Mathematical Systems
GE 112
Mathematics in the Modern World
Faculty, Math & Stat Department
College of Arts & Sciences
University of Southeastern Philippines
Modular Arithmetic
Clock Arithmetic
Many clocks have the familiar 12-hour
design. We designate whether the time is
before noon or after noon by using the
abbreviation A.M and P.M.
A reference to 7:00 A.M. means 7 hours
after 12:00 midnight; a reference to 7:00 P.M.
means 7 hours after 12:00 noon. In both
cases, once 12 is reached on the clock, we
begin again with 1.
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Modular Arithmetic
We will use the symbol ⊕ to denote addition
and the symbol ⊖ to denote subtraction on a
12-hour clock.
Examples
1. Determine the time 8 hours after 3 o’clock.
Solution: Because we did not pass 12 o’clock,
we add 3 and 8. Thus, the time is 11 o’clock.
That is, 3 ⊕ 8 = 11.
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Modular Arithmetic
2. Determine the time 8 hours after 9 o’clock.
Solution: Since we passed 12 o’clock, we begin
again with 1. Therefore, the time is 5 o’clock.
Another Method: When 9 + 8 = 17 is divided by
12 the remainder is 5. Therefore, 9 ⊕ 8 = 5.
3. If the time now is 10 o’clock, then 7 hours ago
the time was 3 o’clock. That is, 10 ⊝ 7 = 3.
4. If the time now is 3 o’clock, then 7 hours ago
the time was 8 o’clock. That is, 3 ⊝ 7 = 8
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Modular Arithmetic
If the difference is negative add it to 12, then
we have 12 + −4 = 8
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Modular Arithmetic
Day-of-the-Week Arithmetic
A similar example involves day-of-the week
arithmetic. If we associate each day of the week with
a number,
Monday=1 Tuesday=2 Wednesday=3 Thursday=4
Friday=5 Saturday=6 Sunday=7
then 6 days after Friday is Thursday since when 5 +
6 = 11 is divided by 7 the remainder is 4. Thus, 5 ⊞
6 = 4.
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Modular Arithmetic
Examples
1. 6 days after Friday is Thursday
Solution: When 5 + 6 = 11 is divided by 7 the
remainder is 4, the number associated with
Thursday. Thus, 5 ⊞ 6 = 4.
2. 16 days after Monday is Wednesday
Solution: When 1 + 16 = 17 is divided by 7 the
remainder is 3, the number associated with
Wednesday. Thus, 1 ⊞ 16 = 3.
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Modular Arithmetic
Arithmetic Modulo n
Two integers a and b are said to be congruent
modulo 𝑛, where 𝑛 is a natural number, if
𝑎−𝑏
is an integer.
𝑛
In this case, we write 𝑎 ≡ 𝑏 𝑚𝑜𝑑 𝑛. The number 𝑛
is called the modulus. The statement
𝑎 ≡ 𝑏 𝑚𝑜𝑑 𝑛
is called a congruence.
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Modular Arithmetic
Examples.
Determine
congruence is true.
1. 29 ≡ 8 𝑚𝑜𝑑 3
whether
the
29−8
Solution: Since
= 7 and 7 is an integer,
3
29 ≡ 8 𝑚𝑜𝑑 3 is a true congruence.
2. 15 ≡ 4 𝑚𝑜𝑑 6
15−4
11
=
6
6
Solution: Since
is not an integer,
15 ≡ 4 𝑚𝑜𝑑 6 is not a true congruence.
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Modular Arithmetic
3. July 4, 2010, was a Sunday. What day of
the week is July 4, 2015?
Solution: There are 5 years between the two
dates. Each year has 365 days except 2012,
which has extra day because it is a leap year.
So, the total number of days between the two
dates is
5 365 + 1 = 1826.
Because 1826 ÷ 7 = 260 remainder 6,
1826 ≡ 6 𝑚𝑜𝑑 7.
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Modular Arithmetic
Let us check if it is a true congruence.
We have 𝑎 = 1826, 𝑏 = 6 and 𝑛 = 7, so,
𝑎 − 𝑏 1826 − 6 1820
=
=
= 260
𝑛
7
7
Since 260 is an integer, hence,
1826 ≡ 6 𝑚𝑜𝑑 7 is a true congruence.
Now, we can say that any multiple of 7
days past a given day will be the same day of
the week. So, the day of the week 1826 days
after July 4, 2010, will be the same as the day 6
days after July 4, 2010. Thus, July 4, 2015 will
be a Saturday.
