Notable Properties of Specific Numbers
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A product of 3 consecutive primes (see also 1001).
A permutation of the digits 1,2,3,4 that is also a multiple of 11 (see 132 for more about this; see also 163 and 10^{1.0979×1019}).
2520 = 2^{3}×3^{2}×5×7 = 5 × 7 × 8 × 9
2520 is the smallest number that is divisible by all the numbers from 1 to 10.
How do you find such a number? It doesn't need to be 1×2×3×4×5×6×7×8×9×10, there are many smaller answers. To consider a simpler example, 12 is the smallest number divisible by 1, 2, 3 and 4 but it is smaller than 1×2×3×4. There are different ways to find the answer, all of which amount essentially to looking at the prime factorizations of all the numbers and keeping the highest exponents of each prime that occur. The simplest way to describe the answer is as follows: Keep the highest power of each prime, and throw away all the rest. For example, in the numbers 1 through 10, the highest prime powers are 8=2^{3}, 9=3^{2}, 5=5^{1} and 7=7^{1}. All the other numbers are composite (e.g. 6=2×3 and 10=2×5) or smaller powers of the primes already mentioned (2=2^{1}, 4=2^{2} and 3=3^{1}).
This number is also related to 10 in another way: it is 10 times the central number (252) on row 10 of Pascal's Triangle. This is sort of related, because 252 = 10!/(5!×5!); see 5040 and 3628800.
This is 72×1000/25.4, the number of pixels per meter corresponding to 72 pixels per inch. A PNG image (such as a screen shot) created on a computer that uses the "72 pixels per inch" standard (e.g. many Apple systems going back to the original Macintosh) will often have 2835 in the "Pixels per Meter" attrbutes.
See 2834.64566929....
2592 = 2^{5}×9^{2}, the only 4digit number of the form ABCD = A^{B}×C^{D}^{66}. (Thanks to Jim Cook for this one). The only other (known) similar number is 24547284284866560000000000.
See also 25920.
3003 occurs at 6 different places in Pascal's Triangle: in rows 14, 15 and 78, because 3003 = 14!/(6!×8!) = 15!/(5!×10!) = 78×77/2. This is related to the fact that 6=F_{3}F_{4} and 15=F_{4}F_{5} are two consecutive golden rectangle numbers. Numbers that occur at least 5 times in Pascal's Triangle include: 120, 210, 1540, 3003, 7140, 11628, 24310, 61218182743304701891431482520, ... (Sloane's A3015).
3003 is also a palindrome, and the 77^{th} triangular number. It shares this property with 1, 3, 6, 55, 66, 171, 595, 666, and many others (Sloane's A3098). Note that the index (77) is itself a palindrome. Possibly because of the simple formula T_{n} = n(n+1)/2, there are also at least two cases with a repunit index: T_{1111}=617716 and T_{111111}=6172882716.
This is 6×7×8×9, which also happens to be equal to the sum 5×6×7×8 + 4×5×6×7 + 3×4×5×6 + 2×3×4×5 + 1×2×3×4. A little investigation shows that this is one of a family of similar patterns: 1+2=3, 1×2+2×3+3×4=4×5, 1×2×3+2×3×4+3×4×5+4×5×6=5×6×7, and so on. The sums/products for the sequence: 1, 3, 20, 210, 3024, 55440, 1235520, 32432400, ... (Sloane's A6963) which counts "planar embedded trees of N nodes".
The solution to the "Monkey and Coconuts problem" (1926 Saturday Evening Post version), which according to Martin Gardner^{62},^{63} is "probably the most worked on and least often solved of all" Diophantine equations. Here is the problem (in my words):
Five sailors, stranded on a desert island, spend the day
gathering coconuts and then go to sleep, agreeing to divide them up
in the morning.
After a while, one sailor wakes up, concerned that he might
not get his fair share. He gives one coconut to the monkey to keep
him quiet, then divides the pile into five parts, and finds
that it divides evenly. He hides one fifth and puts the other 4/5 back
into one pile (the monkey keeps his).
One after another, each of the other four sailors does the
same thing — wakes up, gives a coconut to the monkey, divides the
rest into five parts, which comes out evenly, hides onefifth and
puts the remaining 4/5 back into one pile.
