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radclyffe59
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20 May 2008, 1:26 pm

Does anyone here play any sorts of number games in their head... just at any time, anywhere. For example, whenever I see a number that has more than 1 digit in it (a number plate or a price or anything at all), I have to "add" up the digits in the number until I can bring them down to 1 digit again - like this -

Say the number I see is 87452
In my head I go 8+7+4+5+2=26
Then 2+6=8

Except I do it quicker.

I don't mean to make it look like I'm bragging about anything, I just wondered if anyone did that sort of thing, and whether they've noticed number patterns by doing it - thereby enabling them to work it out quicker.

ie you can ignore 9 and anything that adds up to 9 - so I can ignore the 7 and the 2, and the 5 and the 4, and know that the answer is 8 straight away.

I know it's completely useless, and probably quite obvious too, but it's just what I do...



TallyMan
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20 May 2008, 1:34 pm

I think you'll enjoy it here then :D
If you combine a love of numbers with a little OCD you end up playing the "count to a million" game in the "Off the wall" forum - you can even take delight in pointing out prime numbers and factorising too :D

Welcome to WP.



radclyffe59
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20 May 2008, 1:36 pm

Oo oo *goes off to have a look*



Alaspi
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20 May 2008, 1:48 pm

I love numbers too. I like to add the digits on car plates and then add that total to the next car's total and see how big a number I can get. :D


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Social_Fantom
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20 May 2008, 1:49 pm

I like to add the digits on digital clocks and other devices with numbers. I especially like it when they add up to multiples of 10. :D


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TallyMan
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20 May 2008, 2:02 pm

I like the trick of adding the digits together and if you end up with 3, 6 or 9 the number is divisible by 3. If the addition results in more than two digits then add those together as well until you end up with one digit. Also if the number is even then it automatically means the number is divisible by 6. Example.

729,382,710 = 39 = 12 = 3 therefore the number is divisible by 3 and 6.

Isn't there a similar trick to know if a number is divisible by 11? I've forgotten.



twoshots
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20 May 2008, 2:13 pm

radclyffe59 wrote:
Does anyone here play any sorts of number games in their head... just at any time, anywhere. For example, whenever I see a number that has more than 1 digit in it (a number plate or a price or anything at all), I have to "add" up the digits in the number until I can bring them down to 1 digit again - like this -

Say the number I see is 87452
In my head I go 8+7+4+5+2=26
Then 2+6=8

Except I do it quicker.

I don't mean to make it look like I'm bragging about anything, I just wondered if anyone did that sort of thing, and whether they've noticed number patterns by doing it - thereby enabling them to work it out quicker.

ie you can ignore 9 and anything that adds up to 9 - so I can ignore the 7 and the 2, and the 5 and the 4, and know that the answer is 8 straight away.

I know it's completely useless, and probably quite obvious too, but it's just what I do...

You're actually calculating 87452 modulo 9, which is what the remainder of 87452 is when divided by 9. This is because
87452 = 8(10^4) + 7(10^3) + 4(10^2) + 5(10) + 2
10 = 1 mod 9 (because 10 = 9 + 1 so 1 is the remainder of 10 when it's divided by 9), so any power of 10 is congruent to 1 modulo 9, so
87542 = 8 + 7 + 4 + 5 + 2 = 26 mod 9
Apply again to get 26 = 8 mod 9 so 87452 = 8 mod 9.
This is also why you can ignore any multiple of 9; 9 = 0 mod 9.

If you like doing stuff with numbers number theory is a fun topic. (and by the way, I don't think modular arithmetic is at all useless...)
tallyman wrote:
Isn't there a similar trick to know if a number is divisible by 11? I've forgotten.

It isn't quite as simple for 11. Take the case of 87452 = 8(10^4) + 7(10^3) + 4(10^2) + 5(10) + 2
10 = -1 mod 11, so substitute that in to get
8(10^4) + 7(10^3) + 4(10^2) + 5(10) + 2 = 8 - 7 + 4 - 5 + 2 = 2 mod 11
Hence 87452 is not divisible by 11 (for n to be divisible by 11, it would have to be the case that n = 0 mod 11)
Take instead the case of 2310 = 0 - 1 + 3 - 2 = 0 mod 11, so we can tell that 2310 is divisible by 11.
That is, starting from the rightmost digit, alternate adding and subtracting each digit, and if the end result is divisible by 11, then the whole number is.

Divisibility checks that work for other numbers work on similar principles. There's actually one that works for 7, 11, and 13 that involves alternating adding an subtracting 3 digits at a time rather than just 1.


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20 May 2008, 2:38 pm

I'm not good at doing arithmetic in my head. I can do it fairly well with a pencil and paper, though I usually prefer to use a calculator or computer for the arithmetic when I do algebra, calculus, trigonometry,or statistics. As for numbers themselves, I like e, π, i (the imaginary # that represents the square root of 1), and the square root of 0.5. I also like 2, since it has the same result if you add it to itself, multiply by itself, or raise it to its own power. And I like e^(sqrt(163)π) since it is a product of irrational numbers that has so many 9s after the decimal point (12 to be exact) that it was once mistaken for an integer.



Last edited by archdude on 21 May 2008, 10:25 am, edited 1 time in total.

radclyffe59
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20 May 2008, 3:07 pm

Thanks twoshots for explaining the 9 thing to me - I never understood why I could ignore the 9s always, but what you said made sense, and made me even more interested and obsessed!