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Modular Arithmetic
Arithmetic Operations Modulo n
Procedure: Perform the arithmetic operation and
then divide by the modulus. The answer is the
remainder.
Addition Modulo n
Examples:
1. Evaluate: 23 + 38 𝑚𝑜𝑑 12.
Solution: Note that 𝑚 𝑚𝑜𝑑 𝑛 means the
remainder when m is divided by n. Thus, we
must find the remainder of the sum of 23+38
when divided by 12.
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Modular Arithmetic
Since 23 + 38 = 61 and the remainder when
61 divided by 12 is 1.
61 ÷ 12 = 5 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1
So, the answer is 1.
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Modular Arithmetic
2. Evaluate: 51 + 72 𝑚𝑜𝑑 4.
Solution:
Since 51 + 72 = 123 and the remainder when
123 divided by 4 is 3.
123 ÷ 4 = 30 𝑟3
Thus, (51 + 72)𝑚𝑜𝑑 4 = 3.
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Modular Arithmetic
Subtraction Modulo n
Examples:
1. Evaluate: 33 − 16 𝑚𝑜𝑑 6
Solution: Subtract 33 − 16 = 17. The result is
positive. Divide the difference by the modulus,
6.
17 ÷ 6 = 2 𝑟5
The answer is the remainder which is equal to
5.
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Modular Arithmetic
2. Evaluate: 14 − 27 𝑚𝑜𝑑 5
Solution: Subtract 14 − 27 = −13.
Because the answer is negative, we must find 𝑥 so that
−13 ≡ 𝑥 𝑚𝑜𝑑 5.
Thus we must find 𝑥 so that the value of
−13−𝑥
−(13+𝑥)
=
is an integer.
5
5
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Modular Arithmetic
Trying the whole number values of
𝑥 less than 5, the modulus, we find that
𝑥 = 1,
𝑥 = 2,
𝑥 = 3,
𝑥 = 4,
−(13+1)
−14
=
,
5
5
−(13+2)
−15
=
= −3,
5
5
−(13+3)
−16
=
,
5
5
−(13+4)
−17
=
,
5
5
Therefore, 14 − 27 𝑚𝑜𝑑 5 ≡ 2.
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Modular Arithmetic
Alternative way
Solution:
14 − 27 𝑚𝑜𝑑 5
−13 𝑚𝑜𝑑 5
-13 divided 5 is -2 remainder -3
−13 ÷ 5 = −2 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 − 3
Since the remainder is negative add it to 5
5 + −3 = 2
Hence, 14 − 27 𝑚𝑜𝑑 5 ≡ 2
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Modular Arithmetic
3. Disregarding A.M. or P.M., if it is 5 o’clock now, what time was it 57
hours ago?
Solution:
The time can be determined by calculating 5 − 57 𝑚𝑜𝑑 12. Because 5 −
−52−𝑥
57 = −52 is a negative number, find a whole number 𝑥 so that
=
12
−(52+𝑥)
is an integer. Evaluating the expression for whole number values
12
of 𝑥 less than 12,
𝑥 = 1,
𝑥 = 2,
𝑥 = 3,
𝑥 = 4,
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−(52 + 1) −53
=
,
12
12
−(52 + 2) −54
=
,
12
12
−(52 + 3) −55
=
,
12
12
−(52 + 4) −56
=
,
12
12
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Modular Arithmetic
𝑥 = 5,
𝑥 = 6,
𝑥 = 7,
−(52 + 5) −57
=
,
12
12
−(52 + 6) −58
=
,
12
12
−(52 + 7) −59
=
,
12
12
−(52 + 8) −60
𝑥 = 8,
=
= −5,
12
12
−(52 + 9) −61
𝑥 = 9,
=
,
12
12
−(52 + 10) −62
𝑥 = 10,
=
,
12
12
−(52 + 11) −63
𝑥 = 11,
=
.
12
12
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Modular Arithmetic
When 𝑥 = 8,
−(52+8)
−60
=
= −5 and −5 is an integer.
12
12
Thus, 5 − 57 𝑚𝑜𝑑 12 ≡ 8. Therefore, if it is 5 o’clock now, 57 hours
ago it was 8 o’clock.
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Modular Arithmetic
Alternative way
Solution:
5 − 57 𝑚𝑜𝑑 12
−52 𝑚𝑜𝑑 12
-52 divided 12 is -4 remainder -4
−52 ÷ 12 = −4 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 − 4
Since the remainder is negative add it to 12
12 + −4 = 8
Hence, 5 − 57 𝑚𝑜𝑑 12 ≡ 8.