In the morning, the men together divide the remaining
pile and find it divides equally into five parts. How many coconuts
were there in the beginning?
A little trialanderror reveals that the answer is probably somewhat large. One can work it out methodically by working backwards from the end: The final pile F must be a multiple of both 4 and 5, so it can be any number in the sequence {20, 40, 60, 80, ...}, the pile prior to that has to have been ^{5}/_{4}F+1, which admits the numbers in the sequence {76, 176, 276, ...}, etc. There are also more elaborate solution techniques, some of which are outlined in an excellent Numberphile video. There is more than one answer, but the smallest is 3121. The general solution for n sailors is n^{n}n+1 for odd n, and (2n1)n^{n}n+1 for even n.^{64}
See also 7.76×10^{202544}.
3435 = 3^{3}+4^{4}+3^{3}+5^{5}, and is the only number in base 10 (apart from the trivial 1^{1}=1) formed from the sum of each of its digits raised to its own power. (From van Berkel [194])
This property is similar to properties of 135, 153 and some of the Friedman numbers.
3456 is twice 1728, and has a notable pattern of ascending digits. It also has the following cute relationship to 27: 2^{7} × 27 = 3456. See also 34560.
This is π_{④}e, where _{④} is the lowervalued form of the hyper4 operator. See also π^{e}, 4979.003621... and 11058015.34616.
The number of seconds in an hour (60×60), the number of fingers in a cord (units of length), the number of shekels in a talent (units of weight), the number of years in the (Babylonian) long "saros", and the product of squares of the simplest Pythagorean triangle (3^{2}×4^{2}×5^{2}). All of these were significant to the Babylonians.
Although it is not a divisibility recordsetter itself, 3600 is the square of the popular recordsetter (and base of the Sumerian/Babylonian number system) 60.
The division of the hour into 60 minutes or 3600 seconds is the most familiar relic of the old Sumerian base60 numbering system. It also survived in the divisions of angles into degrees, minutes (short for "minute divisions"), seconds, thirds^{14}, and so on. The division of an hour into 3600 parts also happens to be convenient and useful. The hour had been long established as ^{1}/_{12} part of the daylight period (to such an extent that, in many cultures, the length of the hour increased and decreased with the seasons!). The pace represented by 3600 beats per hour arises naturally because it is close to the frequency of a human heartbeat; most people feel their heartbeat when resting quietly. Since hours and heartbeats were already pretty well established, it was useful to choose a number that was pretty close to the right ratio but also was arithmetically convenient to work with. 3600 was by far the best choice.
A triangular Mersenne number, and the largest. See 91.
The first Fibonacci number with a prime subscript that is not itself prime: F_{19}=4181=37×113.
4879 is a Kaprekar number under the most relaxed Kaprekar rules, in which the point of division can be in a place other than the number of digits in the original number. 4879^2 is 23804641, but instead of dividing it into 2380+4641, it is divided into 238+04641 = 4879. Even though the original number 4879 has four digits and its square has eight, the square is divided unevenly giving three digits to the "left half" and five digits to the "right half" 04681. Under the more strict Kaprekar rules, the division always happens in the place corresponding to the number of digits in the original number.
In 1875 Edouard Lucas considered the problem: "A certain number of cannonballs can be arranged in a square on the ground, and can also be stacked into a square pyramid. How many are there?"
The number of cannonballs in the pyramid is the sum of consecutive squares 1^{2}+2^{2}+3^{2}+...+x^{2}. This problem is formally stated as the diophantine equation x(x+1)(2x+1)=6y^{2}. The only known answers were the trivial case x=y=1 and the intended answer to the problem, x=24, y=70, with y^{2}=4900 cannonballs.
Lucas conjectured that there were no other solutions; it was proven by Watson in 1918, but using a rather complicated method. Interest in the original problem continued, and in 1985 D.G. Ma published the first proof using entirely elementary means.
Amazingly, the fact that 24(24+1)(2×24+1)=70^{2} is connected to 26dimensional Lorentzian spacetime, the Leech lattice, and the Monster group. The connection appears^{96} to be evident in Conway and Sloane [146] and is made more explicit by Borcherds in his 1984 Ph.D. thesis [164].
