A common exercise in number theory is to find the last digits of a large power, like 22009, without using a computer. 22009 is a 605-digit number, so evaluating it by hand is out of the question. So how do you find its last digits — efficiently?
Modular arithmetic, and in particular, modular exponentiation, comes to the rescue. It provides an efficient way to find the last m digits of a power, by hand, with perhaps only a little help from a pocket calculator. All you need to do is compute the power incrementally, modulo 10m.
In this article, I will discuss three methods — all based on modular exponentiation and the laws of exponents — for finding the ending digits of a positive power of two. The techniques I use are easily adapted to powers of any number.
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The positive powers of two — 2, 4, 8, 16, 32, 64, 128, 256, … — follow an obvious repeating pattern in their ending digit: 2, 4, 8, 6, 2, 4, 8, 6, … . This cycle of four digits continues forever. There are also cycles beyond the last digit — in the last m digits in fact — in the powers of two from 2m on. For example, the last two digits repeat in a cycle of length 20 starting with 04, and the last three digits repeat in a cycle of length 100 starting with 008.
In this article, I will show you why these cycles exist, how long they are, how they are expressed mathematically, and how to visualize them.
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In my article “Nines in Binary”, I proved the following: positive integers of the form 10n-1, that is, integers made up of n digits of 9s, have binary representations with exactly n digits of trailing 1s. Pat Ballew made a clever observation, adapting my result to prove an equivalent statement for base 5 (quinary): positive integers of the form 10n-1 have quinary representations that have exactly n digits of trailing 4s. For example, 9999 in decimal is 304444 in quinary.
In “Nines in Binary”, I derived an expression for 10n – 1 that shows its structure as a binary number:
10n – 1 = (5n – 1) 2n + (2n – 1)
Pat derived a similar expression for 10n – 1 that shows its structure as a quinary number:
10n – 1 = (2n – 1) 5n + (5n – 1)
In essence, he swapped the 2s and 5s, making it the “dual” of my formula, if you will.
I’ll show the details of the derivation and prove why the formula works.
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A palindromic number, or number palindrome, is a number like 74347, which is the same written forward and backward.
A number can be palindromic in any base, not just decimal. For example, 101101 is a palindrome in binary. A number can also be palindromic in more than one base, like decimal 719848917, which is 101010111010000000010111010101 in binary and 5272002725 in octal.
An efficient way to find palindromes in a single base is to generate them, iterating through each integer and constructing palindromes from them. An efficient way to find numbers that are palindromic in multiple bases is to take a palindrome in one base and test if it’s a palindrome in one or more additional bases.
In this article, I’ll show you C code I wrote that finds multi-base numeric palindromes. I used this code to generate tables of numbers that are palindromic in decimal and binary, decimal and hexadecimal, and decimal and octal. I also used this code to solve Euler problem 36, which asks for the sum of all numbers, less than one million, that are palindromic in decimal and binary.
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PARI/GP is an open source computer algebra system I use frequently in my study of binary numbers. It doesn’t manipulate binary numbers directly — input, and most output, is in decimal — so I use it mainly to do the next best thing: calculate with powers of two. Calculations with powers of two are, indirectly, calculations with binary numbers.
PARI/GP is a sophisticated tool, with several components — yet it’s easy to install and use. I use its command shell in particular, the PARI/GP calculator, or gp for short. I will show you how to use simple gp commands to explore binary numbers.
PARI/GP Calculator (Sample of Calculations Used to Explore Binary Numbers)
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Pradeep Mutalik of The New York Times recently blogged about a puzzle that is an instance of the Josephus Problem. The problem, restated simply, is this: there are n people standing in a circle, of which you are one. Someone outside the circle goes around clockwise and repeatedly eliminates every other person in the circle, until one person — the winner — remains. Where should you stand so you become the winner?
Here’s an example with 13 participants:
Alternating Elimination with 13 people, order of elimination shown in red (winner is person 11)
As Pradeep and his readers point out, there’s no need to work through the elimination process — a simple formula will give the answer. This formula, you won’t be surprised to hear, has connections to the powers of two and binary numbers. I will discuss my favorite solution, one based on the powers of two.
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The following is a visual depiction of the binary integers 0 through 11111111:
Binary Integers 0-11111111
A nice pattern, right? I generated it based on the image found on page 117 of Stephen Wolfram’s “A New Kind of Science”. I’ll discuss its structure in detail in this article.
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Here is a table you can use to convert small integers — integers between 0 and 255 — directly between decimal and binary (as an alternative to using a decimal/binary converter):
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I discovered a cool property of positive integers of the form 10n-1, that is, integers made up of n digits of 9s: they have binary representations that have exactly n digits of trailing 1s. For example, 9,999,999 in decimal is 100110001001011001111111 in binary.
The property is interesting in and of itself, but what is more interesting is the process I went through to discover it. It’s a small-scale example of experimental mathematics: I observed something interesting, experimented to collect more data, developed a hypothesis, and constructed a proof.
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Wolfram Alpha can do many types of calculations, including conversions between numbers in different bases. I’ll demonstrate by showing examples of decimal to binary and binary to decimal conversion.
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I introduced my mother to binary numbers a few weeks ago when I showed her my One Hundred Cheerios in Binary poster. It shows the decimal number 100 in binary — 1100100. She’s not an engineer but she’s good with numbers, so I knew she would get it — if only I could find the right way to explain it. Two days ago, I found the right way.
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A double-precision floating-point number is represented internally as 64 bits, divided into three fields: a sign field, an exponent field, and a fraction field. You don’t need to know this to use floating-point numbers, but knowing it can help you understand them. This article shows you how to access those fields in C code, and how to print them — in binary or hex.
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