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Modular Arithmetic
Multiplication Modulo n
Example:
1. Evaluate: 15 ∙ 23 𝑚𝑜𝑑 11
Solution: Find the product of 15 and 23
15 ∙ 23 = 345
The result is positive. Divide the product by the
modulus, 11.
345 ÷ 11 = 31 𝑟4
The answer is the remainder which is equal to 4.
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Modular Arithmetic
Solving Congruence Equations
Solving a congruence equation means
finding all whole number values of the
variable for which the congruence is true.
Examples:
1. Solve: 3𝑥 + 5 ≡ 3 𝑚𝑜𝑑 4.
Solution: We search for whole number values
of 𝑥 for which the congruence is true.
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Modular Arithmetic
If we continued trying values, we would find that
10 and 14 are also solutions. Note that the
solutions 6,10, and 14 are all congruent to 2
modulo 4.
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Modular Arithmetic
𝑥 = 2,
𝑥 = 6,
𝑥 = 10,
𝑥 = 14,
1 Topic 1
3 2 + 5 ≡ 3 𝑚𝑜𝑑 4 → 11 ≡ 3 𝑚𝑜𝑑 4
11 − 3 8
= =2
4
4
3 6 + 5 ≡ 3 𝑚𝑜𝑑 4 → 23 ≡ 3 𝑚𝑜𝑑 4
23 − 3 20
=
=5
4
4
3 10 + 5 ≡ 3 𝑚𝑜𝑑 4 → 35 ≡ 3 𝑚𝑜𝑑 4
35 − 3 32
=
=8
4
4
3 14 + 5 ≡ 3 𝑚𝑜𝑑 4 → 47 ≡ 3 𝑚𝑜𝑑 4
47 − 3 44
=
= 11
4
4
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Modular Arithmetic
In general, once a solution is determined,
additional solution can be found by repeatedly
adding the modulus to the original solution.
Thus, the solution of 3𝑥 + 5 ≡ 3 𝑚𝑜𝑑 4 are
2,6,10,14,18,…
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Modular Arithmetic
2. Solve: 2𝑥 + 1 ≡ 3 𝑚𝑜𝑑 10
Solution: We search for whole number values
of 𝑥 for which the congruence is true.
2 1 + 1 ≡ 3 𝑚𝑜𝑑 10
2 2 + 1 ≡ 3 𝑚𝑜𝑑 10
2 3 + 1 ≡ 3 𝑚𝑜𝑑 10
2 4 + 1 ≡ 3 𝑚𝑜𝑑 10
2 5 + 1 ≡ 3 𝑚𝑜𝑑 10
2 6 + 1 ≡ 3 𝑚𝑜𝑑 10
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Modular Arithmetic
If we continued trying values, we would find that 11, 16, 21
and 26 are also solutions.
𝑥 = 11,
2 11 + 1 ≡ 3 𝑚𝑜𝑑 10 → 23 ≡ 3 𝑚𝑜𝑑 10
23 − 3 20
=
=2
10
10
𝑥 = 16,
2 16 + 1 ≡ 3 𝑚𝑜𝑑 10 → 33 ≡ 3 𝑚𝑜𝑑 10
33 − 3 30
=
=3
10
10
𝑥 = 21,
2 21 + 1 ≡ 3 𝑚𝑜𝑑 10 → 43 ≡ 3 𝑚𝑜𝑑 10
43 − 3 40
=
=4
10
10
𝑥 = 26,
2 26 + 1 ≡ 3 𝑚𝑜𝑑 10 → 53 ≡ 3 𝑚𝑜𝑑 10
53 − 3 50
=
=5
10
10
Answer: The solutions are 1,6,11,16,21,26,…
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3 Linear Span, Basis and Dimension
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Modular Arithmetic
Additive Inverse in Modular Arithmetic
The sum of a number and its additive inverse equals
the modulus.
Example:
1. Find the additive inverse of 7 in mod 16 arithmetic.
Solution: In mod 16 arithmetic, 7 + 9 = 16, so the
additive inverse of 7 is 9.
Note: In mod 16 arithmetic, the additive inverse of 9 is
7.
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Modular Arithmetic
Multiplicative Inverse in Modular Arithmetic
To find the multiplicative inverse of
𝑎 𝑚𝑜𝑑 𝑚, solve the modular equation
𝑎𝑥 ≡ 1 𝑚𝑜𝑑 𝑚 for 𝑥.