Mike Hill, instructor of an abstract algebra course at University of Virginia, describes it this way^{97} in course notes on the Monster group:
A path on which the timedistance is always zero in a higher
dimensional (> 4) spacetime (Lorentzian space) yields a perpendicular
Euclidean space of two dimensions lower.
ex. 26dimensional Lorentzian space yields the 24dimensional
Euclidean space which contains the Leech lattice.
The Leech lattice contains a point (0,1,2,3,4,...,24,70)
Time distance from origin in Lorentzian space
0 = 0^{2} + 1^{2} + 2^{2} + ... + 23^{2} + 24^{2}  70^{2}
This point lies on a light ray through the origin
Borcherd said a string moving in spacetime is only nonzero if spacetime is 26dimensional.
A similar description is given by Ronan [186], page 224.
See also 3.377×10^{38}.
This is e_{④}π, where _{④} is the lowervalued form of the hyper4 operator. See also e^{π} 3581.875516... and 4341201053.37.
5040 = 7! = 7×6×5×4×3×2×1 = 10×9×8×7. There are other numbers with a similar property (see 720 and 3628800)^{44}.
5040 is also divisible by all the numbers from 1 to 10, and by 144 and several other useful numbers. Plato^{44} cited this as a reason for 5040 being the ideal number of citizens in a state.
5040 is also a divisibility recordsetter and a member of the longest "chain" (see 293318625600).
5041 = 7!+1 = 71^{2}. This is the highest known case of a square which is one more than a factorial. The other cases are 25 = 4!+1 = 5^{2} and 121 = 5!+1 = 11^{2}. There are several similar special properties of numbers (for examples, see 39, 89, 91 and 51381) where the distribution falls off so quickly that it's difficult to see if there are only a finite number of numbers with the property. In this case for example, the odds of N! + 1 being a square are about 1 in √N!, assuming there is no special relationship between the distribution of squares and factorials. Since the factorials grow very quickly, the infinite sum
SUM [ 1 / √N! ]
converges very quickly, and in fact it's a bit of a surprise that there are as many as three solutions for N! + 1 = M^{2}. The fact that there are three suggests that the distribution of factorials and squares might have a relationship — but one should be wary of the Strong Law of Small Numbers (again see 91).
An approximation of the number of years in the "Mayan long count" calendar, according to some theories. This calendar began at a date that is equivalent to 3114 BC in the Gregorian system; it counts days and has an integer value expressed in a 5digit mixedbase system using bases 13, 20, 20, 18 and 20 in each of the five places. The rightmost two places, counting in cycles of 18×20, correspond to a 360day year, and the other three places count "years" (periods of 360 solar days). The total number of days is 1872000. Expressed more accurately, the "5126 years" is 13×20^{2}×18×20/365.242189670 = 5125.3662719. See 2012.
The use of 13 as the base of the highest place in this calendar's counting system is uncertain, and it could just as likely be 20, which would lead to a "long count" of 18×20^{4}/365.242189670 or about 7885 years.
Number of feet in a mile. 5280 is close to the Roman mile defined as 1000 paces, or 5000 feet, but was adapted to accommodate other units of length including the furlong (660 feet) and chain (66 feet). These numbers (66, 660 and 5280) are all multiples of 11 because of their relation to the old unit of length called a rod, also called a perch or pole, which is the same length as an old farmer's tool called an "ox goad"^{92}, used in plughing. However, the origin appears^{93} to go back to Denmark and Prussia sometime before the 11^{th} century; evidence for this fact lies in the fact that the English rod agrees with the old Danish fod and the Prussian Rheinfuss to within less than half an inch, or 0.19% of the total length.
This is the ending value you get using the "Kaprekar transformation" on 4digit numbers. See 495 for details.
The number of synapses (junctions between neurons) in the nematode worm C. elegans. See 959.
6585.3213142 ≈ 223 × 29.530588853
(the saros cycle in days)
Number of mean solar days in the saros.
