#1




mathematical oddities
I had to work this one out before I believed it......The product of 111,111,111 times 111,111,111 (or 111,111,111 squared) is 12,345,678,987,654,321
Anybody want to post any others?? 
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#2




This actually works for any number of '1's (up to 10):
1 x 1 = 1 11 x 11 = 121 111 x 111 = 12,321 etc.... This finally breaks down when you have 10 '1's: 1,111,111,111 x 1,111,111,111 = 1234567900987654321 
#4




Something else that works for a similar reason:
12345679 x 8 = 98765432 and 12345679 x 9 = 111111111 
#5




Quote:
I like the way that 11,111,111,111 factorises to 21,649 and 513, 239  just two prime factors 5 and 6 digits long! 
#6




If you add odd numbers in sequence you always get a perfect square.
1+3=4 1+3+5=9 1+3+5+7=16 etc. 
#7




142857 x 1 = 142857
142857 x 2 = 285714 142857 x 3 = 428571 142857 x 4 = 571428 142857 x 5 = 714285 142857 x 6 = 857142 The same six digits every time, in the same cyclic order. (BTW, 142857 x 7 = 999999). 
#8




Quote:
(N+1)^{2} = N^{2}+2N+1. So the difference between N^{2} and (N+1)^{2} is just 2N+1. We start off with 1^{2} = 1 So  the difference between 1^{2} and 2^{2} is 2*1+1 = 3. So 2^{2} = 1+3. The difference between 2^{2} and 3^{2} is 2*2+1 = 5. So 3^{2} = 2^{2} + 5, or 1+3+5. Etc.... ad infintum. Effectively, the series 1+3+5+...+(2N+1) is by definition = (N+1)^{2} 
#9




Another one... any number of the form ABC,ABC (e.g., 237,237) will be divisible by 7, by 11, and by 13.
Moreover, the result of dividing such a number by 7, then 11 and then 13 will be ABC. (so, dividing 237,237 by 7, then 11, then 13 will be 237). It's a rather neat parlor trick, if you only perform it once  get sombody to choose the 3digit number, then have another "double" it, a third divide by 7... all the while assuring them it will be divisible, which should seem to be no mean feat in itself! This is because 7*11*13 = 1,001  and the number ABC,ABC is simply ABC*1001. 
#10




The sum of the digits of any integer divisible by 3 will also be divisible by 3.
(393/3=131 3+9+3=15, 15/3=5) Same thing goes for 9. 
#11




Take any fourdigit number in which the digits aren't all the same (e.g. 1112 is OK, but not 1111 or 2222).
Arrange its digits in ascending order and descending order, and subtract the smaller number from the larger. Repeat the process with the new number. Starting with 4698 (for example): 9864  4689 = 5175 Repeat the process with the new number (5175): 7551  1557 = 5994 Repeat with 5994, and keep repeating: 9954  4599 = 5355 5553  3555 = 1998 9981  1899 = 8082 8820  0288 = 8532 8532  2358 = 6174 You always arrive at 6174, whatever number you started with; after which the sequence repeats because 7641  1467 = 6174 again. 
#12




Rules for division:
Division by 1: Any integer Division by 2: The last digit is divisible by two Division by 3: The sum of the digits is divisible by three Division by 4: The last two digits are divisible by four Division by 5: The last digit is either 5 or 0 Division by 6: The last digit is divisible by two, and the sum of the digits is divisible by three Division by 7: Double the last digit and subtract it from the rest of the number. The result is divisible by 7 (e.g., 49. 2*9=18 418=14, divisible by 7). Division by 8: The last three digits are divisible by 8 Division by 9: The sum of the digits are divisible by 9 Division by 10: The last digit is 0 Division by 11: The sum of the even digits minus the sum of the odd digits is divisible by 11 or equals 0. (e.g., 35838. Even digits: 5+3 = 8 Odd digits 3+8+8=19. 198=11). Division by 12: The sum of the digits is divisible by three and the last two digits are divisible by four. Division by 13: Four times the last digit plus the other digits is divisible by 13 (e.g., 455. 4x5= 20. 20+45 = 65. For 65: 5x2 = 20 20+6 = 26) Division by 14: Last digit divisible by two; if so, use the test for division by 7 Division by 15: Last digit 0 or 5, sum of the digits divisible by 3. Division by 16: Last four digits divisible by 16.
__________________
Author of Staroamer's Fate and Syron's Fate, now back in print. 
#13




Quote:
Taker a number, e.g. 85230288 Divide the number into groups of 3 digits, starting at the right, e.g. 85 230 288 Sum the even groups, and sum the odd groups, e.g. 230 and 373 Find the difference between the two sums, e.g. 143 If that difference is divisible by 7, the number is divisible by 7. If that difference is divisible by 13, the number is divisible by 13. (The number is not divisible by 7, but is divisible by 13, because 143 = 13 x 11) (The method relies on 7 and 13 being divisors of 1001) 
#14




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#15




All of this just hits the tip of the iceberg of mathematical oddities. Have a look at David Wells' books The Penguin Dictionary of Numbers and The Penguin Dictionary of Interesting Geometry for all sorts of weird relationships you never knew of and wouldn't expect. Look up Poncelet's Porism and Steiner Chains (I discovered a subset of Poncelet's Porism on my won, but didn't realize how allencompassing it was. Look it up on the Web for some really cute Java Applet animations).

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#19




2 x 2 = 4
2 + 2 = 4 1 + 2 + 3 = 6 1 x 2 x 3 = 6 
#20




sin x
x^{3} the error function  all rather odd. 
#21




12 = 3 * 4
56 = 7 * 8 
#22




My username comes from The Penguin Dictionary of Curious and Interesting Numbers, where 39 is described as the first number with no particularly interesting properties.
I felt sorry for it. My favourite: Between any two rational numbers is an irrational number. Between any two irrational numbers is a rational number. But there are infinitely more irrational numbers! 
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