Consider only natural numbers less than the
modulus.
Example:
1. In mod 7 arithmetic, find the multiplicative
inverse of 2.
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Modular Arithmetic
Solution:
To find the multiplicative inverse of 2, solve the
equation 2𝑥 ≡ 1 𝑚𝑜𝑑 7 by trying different natural
number values of 𝑥 less than the modulus.
In mod 7 arithmetic, the multiplicative inverse of
2 is 4.
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Modular Arithmetic
Computing the Day of the Week
A function that is related to the modulo function
is called the floor function. In modulo function,
we determine the remainder when one number is
divided by another. In the floor function, we
determine the quotient (ignore the remainder)
when one is divided by another.
Examples:
2
10
17
2
= 0,
= 5,
= 8,
=1
3
2
2
2
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Modular Arithmetic
Using the floor function, we can write a
formula that gives the day of the week for any
date on the Gregorian Calendar. The formula,
known as Zeller’s congruence, is given by
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Modular Arithmetic
Zeller’s Congruence:
where
d is the day of the month
m is the month using 1 for March, 2 for April,…,10 for December;
January and February are assigned the values 11 and 12,
respectively
y is the last two digits of the year if the month is March through
December; if the Month is January or February, y is the last
two digits of the year minus 1
c is the first two digits of the year
x is the day of the week (using 0 for Sunday, 1 for Monday, …, 6
for Saturday)
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Modular Arithmetic
Example:
1. Determine the day of the week on July 4,
1776.
Solution: We have 𝑐 = 17, 𝑦 = 76, 𝑑 = 4, 𝑚 =
5. Thus,
𝑥≡
13 5 − 1
76
17
+
+
+ 4 + 76 − 2 17
5
4
4
𝑚𝑜𝑑 7
≡ 12 + 19 + 4 + 4 + 76 − 34 𝑚𝑜𝑑 7
≡ 81 𝑚𝑜𝑑 7
Solving 𝑥 ≡ 81 𝑚𝑜𝑑 7 for x, we get
4. Therefore, July 4, 1776, was a Thursday.
1 Topic 1
𝑥=
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Modular Arithmetic
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Applications of Modular Arithmetic
ISBN
Every book that is cataloged in the
Library of Congress must have an ISBN
(International Standard Book Number).
This 13-digit number was created to help
ensure that orders for books are filled
accurately and that books are cataloged
correctly.
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Applications of Modular Arithmetic
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Applications of Modular Arithmetic
ISBN
The first three digits of an ISBN are 978, the
next digit indicates the country in which the
publisher is incorporated (0, and sometimes
1, for books written in English), the next two
to seven digits indicate the publisher, the next
group of digits indicates the title of the book,
and the last digit (the 13th one) is called the
check digit.
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Applications of Modular Arithmetic
The check digit is used to ensure accuracy.
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Applications of Modular Arithmetic
Examples.
1. Determine the ISBN check digit for the book “The
Equation that Couldn’t Be Solved” by Mario Livio. The
first 12 digits of the ISBN are 978-0-7432-5820-?
Solution: From the formula,
𝑑13 ≡ 10 − [9 + 3 7 + 8 + 3 0 + 7 + 3 4 +
3 + 3 2 + 5 + 3 8 + 2 + 3(0)] 𝑚𝑜𝑑 10
≡ 10 − 97 𝑚𝑜𝑑 10
≡ −87 𝑚𝑜𝑑 10
Thus, the check digit is 3.
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Applications of Modular Arithmetic
−87 𝑚𝑜𝑑 10
-87 divided 10 is -8 remainder -7
−87 ÷ 10 = −8 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 − 7
Since the remainder is negative add it to 10
10 + −7 = 3
Hence, −87 𝑚𝑜𝑑 10 ≡ 3.
Thus, the check digit is 3.
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Applications of Modular Arithmetic
2. Suppose that a bookstore clerk sends an
order for the American Heritage Dictionary
and inadvertently enters the number 978-0395-28517-4, where the clerk transposed 8
and 2 in the five numbers that identify the
book.
