What is $5\times 3-4\times 2$ in
${\mathbb{Z}}_{6}$? (Hint: recall that
the elements of ${\mathbb{Z}}_{6}$
are $\left\{0,\phantom{\rule{0.3em}{0ex}}1,\phantom{\rule{0.3em}{0ex}}2,\phantom{\rule{0.3em}{0ex}}3,\phantom{\rule{0.3em}{0ex}}4,\phantom{\rule{0.3em}{0ex}}5\right\}$.
)
Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is correct!*

$5\times 3-4\times 2=15-8=7=6+1$, and so
the answer in ${\mathbb{Z}}_{6}$
is 1.

*Choice (c) is incorrect*

*Choice (d) is incorrect*

*Choice (e) is incorrect*

*Choice (f) is incorrect*

Find $\left(\left(9\times 2\right)+5\right)\left(3+8+4\right)\left(1+\left(6\times 7\right)\right)$
in ${\mathbb{Z}}_{10}$.
Enter your answer into the answer box.

*Correct!*

We have $\left(\left(9\times 2\right)+5\right)\left(3+8+4\right)\left(1+\left(6\times 7\right)\right)=23\times 15\times 43=3\times 5\times 3=5\phantom{\rule{0.3em}{0ex}}$
in ${\mathbb{Z}}_{10}$.

*Incorrect.*

*Please try again.*

Remember that all computations in ${\mathbb{Z}}_{10}$
should be reduced by subtracting multiples of 10 so that the answer lies from 0 to 9
(inclusive).

Find the dot product $\left[1,3,2\right]\cdot \left[4,5,1\right]$
in ${\mathbb{Z}}_{5}^{3}$.

*Correct!*

$\left[1,3,2\right]\cdot \left[4,5,1\right]=4+15+2=21=1$ in
${\mathbb{Z}}_{5}$.

*Incorrect.*

*Please try again.*

Remember that the dot product is calculated in the usual way
and the answer is then reduced so it is one of the elements of
${\mathbb{Z}}_{5}$.

How many solutions are there in ${\mathbb{Z}}_{6}$
of the equation $2x=0$?
Exactly one option must be correct)
Remember that ${\mathbb{Z}}_{6}$
contains the elements $\left\{0,\phantom{\rule{0.3em}{0ex}}1,\phantom{\rule{0.3em}{0ex}}2,\phantom{\rule{0.3em}{0ex}}3,\phantom{\rule{0.3em}{0ex}}4,\phantom{\rule{0.3em}{0ex}}5\right\}$.
You can use trial and error to find all elements that satisfy the given equation.

*Choice (a) is incorrect*

Remember that
${\mathbb{Z}}_{6}$ contains the
elements $\left\{0,\phantom{\rule{0.3em}{0ex}}1,\phantom{\rule{0.3em}{0ex}}2,\phantom{\rule{0.3em}{0ex}}3,\phantom{\rule{0.3em}{0ex}}4,\phantom{\rule{0.3em}{0ex}}5\right\}$.
You can use trial and error to find all elements that satisfy the given equation.

*Choice (b) is incorrect*

Remember that ${\mathbb{Z}}_{6}$
contains the elements $\left\{0,\phantom{\rule{0.3em}{0ex}}1,\phantom{\rule{0.3em}{0ex}}2,\phantom{\rule{0.3em}{0ex}}3,\phantom{\rule{0.3em}{0ex}}4,\phantom{\rule{0.3em}{0ex}}5\right\}$.
You can use trial and error to find all elements that satisfy
the given equation.

*Choice (c) is correct!*

By trial and error, we find that
$x=0$ and
$x=3$ are the only solutions of
the equation. We have $2\times 0=0$
and $2\times 3=6=0$ in
${\mathbb{Z}}_{6}$.

*Choice (d) is incorrect*

Remember
that ${\mathbb{Z}}_{6}$ contains
the elements $\left\{0,\phantom{\rule{0.3em}{0ex}}1,\phantom{\rule{0.3em}{0ex}}2,\phantom{\rule{0.3em}{0ex}}3,\phantom{\rule{0.3em}{0ex}}4,\phantom{\rule{0.3em}{0ex}}5\right\}$.
You can use trial and error to find all elements that satisfy the given equation.

*Choice (e) is incorrect*

The solutions in ${\mathbb{Z}}_{10}$ of the
equation ${x}^{2}=3x$ are: Exactly one
option must be correct)

*Choice (a) is incorrect*

This is one of the solutions but not the only one. You could try
a systematic approach and find the squares of all the members of
${\mathbb{Z}}_{10}$, then
check to see which of these are equal to three times the original number.

*Choice (b) is incorrect*

These are certainly solutions but they aren’t the only ones. You could
try a systematic approach and find the squares of all the members of
${\mathbb{Z}}_{10}$, then
check to see which of these are equal to three times the original number.

