{{-1}}_{}$](img18.png)
mod26 = 

. The ciphertext F K B H R T M T U has a numerical
equivalent of
5, 10, 1, 7, 17, 19, 12, 19, 20
. Picking three at a
time, we get
The vector
12, 0, 19, 7, 8, 18, 5, 20, 13
corresponds to the
plaintext M A T H I S F U N, or MATH IS FUN.BITMAPSETProbSolvHint0.1807in0.2197in0inq6
- 7.
- Defining
g(x) = x3 + x + 1, we see that
g(1)
mod3 = 0, and hence g(x) is not irreducible (since it has a root in GF3). However, if
f (x) = x3 + 2x + 1, then
f (0)
mod3 = 1,
f (1)
mod3 = 1, and
f (2)
mod3 = 1, and hence f (x) is irreducible. (If f (x) were reducible, it would have a linear factor, and hence a root.) An
element of GF27 can be thought of as a polynomial of degree less than 3
with coefficients in GF3. Given the field elements
2x2 + x + 2 and x + 1, the product is

2x2 + x + 2
2x + 1
mod x3 + 2x + 1
mod3 = x2 + 1, and the sum is
given by
2x2 + x + 2
+
2x + 1
mod3 = 2x2.BITMAPSETProbSolvHint0.1807in0.2197in0inq7
- 8.
- The objective function is 2.3h + 3b. The constraints
are
4h + 5b≤8000,
75h + 100b≤150000, b≥ 0, and h≥ 0. Apply
Maximize from the Simplex submenu to the system
2.3h + 3b |
5h + 4b≤8000 |
75h + 100b≤150000 |
b≥ 0 |
h≥ 0 |
to get the result: Maximum is 4550 at
h = 1000, b = 750.
BITMAPSETProbSolvHint0.1807in0.2197in0inq8
Subsections