Correct ISBN: 978-0-395-82517-4
Incorrect ISBN: 978-0-395-28517-4
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Applications of Modular Arithmetic
Let us now calculate the check digit, we have
𝑑13 ≡ 10 − [9 + 3 7 + 8 + 3 0 + 3 + 3 9 + 5
+ 3 2 + 8 + 3 5 + 1 + 3(7)] 𝑚𝑜𝑑 10
≡ 10 − 124 𝑚𝑜𝑑 10
≡ −114 𝑚𝑜𝑑 10
Solve for the check digit,
−114 𝑚𝑜𝑑 10
-114 divided 10 is -11 remainder -4
−114 ÷ 10 = −11 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 − 4
Since the remainder is negative add it to 10
10 + −4 = 6
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Applications of Modular Arithmetic
Because the check digit is 6 and not 4 as it
should be, the receiving clerk knows that an
incorrect ISBN has been sent.
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Applications of Modular Arithmetic
UPC
Another coding scheme is the UPC
(Universal Product Code).
This number is placed on many items
and is particularly useful in grocery stores.
A check-out clerk passes the product by
a scanner, which reads the number from a bar
code and records the price on the cash
register.
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Applications of Modular Arithmetic
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Applications of Modular Arithmetic
UPC
If the price of an item changes for a
promotional sale, the price is updated in the
computer, thereby relieving a clerk of having
to reprice each item.
In addition to pricing items, the UPC
gives the store manager accurate information
about inventory and the buying habits of the
store’s customers.
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Applications of Modular Arithmetic
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Applications of Modular Arithmetic
Comparison of UPC and ISBN
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Applications of Modular Arithmetic
Example. Find the check digit for the DVD
release of the film Alice in Wonderland. The
first 11 digits are 7-86936-79798-?
Solution: From the formula,
𝑑12 ≡ 10 − [3 7 + 8 + 3 6 + 9 + 3 3 +
6 + 3 7 + 9 + 3 7 + 9 + 3 8 ] 𝑚𝑜𝑑 10
≡ 10 − 155 𝑚𝑜𝑑 10
≡ −145 𝑚𝑜𝑑 10
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Applications of Modular Arithmetic
Solve for the check digit,
−145 𝑚𝑜𝑑 10
-145 divided 10 is -14 remainder -5
−145 ÷ 10 = −14 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 − 5
Since the remainder is negative add it to 10
10 + −5 = 5
Hence, the check digit is 5.
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Applications of Modular Arithmetic
Credit Card Numbers
Credit card numbers are normally 13 to 16
digits long. The first one to four digits are used to
identify the card issuer. The table below shows the
identification prefixes used by four popular card
issuers.
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Applications of Modular Arithmetic
Credit Card Numbers
The Luhn algorithm is used to determine whether a
credit card number is valid and is calculated as
follows:
1. Beginning with the next-to-last digit (the last digit
is the check digit) and reading from right to left,
double every other digit.
2. If the digit becomes a two-digit number after
being doubled, treat the number as two
individual digits.
3. Find the sum of the new list of digits; the final
sum must equal 0 mod 10.
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Applications of Modular Arithmetic
Example. Determine whether 5234 8213 3410 1298
is valid credit card number.
Solution: Highlight every other digit, beginning with the
next-to-last digit and reading from right to left:
5 2 3 4 8 2 1 3 3 4 1 0 1 2 9 8
Next, double each of the highlighted digits:
10 2 6 4 16 2 2 3 6 4 2 0 2 2 18 8
Finally, add all digits, treating two-digit numbers as
two single digits:
1+0 +2+6+4+ 1+6 +2+2+3+6+4
+ 2 + 0 + 2 + 2 + 1 + 8 + 8 = 60
Because 60 ≡ 0 𝑚𝑜𝑑 10, this is a valid credit
card number.
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Applications of Modular Arithmetic
Verify if it is true congruence. Now,
60 − 0 60
=
= 6.
10
10
Since, 6 is an integer. Thus, it is true
congruence.Therefore,60 ≡ 0 𝑚𝑜𝑑 10, this is
a valid credit card number.
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Applications of Modular Arithmetic
2. Is 6011 0123 9145 2317 a valid credit card
number?
Solution: Highlight every other digit, reading
from right to left:
6 0 1 1 0 1 2 3 9 1 4 5 2 3 1 7
Double the highlighted digits:
12 0 2 1 0 1 4 3 18 1 8 5 4 3 2 7
Add all the digits, treating two-digit numbers
as two single digits, we get 53.
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Applications of Modular Arithmetic
1+2 + 0+2+1+0+1+4+3+ 1+8
+ 1 + 8 + 5 + 4 + 3 + 2 + 7 = 53
Since
53 − 0 53
=
not an integer
10
10
53 ≢ 0 𝑚𝑜𝑑 10,
this is not a valid credit card number.
