This may have been a problem I saw in some math book, whose details I don’t remember.
A cow is tied to a rope which is attached to a silo. The silo has a diameter of 1 unit. The rope is pi/2 units long. What is the exact area the cow can roam?
I haven’t figured this out. I’ve tried polar coordinates, parametric equations, etc. This is probably going to turn out like part 2 of the circles problem…
Many numbers can be expressed as the sum of two palindromes. 389 for instance is 383 + 6 as well as 323 + 66.
What three-digit number cannot be expressed as the sum of two palindromes?
(For the sake of this problem, palindromes with leading zeroes [like 030 and 0550] are not allowed, and neither are negative numbers or non-integers. Single digits are allowed.)
This problem’s probably shown up before by a different name or different variation, but I thought it was interesting. It goes like this.
I have a block of wood with integer dimensions. If I saw off two units from each dimension, the volume of the woodblock reduces by half. How many possible sets of dimensions are there for the original block? Or, what integers satisfy 2xyz = (x+2)(y+2)(z+2)?
You can see a program generate them here. There’s eighteen but I could have sworn there were twenty, when I made a similar program some years ago.
The problem changes when you vary the proportion and the number of units cut off from each dimension. (If the proportion is p, 1/p is the how much of the volume is left after cutoff.)
When c=1 and as p increases, there are less and less solutions. 2,3,4 (1,2,3) is the only solution when c=1 and p=4. No solutions when c=1 and p>=5.
When p=2 and as c increases, there are more and more and larger solutions.
It appears that the trend when p=c and both increase is that there are less solutions, though since they become large, and because waiting for it to process is a pain for high maximum values, I’m not sure about this one. With max of 1000 and p=c=11, it didn’t find any solutions.
Let me know if you find anything interesting. :)
Contrived situation time!
Five bands, inconveniently named 1st, 2nd, 3rd, 4th and 5th, are scheduled to play at a festival in a certain order. One manager was looking over this schedule. It had two columns, each with a 1st, 2nd, 3rd, 4th and 5th, but no column titles. So he couldn’t tell which column was the band names, and which one was the order in which they played.
He asked the scheduling manager, who told him that it made no difference which was which.
How many different possible orders are there, in which the bands play?
Extra credit: Same problem, but with 8 bands.
a and b are integers. c and d are the arithmetic and geometric means of a and b, respectively. Likewise, e and f are the arithmetic and geometric means of c and d, respectively. Do there exist a and b such that c,d,e and f are all integers?
So for example, 4 and 16 have geometric mean 8 and arithmetic mean 10. But 8 and 10, while averaging 9, don’t give a neat geometric mean.
Javascript didn’t find anything for a < b < 10000.
Get a calculator, and input any number. It can be fractional, irrational, or even imaginary or complex if possible.
Okay, this has been bugging me for days. See if you can figure it out.
This problem is the same as the last one, except with five surrounding circles instead of three.
Also, as noted in the comments, finding the ratio of two radii is made simpler by assuming the second radius is 1. So:
Update: I found an answer, but it’s an ugly cubic root. Nothing too interesting.
Draw a circle X of radius x, tangent to a line. Then draw three circles of radius y such the first and second are tangent to both X and the line, and the third is tangent to all three circles.
What is y/x?
If x! denotes a factorial, let x? be the smallest number divisible by all integers from 1 to x.
2? = 2;
3? = 6;
4? = 12 rather than 24.
I kept a little table of how many prime factors are conserved, with the ratio (x!/x?). I wondered when that ratio would grow larger than x?, if ever. What I found was even more interesting.
| x | x! | x? | x!/x? |
| 1 | 1 | 1 | 1 |
| 2 | 2 | 2 | 1 |
| 3 | 6 | 6 | 1 |
| 4 | 24 | 12 | 2 |
| 5 | 120 | 60 | 2 |
| 6 | 720 | 60 | 12 |
| 7 | 5040 | 420 | 12 |
| 8 | 40320 | 840 | 48 |
| 9 | 362880 | 2520 | 144 |
| 10 | 3628800 | 2520 | 1440 |
| 11 | 39916800 | 27720 | 1440 |
| 12 | 479001600 | 27720 | 17280 |
| 13 | 6227020800 | 360360 | 17280 |
| 14 | 87178291200 | 360360 | 241920 |
| 15 | 1307674368000 | 360360 | 3628800 |
| 16 | 20922789888000 | 720720 | 29030400 |
15 is the first point when the ratio grows larger than x?, which also happens to be exactly 10 factorial. :o
Start with the product of all the primes: 2305567963945518424753102147331756070
Then just throw in the highest number of each prime required by the other numbers. For example, 64 is 2^6, so multiplying this by 2^5 should cover everything. Also 81 (3^4), 5^2, 7^2, and I think that’s it.
