Hi All:
I have found astonishing math wonders pertaining to Number 7, that I am pleased to put it for challenge for anyone to develop a structure like this one. I could not even get the "HOW" and "WHY" the following results have developed this amazing way.
A = 6555
B = 3450 and C = 3105
D = 3303 and E = 3252
A = B + C
A = D + E
Here are the astonishing results:
34503105 = 7 x 7 x 704145
33033252 = 7 x 7 x 674148
704145674148 = 7 x 100592239164
704145 + 674148 = 7 x 196899
100592239164196899 = 7 x 7 x 2052902840085651
100592239164 + 196899 = 7 x 14370348009
**************************************************************************************************************
Now we come up with more results if we reverse 3303325 (i.e. 25233033):
25233033 = 7 x 3604719
34503105 =7 x 7 x 704145
3604719704145 = 7 x 514959957735
3604719 + 704145 = 7 x 7 x 87936
ISN'T THIS AMAZING???
Bol777
A = 2043
B = 1092
C = 3948
D = 2940
They are mutliples of 3, combining them by putting one after the other, such as 2043 1092 will make multiples of 3 etc etc
Finding a couple of numbers which do that isn't enormously difficult, particularly if you know the little tricks/rules which you can use to work out almost instantly if a number is a multiple of numbers such as 3, 6, 9, 11 etc.
B = 1092
C = 3948
D = 2940
They are mutliples of 3, combining them by putting one after the other, such as 2043 1092 will make multiples of 3 etc etc
Finding a couple of numbers which do that isn't enormously difficult, particularly if you know the little tricks/rules which you can use to work out almost instantly if a number is a multiple of numbers such as 3, 6, 9, 11 etc.
When I was a school we were taught lots of the sort of little tricks Alphanumeric refers to.. all of which I have now forgotten. Does anyone know where they might be found on the web?
-C2.
-C2.
QUOTE (AlphaNumeric+Jul 27 2006, 08:30 PM)
A = 2043
B = 1092
C = 3948
D = 2940
They are mutliples of 3, combining them by putting one after the other, such as 2043 1092 will make multiples of 3 etc etc
Two things:
1. your A + B does not equal C + D.
My numbers were 3450 +3105 = 3303 + 3252
2. We want this based on number 7. It's much easier to develop this with number 3, 9 OR 11 ... Although you still did not satisfy condition 1 above anyway.
TRY AGAIN
B = 1092
C = 3948
D = 2940
They are mutliples of 3, combining them by putting one after the other, such as 2043 1092 will make multiples of 3 etc etc
Two things:
1. your A + B does not equal C + D.
My numbers were 3450 +3105 = 3303 + 3252
2. We want this based on number 7. It's much easier to develop this with number 3, 9 OR 11 ... Although you still did not satisfy condition 1 above anyway.
TRY AGAIN
If "It's much easier to develop this with number 3, 9 OR 11" then does that not mean 3, 9 AND 11 are even more miraculous than 7?
what do those numbers signify?
Does this 7 thing apply to an abitrary set of numbers?
Does this 7 thing apply to an abitrary set of numbers?
Try this.
12345679 x 9 = All the 111's......12345679 x 18 = 222's....continue multplying 12345679 by multiples of 9 and you can get all other numbers up to 999'etc.
Was shown this as a kid.....thought it was cool then
....although seems a bit naff now 
....that's age for you!
12345679 x 9 = All the 111's......12345679 x 18 = 222's....continue multplying 12345679 by multiples of 9 and you can get all other numbers up to 999'etc.
Was shown this as a kid.....thought it was cool then
....although seems a bit naff now ....that's age for you!
7's the key number here. Think about it. 7-Elevens. 7 doors. 7, man, that's the number. 7 chipmunks twirlin' on a branch, eatin' lots of sunflowers on my uncle's ranch. You know that old children's tale from the sea? It's like you're dreamin' about Gorgonzola cheese when it's clearly Brie time, baby. Step into my office, 'cause you're farkin' fired.
- Dave
- Dave
reminds me of the commercial "Make 7 Up Yours" 
√12345678987654321 = 111111111
√12345678987654321 = 111111111
QUOTE (bol777+Jul 27 2006, 10:53 PM)
Two things:
1. your A + B does not equal C + D.
My numbers were 3450 +3105 = 3303 + 3252
Just pick 4 4 digit numbers whose digits sum to 3 and where A+B = C+D
Such as :
A = 1950, B = 6201
C = 2490, D = 5661
So you've found 4 numbers which do that, what underlying principle have you shown? Nothing, you've just found 4 numbers. Is there a general way of finding infinitely many such groups? For any number? That would be a more interesting endeavour rather than sifting through billions of permutations to find another lot.