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Applications of Modular Arithmetic
Cryptology
Cryptology is the study of making and
breaking secret codes.
Plaintext is a message before it is coded.
Ciphertext is the message after it has been
written in code.
The method of changing from plaintext to
ciphertext is called encryption.
To decrypt a message means to take the
ciphertext message and write it in plaintext.
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Applications of Modular Arithmetic
Example.
Plaintext: SHE WALKS IN BEAUTY LIKE THE
NIGHT
Ciphertext: ODA SWHGO EJ XAWQPU HEGA PDA
JECDP
(Each letter in the plaintext was encrypted by
substituting the letter that is 22 letters after that
letter in the alphabet and continue from the
beginning when the end of the alphabet is reached.
This is called cyclical coding scheme.)
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Applications of Modular Arithmetic
How to decrypt the message:
Take a word from the message (usually one
with the longer words) and continue the alphabet for
each letter of the word.
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Applications of Modular Arithmetic
SHE WALKS IN BEAUTY LIKE THE NIGHT
𝑆
𝑇
𝑈
𝑉
𝑊
𝑊
𝑋
𝑌
𝑍
𝐴
𝐻 𝐺 𝑂
𝐼 𝐻 𝑃
𝐽 𝐼 𝑄
𝐾 𝐽 𝑅
𝐿 𝐾 𝑆
Thus, the decrypt message of SWHGO is WALKS
Then,
𝐸 𝐽
𝐹 𝐾
𝐺 𝐿
𝐻 𝑀
𝐼 𝑁
Thus, the decrypt message of EJ is IN
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Applications of Modular Arithmetic
Plaintext: SHE WALKS IN BEAUTY LIKE THE NIGHT
Ciphertext: ODA SWHGO EJ XAWQPU
Find the decrypt message of HEGA.
𝐻 𝐸 𝐺 𝐴
𝐼 𝐹 𝐻 𝐵
𝐽 𝐺 𝐼 𝐶
𝐾 𝐻 𝐽 𝐷
𝐿 𝐼 𝐾 𝐸
Thus, the decrypt message of HEGA is LIKE.
Find the decrypt message of PDA
𝑃 𝐷 𝐴
𝑄 𝐸 𝐵
𝑅 𝐹 𝐶
𝑆 𝐺 𝐷
𝑇 𝐻 𝐸
Thus, the decrypt message of PDA is THE.
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Applications of Modular Arithmetic
Plaintext: SHE WALKS IN BEAUTY LIKE THE NIGHT
Ciphertext: ODA SWHGO EJ XAWQPU HEGA PDA
Find the decrypt message of NIGHT.
𝐽 𝐸 𝐶 𝐷 𝑃
𝐾 𝐹 𝐷 𝐸 𝑄
𝐿 𝐺 𝐸 𝐹 𝑅
𝑀 𝐻 𝐹 𝐺 𝑆
𝑁 𝐼 𝐺 𝐻 𝑇
Thus, the decrypt message of NIGHT is JECDP.
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Applications of Modular Arithmetic
Numerical Equivalents for the Letters of the
Alphabet
If the encrypting code is to shift each letter of
the plaintext message m positions, then the
corresponding letter in the ciphertext
message is given by
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Applications of Modular Arithmetic
𝑐 ≡ 𝑝 + 𝑚 𝑚𝑜𝑑 26
where,
p-the numerical equivalent of the
plaintext
c-the numerical equivalent of the
ciphertext.
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Applications of Modular Arithmetic
Example. Each letter in the previous example was shifted 22
positions (𝑚 = 22) to the right. To code S in the word SHE,
𝑐 ≡ 𝑝 + 𝑚 𝑚𝑜𝑑 26
𝑐 ≡ 19 + 22 𝑚𝑜𝑑 26
𝑐 ≡ 41 𝑚𝑜𝑑 26
41
= 1 remainder15
26
𝑐 = 15
The 15th letter is O. Thus, S is coded as O.