*Choice (c) is correct!*

Strangely, the quadratic equation has four solutions in
${\mathbb{Z}}_{10}$! We
see that

$${0}^{2}=0=3\times 0,\phantom{\rule{1em}{0ex}}{3}^{3}=9=3\times 3,$$

$${5}^{2}=25=5=15=3\times 5,\phantom{\rule{1em}{0ex}}{8}^{2}=64=4=24=3\times 8.$$

*Choice (d) is incorrect*

These are certainly
solutions (${0}^{2}=0=3\times 0,\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{8}^{2}=64=4=24=3\times 8$) but
they aren’t the only ones. You could try a systematic approach and find the squares of all the
members of ${\mathbb{Z}}_{10}$,
then check to see which of these are equal to three times the original number.

Find the solution of the following simultaneous equations in
${\mathbb{Z}}_{12}$.

$$3x=0,\phantom{\rule{2em}{0ex}}2x+4=0.$$

*Correct!*

In ${\mathbb{Z}}_{12}$,
we see by trial and error that there are three values of
$x$ satisfying
$3x=0$, namely
$x=0,\phantom{\rule{0.3em}{0ex}}x=4,\phantom{\rule{0.3em}{0ex}}x=8$. Neither
$x=0$ nor
$x=8$ satisfies
$2x+4=0$, since
$2\times 0+4=4$ and
$2\times 8+4=20=8$.
However, $2\times 4+4=8+4=12=0.$
Therefore $x=4$.

*Incorrect.*

*Please try again.*

One method is to find all solutions to $3x=0$
in ${\mathbb{Z}}_{12}$ by trying all possible
values of $x$ from 0 to 11. Then
see if any of these satisfy $2x+4=0$.

How many values of $a$
are there in ${\mathbb{Z}}_{4}$ for
which we can find an $x$
in ${\mathbb{Z}}_{4}$ satisfying
the equation $ax=1$?
Exactly one option must be correct)

*Choice (a) is incorrect*

Work systematically. First let $a=0$
and see if $x$ can be
found to satisfy $0\times x=1$.
Then let $a=1$
and repeat the process, etc.

*Choice (b) is incorrect*

Work systematically. First let $a=0$
and see if $x$ can be
found to satisfy $0\times x=1$.
Then let $a=1$
and repeat the process, etc.

*Choice (c) is correct!*

We can’t find any solution to $ax=1$
when $a=0$, for then the
left hand side is $0$.
When $a=1$ we see
that $x=1$ is a solution.
When $a=2$ we see that
$2x$ is either 0 or 2
but never 1. When $a=3$
we see that $3x=1$ has
solution $x=3$. So
$ax=1$ can be solved
only when $a=1$
or $a=3$.

*Choice (d) is incorrect*

Work systematically. First let $a=0$
and see if $x$ can be
found to satisfy $0\times x=1$.
Then let $a=1$
and repeat the process, etc.

How many of the following three ISBNs is/are correct? (Recall that the check vector for ISBNs is
$\mathbf{c}=\left[10,9,8,7,6,5,4,3,2,1\right]$ and that operations
are performed in ${\mathbb{Z}}_{11}$.)
$$\mathbf{u}=\left[0,1,9,2,8,6,1,0,3,4\right]$$
$$\mathbf{v}=\left[0,1,4,0,0,8,3,7,1,3\right]$$
$$\mathbf{w}=\left[0,5,8,2,7,0,0,5,5,2\right]$$
Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

There is a single correct ISBN, namely
$\mathbf{u}$. This is the only one
whose dot product with $\mathbf{c}$
equals zero in ${\mathbb{Z}}_{11}$.

*Choice (d) is incorrect*

What is the value of the GTIN-13 check digit
$d$ in
$\mathbf{u}=\left[4,2,3,4,1,8,6,2,2,6,7,5,d\right]$?
(Recall that the check vector for GTIN-13s is
$\mathbf{c}=\left[1,3,1,3,1,3,1,3,1,3,1,3,1\right]$ and that operations
are performed in ${\mathbb{Z}}_{10}$.)

*Correct!*

The dot product of $\mathbf{u}$
and $\mathbf{c}$
equals

$$\left(4\times 1\right)+\left(2\times 3\right)+\left(3\times 1\right)+\left(4\times 3\right)+\left(1\times 1\right)+\left(8\times 3\right)+\left(6\times 1\right)+\left(2\times 3\right)$$

$$+\left(2\times 1\right)+\left(6\times 3\right)+\left(7\times 1\right)+\left(5\times 3\right)+\left(d\times 1\right)$$

$$=4+6+3+12+1+24+6+6+2+18+7+15+d$$

$$=4+d.$$

Thus $d=6$.

*Incorrect.*

*Please try again.*

The dot product of the check vector and the GTIN-13 vector must equal 0 in
${\mathbb{Z}}_{10}$.
Calculate the dot product - it will contain a term in
$d$. Select the
correct value of $d$
to make the dot product a multiple of 10 (that is, equal to 0 in
${\mathbb{Z}}_{10}$).