Amazing! Now number 7 made him disappear!
Hi there ... Sorry i didn't check the forum on the weekend!
I am still here, but have not seen my structure developed by using number 7. Again, I said number 3 was much easier to get structure with, now who's the champion to do this (4 four or 4 five digit numbers) with number 7 ... this is where I promised the laptop, not using number 3, check my quote. "If anyone gets the same results as my original post, based on number 7 ... I'll be glad to give the winner any laptop (or other prize), of his choice....... How is that for a challenge???"
1. your A + B does not equal C + D.
My numbers were 3450 +3105 = 3303 + 3252
Just pick 4 4 digit numbers whose digits sum to 3 and where A+B = C+D
Such as :
A = 1950, B = 6201
C = 2490, D = 5661
QUOTE
2. We want this based on number 7. It's much easier to develop this with number 3, 9 OR 11 ... Although you still did not satisfy condition 1 above anyway.
I was demonstrating that there isn't anything amazing about 7, that such "Oh, interesting" collections of numbers can be found based on most, if not all, numbers. So you've found 4 numbers which do that, what underlying principle have you shown? Nothing, you've just found 4 numbers. Is there a general way of finding infinitely many such groups? For any number? That would be a more interesting endeavour rather than sifting through billions of permutations to find another lot.
QUOTE (->
| QUOTE |
| 2. We want this based on number 7. It's much easier to develop this with number 3, 9 OR 11 ... Although you still did not satisfy condition 1 above anyway. |
I was demonstrating that there isn't anything amazing about 7, that such "Oh, interesting" collections of numbers can be found based on most, if not all, numbers.
So you've found 4 numbers which do that, what underlying principle have you shown? Nothing, you've just found 4 numbers. Is there a general way of finding infinitely many such groups? For any number? That would be a more interesting endeavour rather than sifting through billions of permutations to find another lot.
If "It's much easier to develop this with number 3, 9 OR 11" then does that not mean 3, 9 AND 11 are even more miraculous than 7?
So you've found 4 numbers which do that, what underlying principle have you shown? Nothing, you've just found 4 numbers. Is there a general way of finding infinitely many such groups? For any number? That would be a more interesting endeavour rather than sifting through billions of permutations to find another lot.
If "It's much easier to develop this with number 3, 9 OR 11" then does that not mean 3, 9 AND 11 are even more miraculous than 7?
No, it's based on our totally arbitrary choice of a base 10 maths system. 9 and 11 are 10-1 and 10+1, so they create patterns in the digits of numbers divisible by them. Similarly 3 because 3*3 = 9.
If we worked in base 16, you'd see the same kinds of patterns with 15 and 17 as we see at the moment with 9 and 11. There's nothing really fundamental about it because it depends on the choice of how to represent the numbers.
If we worked in base 16, you'd see the same kinds of patterns with 15 and 17 as we see at the moment with 9 and 11. There's nothing really fundamental about it because it depends on the choice of how to represent the numbers.
QUOTE (AlphaNumeric+Jul 28 2006, 01:01 PM)
Just pick 4 4 digit numbers whose digits sum to 3 and where A+B = C+D
Such as :
A = 1950, B = 6201
C = 2490, D = 5661
I was demonstrating that there isn't anything amazing about 7, that such "Oh, interesting" collections of numbers can be found based on most, if not all, numbers.
Again two things:
1) your four numbers do not work as supposed to be in my original post.
1950 + 6201 = 3 x 2717 (my first level numbers divided by 7 twice, not once)
2490 + 5661 = 3 x 2717 (my first level numbers divided by 7 twice, not once)
Then let's work your remainders:
2717 + 2717 is not divisable by 3
27172717 is also not divisable by 3
So, basically, you didn't even get similar first level results
2. With number 2 or 3 which are probably the easiest (smallest) numbers, you can develop my structure based on, yet you'll find it IMPOSSIBLE even based on 2 or 3
I know it's impossible to do this with any other 4 - four, digit numbers based on number 7. It could be even impossible for any different number of digits as well.
If anyone gets the same results as my original post, based on number 7 ... I'll be glad to give the winner any laptop (or other prize), of his choice....... How is that for a challenge???
Note: I said 2 or 3 are the easiest to develop the structure with because there are more than twice more numbers divisable by 3 than 7 (For example from 1 to 75, you find there are 25 numbers divisable by 3, and only 10 numbers divisable by 7).
AGAIN THE CHALLENGE IS OPEN TO ANYBODY ... EVEN BY BASING YOUR STRUCTURE ON NUMBER 2 OR 3. (NOT THE MORE DIFFICULT 7)
Such as :
A = 1950, B = 6201
C = 2490, D = 5661
I was demonstrating that there isn't anything amazing about 7, that such "Oh, interesting" collections of numbers can be found based on most, if not all, numbers.