That * 2^5 * 3*3 * 5 * 7 = 69720375229712477164533808935312303556800. Still, it’s over a hundred digits shorter than 100!.
I didn’t find this on Wolfram so this feels slightly more like I’ve done something with some practical application. WolframAlpha, however, helped in calculating. :D
Two points A and B are chosen at random in the plane. A ray is drawn in a random direction from each point (each direction has an equal chance). What is the probability that the two rays intersect?
StevenRoy gave a very nice solution to the 6742816 problem here.
I thought about how it might be extended to three digit numbers. While that’s a much larger field of possibilities, the rules of the problem still restrict them quite a bit. C++ gave me this list of three-digit numbers that survived one multiplication:
2214
31515
35115
4114
499324
555125
61212
61954
62112
71428
72228
74128
899648
I continued them and found:
2214832
3151525
351155250
4114416
4993243
555125
612124864
619545
621122416
7142864
7222832
7412816
8996486
So it appears 351, 612 and 621 tie for the longest three-digit starters for this problem.
Where that might be considered a “strict” multiplier chain, let’s call the following one a “relaxed” one:
“You can make a chain by starting with a number, multiplying its digits, multiplying the result’s digits, etc. The chain ends when it has decreased to one digit. What starter number for x digits results in the longest chain?”
This is tough to solve with anything other than brute force, so with my new Javascript powers, I made this. (It may take a minute to generate)
Both answers for the three digit (and their permutations, by consequence) have digits that add to 22. :D
Fascinated by the emerging patterns, I did a google search and found that this kind of thing had already been investigated and was called “Multiplicative persistence.” Good band name perhaps?
I didn’t find much on the internet about the 6742816 problem though. Mine! :D
Finding four-digit starters for strict multiplier chains definitely needs computer help. Here’s (hopefully) the program that generates the numbers that work for the first two multiplications.
These definitely work for at least one step. Notice the 767720588 – it contains a zero and works, but it won’t get much further. If there aren’t any longer ones, this may take it.
I’ve spent literally all day on this and don’t feel any closer to a solution. :< Anyone want to tackle it?
Update! This script should do the trick. Check out the source code too, for a good laugh. :D
Also, this script solves it for X digits. Right now it’s set to 5, and I haven’t tried 6; it would probably take minutes to process it. I would hazard a guess though, that the longest chains in 6+ digits will start with 315, 612, and 621 (which all have digits that sum to 9.)
Fractals = math + art = things that I find awesome
The last one “Violins” is my favorite, and my current desktop.
I could say I created these, but apart from a few clicks and keystrokes, the program Apophysis did most of the creating. I just picked out which ones looked coolest and messed with them.
But I’ve liked fractals for a long time, and can create some from scratch. The ones in Tegaki for instance, each followed a certain pattern, which is easier to explain in pictures than in words:
Other patterns lead to these. Working with the first one nearly crashed my computer.
Here I showed you a creature from my fractal-animal theme. Here’s the complete series, when I was just getting the hang of coloring lineart:
I extended this theme into general creatures, and called them fractures. I’m probably not the first to come up with this, so if you see this idea elsewhere, show me. :)
This puzzle is still up for those who want to try it. :)
I noticed that when I list out all the 2-digit squares, there’s a way to fuse them end-to-end to form a chain.
81, 16, 64, 49… 81649.
I noticed also that that starts with the same three digits that end the other unique number mentioned above, 6742816. Linking those together we have:
674281649.
It was in Calc II one day that I wondered if this number was prime. So I plugged it into the factorization program on my calculator (which works very similarly to this) and let it churn.
Now that calculator, a TI-83 which I still use to this day, is the same one that I’ve had since I was in middle school, which in turn belonged to my older sister when she was in high school. It is not fast. I was about to give up hope and start paying attention when -!
12149
Immediately I ended the program and checked it; sure enough, 674281649 = 12149 * 55501.