6939.60160373 = 19 × 365.242189670
(the metonic cycle)
Number of mean solar days in the metonic cycle of 19 tropical years or 235 synodic months. Its integer approximation (6940) can be used as the basis of a lunisolar calendar that repeats every 19 years, and with some effort you can also make the months fairly regular. For example, every year can have 12 months each of which has a specific name, 6 with 29 days and 6 with 30 days; 7 out of 19 years have a 13th month of 30 days, and 4 of those 7 years also add an extra day to one of the normal 29day months. Other similar systems are possible but all solutions have equal complexity; actual lunisolar calendars (like the Hebrew calendar) are more complex but achieve greater accuracy.
Although it is complex, a lunisolar calendar has very strong practical motivations. The importance of the solar day is obvious; the tropical year is important to anyone living in a climate with seasons. And the synodic month tells us when it is possible to see at night by the light of the moon, and what time of day the high and low tides will take place. These things are important even in modern urban society. Under a lunisolar calendar you can agree to meet outdoors at 8PM every month on the 15th of the month and know that there will be moonlight (weather permitting), or you can agree to go fishing every 10th of the month at 5AM and know there will be a low tide (assuming that's how the tide lines up at your location).
The smallest number with a persistence of 6: 6×7×8×8 = 2688; 2×6×8×8 = 768; 7×6×8 = 336; 3×3×6 = 54; 5×4 = 20; 2×0 = 0.
The integer approximation of 6939.60160373. There would be this many days in 19 years, if the 19year metonic cycle were exact. See also 20819 and 161178.
7920 is the ratio between 11!=39916800 and 7!=5040, and the ratio 7920/5040 = 11/7 = 1.571428571428... is kind of an approximation to π/2.
One of the more memorable small powers. While I was 10 and 11 years old I memorized integer exponents of integers just for fun. I still know all of these by heart:

7776 is also a Kaprekar number for 5^{th} powers: 7776^{5}=28430288029929701376, and 2843+0288+0299+2970+1376=7776.
(And better still, 7+7+7+6 = 27.)
7980 = 15×19×28, the product of three numbers that have an important relation to calendars in the Roman, Byzantine and Christian worlds. 28 years is the "solar cycle", 4×7, the number of years it takes before any date falls on the same day of the week again (that's 4 years per leapyear cycle, and 7 days per week). 19 is the length of the metonic cycle, see 19 for more. 15 years is the "indiction cycle", a period related to certain phenomena such as taxation. Since all three numbers are relatively prime to each other, the least common multiple is 7980. The "Julian day number", used in astronomy, is based on a system proposed in 1583 just after the adoption of the Gregorian calendar. Julian day 1 is January 1^{st} 4713 BC. The year 4713 BC happened to be the most recent time that all three cycles (the 15 18 and 28year cycles) were aligned. It is convenient primarily because it predates all recorded history, even in China, Egypt, Greece and Mesopotamia.
8000 = 20^{3} = 11^{3}+12^{3}+13^{3}+14^{3}, the smallest cube that can be expressed as the sum of 4 consecutive cubes. See also 216. See also 246924.
This number has a property related to vampire numbers: if you take the first two digits and multiply by the other two, the product has the same four digits (but in a different order): 81×27=2187.
See also 8712.
(Personal: For a while during my childhood the numbers 7, 27 and 127 were my favorite 1 2 and 3digit numbers. I have since forgotten why the properties of 127 appealed to me, but I suspect it was partly because it ends in "27". If I had continued the series, I probably would have picked 8127, because it is divisible by 7 and by 27, and consists of the cubes 8, 1 and 27 strung together; also the "81" is 3×27.)
The fourth perfect number, and of the form 2^{(p1)}(2^{p}1) where p=7.
A Mersenne prime, 2^{13}1 = 8127.
A narcissistic number, 8208 has 4 digits and is equal to 8^{4}+2^{4}+0^{4}+8^{4}.
This number appeared along with 8128 and 8191 in the 2006 Simpsons episode "Marge and Homer Turn a Couple Play" [216].