Again two things:
1) your four numbers do not work as supposed to be in my original post.
1950 + 6201 = 3 x 2717 (my first level numbers divided by 7 twice, not once)
2490 + 5661 = 3 x 2717 (my first level numbers divided by 7 twice, not once)
Then let's work your remainders:
2717 + 2717 is not divisable by 3
27172717 is also not divisable by 3
So, basically, you didn't even get similar first level results
2. With number 2 or 3 which are probably the easiest (smallest) numbers, you can develop my structure based on, yet you'll find it IMPOSSIBLE even based on 2 or 3
I know it's impossible to do this with any other 4 - four, digit numbers based on number 7. It could be even impossible for any different number of digits as well.
If anyone gets the same results as my original post, based on number 7 ... I'll be glad to give the winner any laptop (or other prize), of his choice....... How is that for a challenge???
Note: I said 2 or 3 are the easiest to develop the structure with because there are more than twice more numbers divisable by 3 than 7 (For example from 1 to 75, you find there are 25 numbers divisable by 3, and only 10 numbers divisable by 7).
AGAIN THE CHALLENGE IS OPEN TO ANYBODY ... EVEN BY BASING YOUR STRUCTURE ON NUMBER 2 OR 3. (NOT THE MORE DIFFICULT 7)
A = 2187 B = 6561
C = 2178 D = 6570
A+B = C+D
21876561 = 3*3*2430729
21786570 = 3*3*2420730
24307292420730 = 3*8102430806910 = 3*3*2700810268970
2430729+2420730 = 3*1617153 = 3*3*539051
81024308069101617153 = 3*27008102689700539051
8102430806910+1617153 = 3*2700810808021
Now lets reverse 21876561 to get 16567821
16567821 = 3*3*1840869
21786570 = 3*3*2420730
18408692420730 = 3*3*2045410268970
1840869 + 2420730 = 3*3*473511
Done. In some places I've more divisibility by 3 than you did by 7 too.
I would like to point out that was my first attempt at finding such numbers and all I did was pick the two 4 digit powers of 3 for A and B and add/subtract 9 from them and go from there. Easy when you know how.
Still sure you want to give out a laptop?
C = 2178 D = 6570
A+B = C+D
21876561 = 3*3*2430729
21786570 = 3*3*2420730
24307292420730 = 3*8102430806910 = 3*3*2700810268970
2430729+2420730 = 3*1617153 = 3*3*539051
81024308069101617153 = 3*27008102689700539051
8102430806910+1617153 = 3*2700810808021
Now lets reverse 21876561 to get 16567821
16567821 = 3*3*1840869
21786570 = 3*3*2420730
18408692420730 = 3*3*2045410268970
1840869 + 2420730 = 3*3*473511
Done. In some places I've more divisibility by 3 than you did by 7 too.
I would like to point out that was my first attempt at finding such numbers and all I did was pick the two 4 digit powers of 3 for A and B and add/subtract 9 from them and go from there. Easy when you know how.
Still sure you want to give out a laptop?
QUOTE (bol777+Jul 27 2006, 04:27 PM)
I have found astonishing math wonders pertaining to Number 7
7 is first among so called periodic (cyclic) numbers
1/7 = 1.142857142...
Cycle is 142857 where 0, 3, 6, 9 are missing.
Well known Ahmes' rounding
Pi ~ 2*11/7
but also primes for e
e ~ 19/7
13 is next cyclic number
7 is first among so called periodic (cyclic) numbers
1/7 = 1.142857142...
Cycle is 142857 where 0, 3, 6, 9 are missing.
Well known Ahmes' rounding
Pi ~ 2*11/7
but also primes for e
e ~ 19/7
13 is next cyclic number
QUOTE (AlphaNumeric+Jul 29 2006, 11:57 AM)
Still sure you want to give out a laptop?
Amazing! Now number 7 made him disappear!
Amazing! Now number 7 made him disappear!
QUOTE (Upisoft+Jul 29 2006, 09:56 PM)
Amazing! Now number 7 made him disappear!
Hi there ... Sorry i didn't check the forum on the weekend!
I am still here, but have not seen my structure developed by using number 7. Again, I said number 3 was much easier to get structure with, now who's the champion to do this (4 four or 4 five digit numbers) with number 7 ... this is where I promised the laptop, not using number 3, check my quote. "If anyone gets the same results as my original post, based on number 7 ... I'll be glad to give the winner any laptop (or other prize), of his choice....... How is that for a challenge???"
