Giant Numbers and Modulo Arithmetic

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Q: What is the last digit of the giant number \(7^{2000}\).

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A: At first blush this is an impossible task. The above number is impossible to compute. However, using some simple rules from modulo arithmetic, it is possible to answer this.

Let us first try to solve this the long way and run some experiments with 7.
$$
7^{2} = 49 \text{ last digit is 9}\\
7^{3} \equiv 49 \times 7  \equiv \text{ last digit is 3 as } 9\times7 \equiv \text{ last digit is 3}\\
$$
Continuing in the same fashion
$$
7^{4} \equiv 3 \times 7 \equiv \text{ last digit is 1 as } 3 \times 7 = 21
$$
Now \(7 ^ {5}\) would have its last digit as 7, because 21's last digit is 1 and when multiplied by 7 yields 7. But this is the same as \(7^{1}\). This resets the entire process. We can conclude that as we multiply into higher powers of 7, the last digits cycles (of length 4) as \({7,9,3,1,...}\). This implies, \(7^{2000}\) would have the units place as 1, as 2000 divides 4 with no remainders.

The problem can be solved in a terser way if we use rules from modulo arithmetic. A refresher on modulo arithmetic. When we write \( a \equiv b (\textrm{mod}\ c)\) we mean that \(a-b\) is divisible by \(c\) without leaving any remainder.

The "product rule" for modulo arithmetic states, if
$$
a \equiv b \textrm{( mod}\ n ) \\
c \equiv d \textrm{( mod}\ n )
$$
then
$$
a\times c \equiv b\times d \textrm{( mod}\ n )
$$
\(7^{2000}\) is really just 7 multiplied 2000 times. So we can write \(7 \equiv -1 \textrm{( mod} 4) \)
$$
7^{2000} \equiv (-1)^{2000}\textrm{( mod} 4)
$$
which in turn implies that the only number we need to compute is \((-1)^{2000}\) which is 1.

If you are looking to buy some books in probability here are some of the best books to learn the art of Probability


Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)
This book is a great compilation that covers quite a bit of puzzles. What I like about these puzzles are that they are all tractable and don't require too much advanced mathematics to solve.

Introduction to Algorithms
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Introduction to Probability Theory
Overall an excellent book to learn probability, well recommended for undergrads and graduate students

An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition
This is a two volume book and the first volume is what will likely interest a beginner because it covers discrete probability. The book tends to treat probability as a theory on its own

The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else!)
A good book for graduate level classes: has some practice problems in them which is a good thing. But that doesn't make this book any less of buy for the beginner.

Introduction to Probability, 2nd Edition
A good book to own. Does not require prior knowledge of other areas, but the book is a bit low on worked out examples.

Bundle of Algorithms in Java, Third Edition, Parts 1-5: Fundamentals, Data Structures, Sorting, Searching, and Graph Algorithms (3rd Edition) (Pts. 1-5)
An excellent resource (students, engineers and even entrepreneurs) if you are looking for some code that you can take and implement directly on the job

Understanding Probability: Chance Rules in Everyday Life
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Data Mining: Practical Machine Learning Tools and Techniques, Third Edition (The Morgan Kaufmann Series in Data Management Systems)
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Discovering Statistics Using R
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A Course in Probability Theory, Third Edition
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Probability and Statistics (4th Edition)This book has been yellow-flagged with some issues: including sequencing of content that could be an issue. But otherwise its good