The smallest number that is not a palindrome, and is divisible by the number you get when you reverse its digits: 8712 = 4×2178. The next number like this is 9801=99^{2}. The numbers with this property comprise Sloane's A31877 and their factors are A8919. (Thanks to Tavi Laiu for this one). See also 1089. Another related property: 8712 × 2178 = 66^{4}.
9000 was used as the starting value in an xkcd forum discussion/contest "My number is bigger!":
I'll start us off with nine thousand. Because we all know that anything over nine thousand is f***ing unimaginably huge.
This was a reference to the "over 9000" meme, see 9001.
An internet meme: "Anything over 9000 is L337". The origin seems to be a Dragonball Z line 八千以上だ！ "it's higher than 8000!" mistranslated into English as "It's over 9000!!!" (see Dragon Ball wiki, Over 9000), and then spread through videogame culture. There is no relation to HAL 9000 or ISO 9001. See also 9000 and 900901.
See also 573, 73735963, and 1597463007.
9801 = 99^{2} and has a curious reciprocal: 1/9801 = 0.00010203040506070809101112131415161718192021222324252627... For more on this see 997002999.
9801 = 9×1089, a multiple of its "reversal". See 1089 and 8712.
(a myriad)
(basis of the classical Greek large number system)
From the 3^{rd} century BCE to as late as the 12^{th} century, in Greek, Coptic and Armenian texts, letters of the alphabet were sometimes used to represent numbers^{122}:
Α=1 Β=2 Γ=3 Δ=4 Ε=5 ς=6 Ζ=7 Η=8 Θ=9 Ι=10 Κ=20 Λ=30 Μ=40 Ν=50 Ξ=60 Ο=70 Π=80 Ϙ=90 Ρ=100 Σ=200 Τ=300 Υ=400 Φ=500 Χ=600 Ψ=700 Ω=800 Π=900 ͵Α=1000 ͵Β=2000 ͵Γ=3000 ͵Δ=4000 ͵Ε=5000 ͵Δ=6000 ͵Ζ=7000 ͵Η=8000 ͵Θ=9000
The necessary letters are combined to make a quantity, so for example the number 8127 would be ͵ΗΡΚΖ. For numbers above this they used a capital letter mu (Μ) to represent 10000, whose name in Greek is myriad. To distinguish it from Μ=40, a small alpha α was written above the Μ. The precise way in which larger numbers were handled differed through the centuries in ancient Greece. In the most ambitious system, a number (like ͵ΗΡΚΖ) was written in small letters (in this case ͵ηρκζ) directly above a letter Μ to represent 10000 raised to that power; see 10^{40000}. However, in much more common usage the small number placed above the letter Μ merely multiplies the value; see 10^{8}.
10000 is 万 in China (pronounced wàn) and Japan (pronounced man); the traditional form is 萬. Myriads and powers of a myriad are used in several east Asian systems for naming large numbers; see 100000000 and 10^{4096} and my article on large numbers in Japanese. See also the Knuth yllion number names.
See also 100000, 500000 and 10000000.
Generalised Fermat numbers
10001 is a "generalised Fermat number", of the form a^{2n} + 1, with a=10 and n=2. They include the normal Fermat numbers (when a=2 or 4), OEIS sequence A78303 when a=6, A152581 when a=8, A80176, A178426, A152587, etc. If a is odd they are even (and thus composite); when a is even they have some of the properties of the normal Fermat numbers, such as being prime a little more often than one might expect, or having reasonably large prime factors. For example, 10001 is 73×137, and 10^{16}+1 = 353×449×641×1409×69857. The "generalised Fermat primes" would be all primes expressible as a^{2n} + 1, which is the same as OEIS sequence A2496, primes of form n^{2} + 1.
10080 = 60 × 24 × 7, the number of minutes in a week. 10080 is also a divisibility recordsetter with 72 divisors. Thus, there are 72 different ways to divide a week into equal parts that are whole multiples of minutes. 10080 is also 2×7!, see 40320 and 604800.
In France just after the revolution they switched to the nowstandard system of weights and measures based on powers of 10 (SI, systeme internationale, or the "metric system"). They also set up (for a while) a calendar involving a 10day week, and dividing the day into powers of 10. It didn't last long, and France switched back to the Gregorian calendar a few years later.
