Which is bigger, 64! or 264? 64! is, because it follows from a proof by induction for any integer n greater than or equal to 4. It’s also easy to just reason that 64! is bigger: 264 is 64 factors of 2, whereas 64! has 64 factors, except all but one of them (1) are 2 or greater.
When I saw this problem though I wondered if I could solve it in another way: Could the factors of two alone in 64! be greater than 264? As it turns out, almost.
Any double-precision floating-point number can be identified with at most 17 significant decimal digits. This means that if you convert a floating-point number to a decimal string, round it (to nearest) to 17 digits, and then convert that back to floating-point, you will recover the original floating-point number. In other words, the conversion will round-trip.
Sometimes (many) fewer than 17 digits will serve to round-trip; it is often desirable to find the shortest such string. Some programming languages generate shortest decimal strings, but many do not.1 If your language does not, you can attempt this yourself using brute force, by rounding a floating-point number to increasing length decimal strings and checking each time whether conversion of the string round-trips. For double-precision, you’d start by rounding to 15 digits, then if necessary to 16 digits, and then finally, if necessary, to 17 digits.
There is an interesting anomaly in this process though, one that I recently learned about from Mark Dickinson on stackoverflow.com: in rare cases, it’s possible to overlook the shortest decimal string that round-trips. Mark described the problem in the context of single-precision binary floating-point, but it applies to double-precision binary floating-point as well — or any precision binary floating-point for that matter. I will look at this anomaly in the context of double-precision floating-point, and give a detailed analysis of its cause.
In decimal, “0.9 repeating”, or 0.9, equals 1. In binary, a similar thing is true: “0.1 repeating”, or 0.1, equals 1. I’ll show you three ways to prove it, using the three bicimal to fraction conversion algorithms I described recently.
In my article “Counting Binary and Hexadecimal Palindromes” I derived formulas for counting binary palindromes and hexadecimal palindromes. For each type of palindrome, I derived two pairs of formulas: one pair to count n-digit palindromes, and one pair to count palindromes of n digits or less.
Part 1 shows that the powers of five mod 2m cycle with period 2m-2, m ≥ 2, starting at 50.
Part 2 shows that the powers of five mod 10m cycle with the same period as the powers of five mod 2m, starting at 5m.
The highlight of my proof is in part 1, where I derive a formula to show that the period, or order, of 5 mod 2m is 2m-2. While it is in general not possible to derive a formula for the order of a number, I’ll show it is possible for the powers of five mod 2m — due to a hidden, binary structure I’ve uncovered.
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.
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:
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.
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.