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Applications of Modular Arithmetic
Examples. Use cyclical alphabetic encrypting
code that shifts each letter 11 positions to
a. Code CATHERINE THE GREAT
1 Topic 1
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Applications of Modular Arithmetic
Plaintext
𝒄 ≡ 𝒑 + 𝒎 𝒎𝒐𝒅 𝟐𝟔
Ciphertext
C
𝑐 ≡ 3 + 11 𝑚𝑜𝑑 26 = 14 𝑚𝑜𝑑 26 = 14
N
A
𝑐 ≡ 1 + 11 𝑚𝑜𝑑 26 = 12 𝑚𝑜𝑑 26 = 12
L
T
𝑐 ≡ 20 + 11 𝑚𝑜𝑑 26 = 31 𝑚𝑜𝑑 26 = 5
E
H
𝑐 ≡ 8 + 11 𝑚𝑜𝑑 26 = 19 𝑚𝑜𝑑 26 = 19
S
E
𝑐 ≡ 5 + 11 𝑚𝑜𝑑 26 = 16 𝑚𝑜𝑑 26 = 16
P
R
𝑐 ≡ 18 + 11 𝑚𝑜𝑑 26 = 29 𝑚𝑜𝑑 26 = 3
C
I
𝑐 ≡ 9 + 11 𝑚𝑜𝑑 26 = 20 𝑚𝑜𝑑 26 = 20
T
N
𝑐 ≡ 14 + 11 𝑚𝑜𝑑 26 = 25 𝑚𝑜𝑑 26 = 25
Y
E
𝑐 ≡ 5 + 11 𝑚𝑜𝑑 26 = 16 𝑚𝑜𝑑 26 = 16
P
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Applications of Modular Arithmetic
Plaintext
𝒄 ≡ 𝒑 + 𝒎 𝒎𝒐𝒅 𝟐𝟔
Ciphertext
T
𝑐 ≡ 20 + 11 𝑚𝑜𝑑 26 = 31 𝑚𝑜𝑑 26 = 5
E
H
𝑐 ≡ 8 + 11 𝑚𝑜𝑑 26 = 19 𝑚𝑜𝑑 26 = 19
S
E
𝑐 ≡ 5 + 11 𝑚𝑜𝑑 26 = 16 𝑚𝑜𝑑 26 = 16
P
G
𝑐 ≡ 7 + 11 𝑚𝑜𝑑 26 = 18 𝑚𝑜𝑑 26 = 18
R
R
𝑐 ≡ 18 + 11 𝑚𝑜𝑑 26 = 29 𝑚𝑜𝑑 26 = 3
C
E
𝑐 ≡ 5 + 11 𝑚𝑜𝑑 26 = 16 𝑚𝑜𝑑 26 = 16
P
A
𝑐 ≡ 1 + 11 𝑚𝑜𝑑 26 = 12 𝑚𝑜𝑑 26 = 12
L
T
𝑐 ≡ 20 + 11 𝑚𝑜𝑑 26 = 31 𝑚𝑜𝑑 26 = 5
E
Therefore, The Code CATHERINE THE GREAT is NLESPCTYP ESP RCPLE.
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Applications of Modular Arithmetic
Examples. Use cyclical alphabetic encrypting
code that shifts each letter 11 positions to
b. Decode TGLY ESP EPCCTMWP
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Applications of Modular Arithmetic
Ciphertext
𝒄 ≡ 𝒑 − 𝒎 𝒎𝒐𝒅𝟐𝟔
Plaintext
T
𝑐 ≡ 20 − 11 𝑚𝑜𝑑 26 = 9 𝑚𝑜𝑑 26 = 9
I
G
𝑐 ≡ 7 − 11 𝑚𝑜𝑑 26 = −4 𝑚𝑜𝑑 26 = 22
V
L
𝑐 ≡ 12 − 11 𝑚𝑜𝑑 26 = 1 𝑚𝑜𝑑 26 = 1
A
Y
𝑐 ≡ 25 − 11 𝑚𝑜𝑑 26 = 14 𝑚𝑜𝑑 26 = 14
N
E
𝑐 ≡ 5 − 11 𝑚𝑜𝑑 26 = −6 𝑚𝑜𝑑 26 = 20
T
S
𝑐 ≡ 19 − 11 𝑚𝑜𝑑 26 = 8 𝑚𝑜𝑑 26 = 8
H
P
𝑐 ≡ 16 − 11 𝑚𝑜𝑑 26 = 5 𝑚𝑜𝑑 26 = 5
E
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Applications of Modular Arithmetic
Ciphertext
𝒄 ≡ 𝒑 − 𝒎 𝒎𝒐𝒅𝟐𝟔
Plaintext
E
𝑐 ≡ 5 − 11 𝑚𝑜𝑑 26 = −6 𝑚𝑜𝑑 26 = 20
T
P
𝑐 ≡ 16 − 11 𝑚𝑜𝑑 26 = 5 𝑚𝑜𝑑 26 = 5
E
C
𝑐 ≡ 3 − 11 𝑚𝑜𝑑 26 = −8 𝑚𝑜𝑑 26 = 18
R
C
𝑐 ≡ 3 − 11 𝑚𝑜𝑑 26 = −8 𝑚𝑜𝑑 26 = 18
R
T
𝑐 ≡ 20 − 11 𝑚𝑜𝑑 26 = 9 𝑚𝑜𝑑 26 = 9
I
M
𝑐 ≡ 13 − 11 𝑚𝑜𝑑 26 = 2 𝑚𝑜𝑑 26 = 2
B
W
𝑐 ≡ 23 − 11 𝑚𝑜𝑑 26 = 12 𝑚𝑜𝑑 26 = 12
L
P
𝑐 ≡ 16 − 11 𝑚𝑜𝑑 26 = 5 𝑚𝑜𝑑 26 = 5
E
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Applications of Modular Arithmetic
b. Decode TGLY ESP EPCCTMWP
Answer:
b. IVAN THE TERRIBLE
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Applications of Modular Arithmetic
Use the congruence 𝑐 ≡ 5𝑝 + 2 𝑚𝑜𝑑 26 to encode the
message LASER PRINTER.
