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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

i found this in a book:

11826²=139854276

30384²=923187456

which are all the digits from 1 to 9

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

I found this in a book...

1 + 6 + 7 + 17 + 18 + 23 = 2 + 3 + 11 + 13 + 21 + 22;

1^2 + 6^2 + 7^2 + 17^2 + 18^2 + 23^2 = 2^2 + 3^2 + 11^2 + 13^2 + 21^2 + 22^2;

1^3 + 6^3 + 7^3 + 17^3 + 18^3 + 23^3 = 2^3 + 3^3 + 11^3 + 13^3 + 21^3 + 22^3;

1^4 + 6^4 + 7^4 + 17^4 + 18^4 + 23^4 = 2^4 + 3^4 + 11^4 + 13^4 + 21^4 + 22^4;

1^5 + 6^5 + 7^5 + 17^5 + 18^5 + 23^5 = 2^5 + 3^5 + 11^5 + 13^5 + 21^5 + 22^5.

2 + 2 = 5, for large values of 2.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

What an amazing coincidence! It must be coincidence because it stops working when you use any power higher than 5.

Why did the vector cross the road?

It wanted to be normal.

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

mathsyperson wrote:

What an amazing coincidence! It must be coincidence because it stops working when you use any power higher than 5.

It's a consequence of the following even bigger coincidence!

a^n + (a + 4b + c)^n + (a + b + 2c)^n + (a + 9b + 4c)^n + (a + 6b + 5c)^n + (a + 10b + 6c)^n = (a + b)^n + (a + c)^n + (a + 6b + 2c)^n + (a + 4b + 4c)^n + (a + 10b + 5c)^n + (a + 9b + 6c)^n,

where a, b, c are any positive integers and n can be 1, 2, 3, 4, or 5.

See also http://nrich.maths.org/askedNRICH/edited/412.html and http://www.primepuzzles.net/puzzles/puzz_065.htm

*Last edited by NIH (2005-08-13 11:39:19)*

2 + 2 = 5, for large values of 2.

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Here's an interesting one:

111111111² = 12345678987654321.

Basically, when any number composed entirely of n ones is squared, the result will be 12...n...21. Well, at least when n < 10, once we get to 10 the middle gets screwy.

*Last edited by Zhylliolom (2005-09-05 12:33:34)*

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