I saw that 12149 is two (and also three) squares placed end-to-end, and that if you took 100000 – 55501, you get 44499 which is five (and also four) squares placed end-to-end.
121 | 49
121 | 4 | 9
4 | 4 | 4 | 9 | 9
4 | 4 | 49 | 9
Finally, 12149 + 55501 is 67650, whose first three digits are a perfect square as well as a palindrome.
Numbers are cool, you guys. I’ve collected tons of little tricks and curiosities and I love to share them. :)
There apparently exist divisibility rules for every number, not just 2-10. I don’t know them all, but thankfully, one doesn’t have to. :)
Suppose we’re faced with finding out what numbers evenly divide 923, if any.
Right off we can tell that 2 does not (it’s odd) and neither does 3 (the digits add to 14, which is not divisible by 3).
And we can skip 4 because if it wasn’t divisible by 2, it certainly won’t be divisible by 4.
In fact, we can skip every composite number, because every composite number is a product of primes less than it. (Fundamental theorem of arithmetic)
It’s still useful to know the rules for 4, 6, and 9. Just notice that 6 depends on the rules for both 2 and 3 :)
So when prime factorizing, all we need to test for divisibility is the primes: 2,3,5,7,11,13,17,19, etc.
And if you’re ever unsure about what’s a prime and what isn’t, you can test them by doing the divisibility rules for the primes underneath it! :D
Ex: 39 looks like a prime at first, but a check for 3 shows that it is 3*13.
41, though, is not divisible by 2, 3, 5, or 7, so it’s prime. (Why do we only have to test up to 7? We’ll see later.)
So we should use 41 as a divisor for larger numbers, but we don’t need 39 if we’ve already tested for 3 and 13.
Continuing, 923 does not end in 5 or 0, therefore it is not divisible by 5.
7 is tricky. Here are some methods for it.
But once we start getting into these higher primes, it’s daunting to try and memorize rules for each individual one. But once again, we don’t have to. The process I use works for all primes, and makes use of properties of modular arithmetic and a little creativity.
One property is that if x is divisible by d (or in shorthand, d|x or “d divides x”) and d also divides y, then d|(x+y). In other words, if two numbers are both divisible by the same number (they are both multiples of d), then their sum will also be divisible by d.
Ex: We know that 36 is divisible by 3. So is 36+3 = 39, 36+30 = 66, 36-12 = 24 and 3000-36 = 2974.
Also, if d|x and n is any integer, then d|(x*n).
Ex: 2 divides 16, so 2 will also divide 16*4 = 64, 16*7 = 112, and 16*100 = 1600.
On the other hand, if x is NOT divisible by d (we’ll say “d!x”), while d|y, then then d!(x+y).
Ex: 24 is not divisible by 5. Neither is 24+5 = 29, 24+25 = 49, 24-10 = 14, or 50-24 = 26.
And if d!x and d!n, then d!(x*n).
Ex: 4 is not divisible by 3. So of 4*10=40, 4*100=400 and 4*1000=4000, none are divisible by 3. 4*24=96 is though, because 3|24.
It can also be shown that if d|(x*y) and d is prime, then d divides either x or y or both.
(Be careful – this does not always work if d is composite. It’s true that 6|210, and that 6|(21*10). But 6 divides neither factor.)
Ex: 13|5252, therefore 13 will always divide at least one of the several ways 5252 can be split into two factors: 2626*2, 1313*4, 404*13, etc. This will be very useful. :D
So 923 is either divisible by 7 or it isn’t.
Whichever one it is, if we stick to those rules of modular arithmetic, we can do all kinds of things to 923 and leave its (in)divisibility intact.
We could start by adding 7 to get 930.
If 7|930, then by the last rule, 7|93. (It doesn’t divide 10, so it must divide 93.)
But does 7|93? Let’s keep going.
Add 7 to 93 and we get 100.
Unfortunately the first dozen or so multiples of 7 skips right over 100. So 100 is not divisible by 7, and so, neither is 923.
The next prime is 11. There’s quite a few easy-to-remember rules for this.
If we try the last-digit one, we see that 92-3 = 89, just a little over 88. So it’s not divisible by 11.
Now let’s try 13. 923-13 = 910. What’s true of 910 in this case is also true of 91, by our last rule.
13, 26, 39, 52, 65, 78, 91! 923 seems to be divisible by 13. A calculator confirms that 923 = 13*71.
So 923 is divisible by both 13 and 71. But are we done? Is 71 prime?