But there have been other times since then that people have considered the idea of measuring time in powers of 10. Clearly this "metric time" idea has some appeal, because there are dozens of web sites about it (do a search for "metric time"). Perhaps bestknown is Swatch's "Internet time", based on the division of the day into 1000 parts called "beats" (with 86.4 seconds per "beat"). For subdivisions of the day it is easy, the hours and minutes we have now are just arbitrary divisions and there is no conflict with the other important environmental cycles.
When considering a metric replacement for the synodic month and tropical year, clearly there is no good practical solution. These are all very important physical cycles and have strange ratios that cannot be changed (at least not yet :) and cannot easily be adapted to work with powers of 10.
Personally, I think the existing standard time system is just as cool as any 10based system would be, because of the utilitarian properties of the factor recordsetters and cute things like the minutes in February and the 86400000 property. However, if you insist on setting up a "metric time" system, you could do worse than to take advantage of the fact that a minute and a week are almost at an exact ratio of 10000. Such a system would leave the definition of "week" unchanged and define a special "minute" which is exactly 7/10000 of a day (which comes out to exactly 60.48 "standard" seconds). Then all weekly events would always start at the same minutemark, which would be a 4digit number from 0000 to 9999. You'd still have to deal with the fact that the number of weeks per year is odd (the fact that it's close to 100/2 makes this a little more bearable), the number of minutes per day is odd, and there is no suitable "month".
See also 3628800 and 86400000.
The 2^{nd} row of Pascal's Triangle shows clearly in the 2^{nd} power of 101. The pattern continues in higher powers, and goes further than for the powers of 11. See 14641 for more.
The number of arcminutes in 180 degrees, and thus the idealized length of a meridian (see 20003931.4585) in nautical miles (see 1852.216). See also 108 and 10080.
The name of the worker who is befriended by Freder in the 1927 movie Metropolis. Also a palindrome, and strobogrammatic.
In 2011 November, this was identified as the "smallest uninteresting number" in an episode of QI, based on its being the smallest (positive) integer not appearing in the listed terms of any sequence in the OEIS. It is important to note that some sequences, such as A0027, contain every integer — so this property of 12407 was only possible because each entry in the OEIS gives only a few lines of initial terms for each sequence. There is a lot more about this in [205].
(Pascal's triangle)
The smaller powers of 11 (121, 1331 and 14641) give rows of Pascal's Triangle. Pascal's Triangle is a rather useful table of numbers (called binomial coefficients) that is formed by continually adding numbers in a cascade starting from a single 1, as shown here:

Each is the sum of the (one or) two numbers above it.
These numbers have many applications because they count "combinations", such as "how many combinations of 3 colors can you make if you have 5 colors to choose from?" (the answer is the 3^{rd} number in row 5 of the triangle: 10). Another question with the same answer is, "If you flip a coin 5 times, how many ways are there to end up getting heads exactly 3 times?" (the answer is also 10).
A general formula that gives the value of item N in row M is:
_{N}C_{M} = M!/((MN)!×N!)
For example, _{3}C_{5} = 5!/((53)!×3!) = 120/(2×6) = 10. However this formula gets rather impractical to use when M is large, even when the value being computed is pretty reasonable. For example, consider _{3}C_{143}=143!/(140!×3!). 143! has 248 digits, and 140!×3! has 242 digits — so you probably couldn't even get the answer on a calculator, as the calculator would overflow. However, most of the 143! in the numerator cancels out with the 140! in the denominator, leaving _{3}C_{143} = 143×142×141/3! = 477191, which you could even calculate by hand if you needed to.
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Quick index: if you're looking for a specific number, start with whichever of these is closest: 0.065988... 1 1.618033... 3.141592... 4 12 16 21 24 29 39 46 52 64 68 89 107 137.03599... 158 231 256 365 616 714 1024 1729 4181 10080 45360 262144 1969920 73939133 4294967297 5×10^{11} 10^{18} 5.4×10^{27} 10^{40} 5.21...×10^{78} 1.29...×10^{865} 10^{40000} 10^{9152051} 10^{1036} 10^{1010100} — — footnotes Also, check out my large numbers and integer sequences pages.
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