By the way, I forgot to point out in my original post that my A+B = C+D = 6555
and 6555 has a special nature that it is the sum of numbers from 1 to 114 (1+2+3+4+5+......+112+113+114 = 6555)
(while the number developed by my freind AlphaNumeric "i.e. 8151, has no significant math representation"
I also noticed that any number divisable by 3, its reverse is always divisable by 3, while this is not the case with 7!!!
and 6555 has a special nature that it is the sum of numbers from 1 to 114 (1+2+3+4+5+......+112+113+114 = 6555)
(while the number developed by my freind AlphaNumeric "i.e. 8151, has no significant math representation"
I also noticed that any number divisable by 3, its reverse is always divisable by 3, while this is not the case with 7!!!
QUOTE (bol777+Jul 31 2006, 07:42 PM)
and 6555 has a special nature that it is the sum of numbers from 1 to 114 (1+2+3+4+5+......+112+113+114 = 6555)
(while the number developed by my freind AlphaNumeric "i.e. 8151, has no significant math representation"
So, my 'special property' is that the A = 3^7 and B = 3^8, so my number is 3^7 + 3^8. Wow, who cares
It seems to me, you just add a 'special feature' every time I come up with an similar thing to you.
Frankly, who cares? This is hardly deep number theory and I'm 100% positive someone with a bit of time and a knack for number bases could come up with plenty of such examples for any number, though I have little doubt you'd come up with a reason why yours is somehow 'superior'.
There are plenty of loops, cycles and tricks that numbers do like this. Google "Amicable numbers", "Happy Numbers", "Harshad Numbers", plenty of things like that. You've discovered nothing deep or magical, unless you can prove it for an n'th formula somehow, instead you've discovered one of probably infinitely many 'slight of hands' which the infinitely many integers have within them.
For instance, consider 'Kaprekar's Process'. Pick any 4 digit number which doesn't have all the same digits, order the digits in descending order, subtract the original number when arranged in asending order and repeat on the new number. You'll ALWAYS get 6174.
Consider 4828. Rearranging into descending order 8842, ascending order is 2488
8842 - 2488 = 6354
6543 - 3456 = 3087
8730 - 0378 = 8352
8532 - 2358 = 6174
Wow, amazing? No, not really. This and about 500 other similar tricks, possibly including yours can be found in "The Penguin Dictionary of Curious and Interesting Numbers" by David Wells.
By the way, this doesn't make you an amazing mathematician. Given enough time, anyone can find or derive such a things. It took me ONE attempt to get very close to yours in complexity. I imagine if I worked in base 7 (or another prime base) instead of base 10, I'd come up with something even better extremely quickly, but why bother? I'd rather spend my time learning applications of differential geometry to supersymmetry and 10 dimensional string theory, that proper maths
(while the number developed by my freind AlphaNumeric "i.e. 8151, has no significant math representation"
So, my 'special property' is that the A = 3^7 and B = 3^8, so my number is 3^7 + 3^8. Wow, who cares
It seems to me, you just add a 'special feature' every time I come up with an similar thing to you.
QUOTE
I also noticed that any number divisable by 3, its reverse is always divisable by 3, while this is not the case with 7!!!
Duh! For the reason I outlined before, it's the sum of a numbers digits which tell you if it's divisible by 3 or 9, and since addition is abelian, reversing the digits doesn't change the fact it's divisible by 3 or 9.Frankly, who cares? This is hardly deep number theory and I'm 100% positive someone with a bit of time and a knack for number bases could come up with plenty of such examples for any number, though I have little doubt you'd come up with a reason why yours is somehow 'superior'.
There are plenty of loops, cycles and tricks that numbers do like this. Google "Amicable numbers", "Happy Numbers", "Harshad Numbers", plenty of things like that. You've discovered nothing deep or magical, unless you can prove it for an n'th formula somehow, instead you've discovered one of probably infinitely many 'slight of hands' which the infinitely many integers have within them.
For instance, consider 'Kaprekar's Process'. Pick any 4 digit number which doesn't have all the same digits, order the digits in descending order, subtract the original number when arranged in asending order and repeat on the new number. You'll ALWAYS get 6174.
Consider 4828. Rearranging into descending order 8842, ascending order is 2488
8842 - 2488 = 6354
6543 - 3456 = 3087
8730 - 0378 = 8352
8532 - 2358 = 6174
Wow, amazing? No, not really. This and about 500 other similar tricks, possibly including yours can be found in "The Penguin Dictionary of Curious and Interesting Numbers" by David Wells.
By the way, this doesn't make you an amazing mathematician. Given enough time, anyone can find or derive such a things. It took me ONE attempt to get very close to yours in complexity. I imagine if I worked in base 7 (or another prime base) instead of base 10, I'd come up with something even better extremely quickly, but why bother? I'd rather spend my time learning applications of differential geometry to supersymmetry and 10 dimensional string theory, that proper maths
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