71 is not divisible by 2, 3, 5, or 7, so it is prime. But once again, why can we stop there? Let’s do an experiment.
1103 is prime. (Pretend we don’t know this.) If we used a calculator to test each divisibility rule, here’s what we would get:
1103 / 2 = 551.5
1103 / 3 = 367.66667
1103 / 5 = 220.6
1103 / 7 = 157.57143
1103 / 11 = 100.27273
…
Eventually the divisor would grow larger than the quotient.
…
1103 / 31 = 35.58065
1103 / 37 = 29.81081
There wouldn’t be much point in testing divisibility any higher than that, because if there was any even division, it would have shown up in the quotient!
So the quitting point, roughly, is where the divisor equals the quotient, the square root. :D
Therefore, when testing any 2-digit number to see whether it is prime, you need only test it for 2,3,5, and 7, as we’ve done above. This is because the square root of any 2-digit number is less than 10.
Sometimes you’ll come across numbers that work very easily with this process, for example, dividing 4108205 by 41; you can knock out the 4100000 then the 8200 and you’re left with 5. But 99% of the time you’ll have numbers that aren’t so cooperative. But no matter what, you can always knock out at least one digit with each step.
The key thing to note is that all primes (or possible primes) of 2 or more digits end in either 1,3,7 or 9. Take 43891 for example. It’s a possible prime, and to test it we end up using primes.
11: Subtract 11 to get 43880, and divide by 10 to work with 4388
13: Add three times 13 (39) to get 43930, and divide by 10 to work with 4393.
17: Subtract three times 17 (51) to get 43840, and divide by 10 to work with 4384.
19: Add 19 to get 43910, and divide by 10 work with 4391.
It turns out that, for all primes, you can “even out the ones” every time by either adding or subtracting either the prime or triple the prime. You can use 17 to even out any number ending in 1,3,7,9, by subtracting triple 17, adding 17, subtracting 17, and adding triple 17, respectively. There are sixteen combinations in all, try them out.
So let’s see this all in action.
900 = 302, so the square root of 887 will be a little under 30. Tests up to 29 should cover it.
Is 887 divisible by:
So 887 is prime. It just thinks it’s so special, doesn’t it?
It’s close to 10000, whose square root is 100. Since 1102 = 12100, we need not go past 110. The closest prime under that is 109.
Is 11647 divisible by:
Phew, I wasn’t looking forward to getting to the triple digits.
So we’ve found a factor. 11647 / 19 = 613.
But is 613 prime?
Here’s something even cooler: we don’t have to start over to find out! If 613 was divisible by anything from 2-17, we would know by now.
We should start over at the factor we just used, because prime factors can be used more than once.
(And if the factor you just used is greater than the square root of the new number in question, you’re done – you know it’s prime!)
So we’ll start over at 19, and we can go on until… calcy says square root of 613 is 24.7, so 23 is where we stop.
Is 613 divisible by:
So 613 is prime and 11647 is completely factorized into 19 and 613.
For five-figure numbers and greater, I can get to about 47 before I lose patience and just use a calculator.
Like many things in math, this is a good party trick.
If I’ve made any errors, please point them out, and if there’s any parts in particular that you’d like made clearer, say so too. :)
This is a difficult, abstract thinker.
Part 1: What property does the number 6742816 have, shared only by less than a hundred other numbers?
Part 2: There is a ten-digit number, the longest number that shares 6742816’s property. What is it?
Can you find six distinct digits (none of them 0) such that
ab + cd = ef?
I only know of one solution.
In Kansas, we have license plates of that form – three letters and three numbers.
The other day, while driving, I saw a license plate where the letters were in consecutive alphabetical order and the numbers were in numerical order.
There are 26*26*26*10*10*10 = 17576000 possible plates, 8 arrangements of consecutive numbers, 24 arrangements of consecutive letters for 8*24=192 “ordered” plates. So the likelihood of seeing one such plate is 192/17576000 = 3/274625 = about 1 in 90000.
I made math a new category because I see so many cool math things on a regular basis. And this being the internet, I think there’s a good enough audience. :D
But I also want to dedicate this category to math tutorials and help. Seriously – I love helping people understand math. I plan to make it my career. Keep this in mind when school starts, and don’t be afraid to comment or email for something you’d like help on. I’ll draw up some comics about anything from long division to differential equations.
Also the party is still on. :)
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