Solution. Replace p by the numerical equivalent of each
letter and determine c. the results for LASER are shown
below.
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Applications of Modular Arithmetic
LASER PRINTER.
Plaintext
𝒄 ≡ 𝟓𝒑 + 𝟐 𝒎𝒐𝒅𝟐𝟔
Ciphertext
P
𝑐 ≡ 5 16 ) + 2 𝑚𝑜𝑑 26 = 82 𝑚𝑜𝑑 26 = 4
D
R
𝑐 ≡ 5 18 ) + 2 𝑚𝑜𝑑 26 = 92 𝑚𝑜𝑑 26 = 14
N
I
𝑐 ≡ 5 9 ) + 2 𝑚𝑜𝑑 26 = 146 𝑚𝑜𝑑 26 = 21
U
N
𝑐 ≡ 5 14 ) + 2 𝑚𝑜𝑑 26 = 72 𝑚𝑜𝑑 26 = 20
T
T
𝑐 ≡ 5 20 ) + 2 𝑚𝑜𝑑 26 = 102 𝑚𝑜𝑑 26 = 24
X
E
𝑐 ≡ 5 5 + 2 𝑚𝑜𝑑 26 = 27 𝑚𝑜𝑑 26 = 1
A
R
𝑐 ≡ 5 18 ) + 2 𝑚𝑜𝑑 26 = 92 𝑚𝑜𝑑 26 = 14
N
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Applications of Modular Arithmetic
LASER PRINTER.
Therefore, the Ciphertext of
PRINTER is JGSAN DNUTXAN.
1 Topic 1
LASER
Page 79
Applications of Modular Arithmetic
Example. Decode the message ACXUT
CXRT, which was encrypted using the
congruence 𝑐 ≡ 3𝑝 + 5 𝑚𝑜𝑑 26.
Solution. Solve the congruence for p.
𝑐 = 3𝑝 + 5
𝑐 − 5 = 3𝑝
9 𝑐 − 5 = 9 3𝑝
9 𝑐 − 5 𝑚𝑜𝑑 26 = 𝑝
Thus, the decoding congruence is
𝑝 ≡ 9 𝑐 − 5 𝑚𝑜𝑑 26.
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Applications of Modular Arithmetic
Using this congruence, we have
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Applications of Modular Arithmetic
ACXUT CXRT
Ciphertext
𝒑 ≡ 𝟗 𝒄 − 𝟓 𝒎𝒐𝒅 𝟐𝟔
Plaintext
C
9 3 − 5 𝑚𝑜𝑑 26 = −18 mod 26 = 8
H
X
9 24 − 5 𝑚𝑜𝑑 26 = 171 mod 26 = 15
O
R
9 18 − 5 𝑚𝑜𝑑 26 = 117 mod 26 = 13
M
T
9 20 − 5 𝑚𝑜𝑑 26 = 135 mod 26 = 5
E
Therefore, we decode the message as PHONE HOME.
1 Topic 1
Page 82
References
Reference: Aufmann, R., Lockwood, J.,
Nation, R. and Clegg, D. (2013).
Mathematical Excursions
(3rd
Edition).
Brooks/Coole: Cengage Learning
1 Topic 1
Page 83