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  #1  
Old 03-19-2007, 04:56 AM
brokespoke brokespoke is offline
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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  
Old 03-19-2007, 05:07 AM
Noone Special Noone Special is offline
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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
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  #3  
Old 03-19-2007, 05:38 AM
Robot Arm Robot Arm is offline
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eiπ = -1
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  #4  
Old 03-19-2007, 05:44 AM
Giles Giles is offline
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Something else that works for a similar reason:

12345679 x 8 = 98765432

and

12345679 x 9 = 111111111
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  #5  
Old 03-19-2007, 06:06 AM
Malacandra Malacandra is offline
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Quote:
Originally Posted by Robot Arm
eiπ = -1
It's even prettier expressed as eip + 1 = 0; five fundamental constants, addition, multiplication, exponentiation and equality all in one bite-size package.

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!
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  #6  
Old 03-19-2007, 06:41 AM
BobLibDem BobLibDem is offline
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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.
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  #7  
Old 03-19-2007, 06:44 AM
Alive At Both Ends Alive At Both Ends is offline
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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).
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  #8  
Old 03-19-2007, 06:56 AM
Noone Special Noone Special is offline
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Quote:
Originally Posted by BobLibDem
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.
Well, if you have N2, what is the square of N+1?

(N+1)2 = N2+2N+1. So the difference between N2 and (N+1)2 is just 2N+1.

We start off with 12 = 1
So -- the difference between 12 and 22 is 2*1+1 = 3. So 22 = 1+3.
The difference between 22 and 32 is 2*2+1 = 5. So 32 = 22 + 5, or 1+3+5.
Etc.... ad infintum. Effectively, the series 1+3+5+...+(2N+1) is by definition = (N+1)2
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  #9  
Old 03-19-2007, 07:00 AM
Noone Special Noone Special is offline
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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 3-digit 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.
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  #10  
Old 03-19-2007, 07:13 AM
Ludovic Ludovic is offline
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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.
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  #11  
Old 03-19-2007, 07:35 AM
Alive At Both Ends Alive At Both Ends is offline
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Take any four-digit 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.
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  #12  
Old 03-19-2007, 08:14 AM
RealityChuck RealityChuck is offline
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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 4-18=-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. 19-8=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.
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  #13  
Old 03-19-2007, 08:44 AM
Giles Giles is offline
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Quote:
Originally Posted by RealityChuck
Rules for division:

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 4-18=-14, divisible by 7).

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)
It's easier to do 7 and 13 together, especially for relatively large numbers:

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)
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  #14  
Old 03-19-2007, 09:26 AM
Noone Special Noone Special is offline
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Quote:
Originally Posted by Giles
It's easier to do 7 and 13 together, especially for relatively large numbers:

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)
I take it, then, that this method will work for 11 as well? (since 1001 = 7*11*13)
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  #15  
Old 03-19-2007, 09:32 AM
CalMeacham CalMeacham is offline
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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 all-encompassing it was. Look it up on the Web for some really cute Java Applet animations).
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  #16  
Old 03-19-2007, 10:12 AM
Giles Giles is offline
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Quote:
Originally Posted by Noone Special
I take it, then, that this method will work for 11 as well? (since 1001 = 7*11*13)
Yes -- but there's a simpler way to test for division by 11.
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  #17  
Old 03-19-2007, 10:16 AM
Noone Special Noone Special is offline
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Quote:
Originally Posted by Giles
Yes -- but there's a simpler way to test for division by 11.
Yes, I know the "classic" division-by-11 test (difference between sums of even- and odd-placed digits); I just liked the idea of completing the 1001 factorization by testing for all three factors
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  #18  
Old 03-19-2007, 10:32 AM
Giles Giles is offline
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Quote:
Originally Posted by Noone Special
Yes, I know the "classic" division-by-11 test (difference between sums of even- and odd-placed digits); I just liked the idea of completing the 1001 factorization by testing for all three factors
Indeed. The most important factor tests are for prime factors. The first three primes (2, 3 and 5) all have easy tests. The next three primes (7, 11 and 13) can be done all together. So base-10 arithmetic helps up to that point: but then it suddely becomes hard. The equivalent test for 17 involves dividing into groups of 8 digits; and for 19, you divide into groups of 9 digits. Unless the number is very large -- say, bigger than 10^20 -- it would be easier just to do the straight long division.
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  #19  
Old 03-19-2007, 12:33 PM
Amp Amp is offline
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2 x 2 = 4

2 + 2 = 4


1 + 2 + 3 = 6

1 x 2 x 3 = 6
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  #20  
Old 03-19-2007, 12:40 PM
panamajack panamajack is offline
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sin x

x3

the error function

- all rather odd.
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  #21  
Old 03-19-2007, 02:25 PM
rowrrbazzle rowrrbazzle is offline
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12 = 3 * 4
56 = 7 * 8
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  #22  
Old 03-19-2007, 02:36 PM
Thirty-Nine Thirty-Nine is offline
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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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