(Harder) Two adjacent entries in a correct ISBN have been transposed in error,
resulting in an incorrect ISBN of
Suppose that the transposed elements are
$a$ and
$b$,
where $a,b$
is the correct order. In the correct order, suppose
$a$ is in the
$\left(n+1\right)$th position from
the right and $b$
is in the $n$th
position from the right. The incorrect ISBN lists
$b,a$ in
these positions, respectively.
In the calculation of the dot product with
$\mathbf{c}$,
the correct ISBN results in the inclusion of terms
$\left(n+1\right)a+nb=na+nb+a$
whereas the incorrect ISBN results in the inclusion of
$\left(n+1\right)b+na=na+nb+b$.
All other terms in the dot product remain the same. Therefore
the incorrect dot product minus the correct dot product equals
$b-a$. Now
calculate the dot product of the check vector with the incorrect ISBN. You will be
very close to arriving at the answer!

$$\left[0,4,9,5,0,4,1,4,2,7\right].$$

Find the incorrect positions (where positions are numbered
from right to left). (Recall that the check vector for ISBNs is
$\mathbf{c}=\left[10,9,8,7,6,5,4,3,2,1\right]$ and that operations
are performed in ${\mathbb{Z}}_{11}$.)
Exactly one option must be correct)
*Choice (a) is incorrect*

Suppose that the transposed
elements are $a$
and $b$,
where $a,b$
is the correct order. In the correct order, suppose
$a$ is in the
$\left(n+1\right)$th position from
the right and $b$
is in the $n$th
position from the right. The incorrect ISBN lists
$b,a$ in
these positions, respectively.
In the calculation of the dot product with
$\mathbf{c}$,
the correct ISBN results in the inclusion of terms
$\left(n+1\right)a+nb=na+nb+a$
whereas the incorrect ISBN results in the inclusion of
$\left(n+1\right)b+na=na+nb+b$.
All other terms in the dot product remain the same. Therefore
the incorrect dot product minus the correct dot product equals
$b-a$. Now
calculate the dot product of the check vector with the incorrect ISBN. You will be
very close to arriving at the answer!

*Choice (b) is incorrect*

Suppose that the transposed elements are
$a$ and
$b$,
where $a,b$
is the correct order. In the correct order, suppose
$a$ is in the
$\left(n+1\right)$th position from
the right and $b$
is in the $n$th
position from the right. The incorrect ISBN lists
$b,a$ in
these positions, respectively.
In the calculation of the dot product with
$\mathbf{c}$,
the correct ISBN results in the inclusion of terms
$\left(n+1\right)a+nb=na+nb+a$
whereas the incorrect ISBN results in the inclusion of
$\left(n+1\right)b+na=na+nb+b$.
All other terms in the dot product remain the same. Therefore
the incorrect dot product minus the correct dot product equals
$b-a$. Now
calculate the dot product of the check vector with the incorrect ISBN. You will be
very close to arriving at the answer!

*Choice (c) is correct!*

Suppose that the transposed elements are
$a$ and
$b$,
where $a,b$
is the correct order. In the correct order, suppose
$a$ is in the
$\left(n+1\right)$th position from
the right and $b$
is in the $n$th
position from the right. The incorrect ISBN lists
$b,a$ in
these positions, respectively.
In the calculation of the dot product with
$\mathbf{c}$, the correct ISBN results in
terms $\left(n+1\right)a+nb=na+nb+a$ whereas the incorrect
ISBN results in terms $\left(n+1\right)b+na=na+nb+b$.
All other terms in the dot product remain the same. Calculating the dot
product of the check vector with the incorrect ISBN gives an answer of 3.
Therefore the incorrect dot product minus the correct dot product equals
$b-a=3-0=3$. We
search for two adjacent entries differing by 3 (modulo 11) in the incorrect
ISBN, finding only one such pair, at the 4th and 5th positions from the right.
The correct ISBN is therefore found by swapping these two entries, to give
$\left[0,4,9,5,0,1,4,4,2,7\right]$.

*Choice (d) is incorrect*

Suppose that the transposed elements are
$a$ and
$b$,
where $a,b$
is the correct order. In the correct order, suppose
$a$ is in the
$\left(n+1\right)$th position from
the right and $b$
is in the $n$th
position from the right. The incorrect ISBN lists
$b,a$ in
these positions, respectively.
In the calculation of the dot product with
$\mathbf{c}$,
the correct ISBN results in the inclusion of terms
$\left(n+1\right)a+nb=na+nb+a$
whereas the incorrect ISBN results in the inclusion of
$\left(n+1\right)b+na=na+nb+b$.
All other terms in the dot product remain the same. Therefore
the incorrect dot product minus the correct dot product equals
$b-a$. Now
calculate the dot product of the check vector with the incorrect ISBN. You will be
very close to arriving at the answer!

*Choice (e) is incorrect*