Post your favorite calculation shortcuts. Any subject or difficulty level.
>inb4 just use a cuckulator XDDD
Calculation Shortcuts
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if an integers added numbers add up to something divisible by 3 the number itself is divisible by 3
ex
[math] 222 = 2 + 2 + 2 = 6 [/math]
222 is evenly divisible by 3
Brilliant!
Every composite number other than 6 has non trivial dubs in some base.
if a^b = c
then b = log c / log a
used this all the time in high school to be quick as fuck
Also works for 9.
Any number divisible by nine has digits that sum to a multiple of nine.
To multiply two numbers between 10 and 20 (let's note them in decimal notation ab and cd)
First add ab and d
Then multiply by ten
Finally add to b*d the result
Example :
16*18 = ?
24
240
288
It is easy to prove but makes life easier on many cases especially if you work with hexadecimal notations
The same trick exists with 11
Can 94578132 be divided by 11 ?
2-3+1-8+7-5+4 - 9 = -11
The answer is yes
>same trick
same? you're subtracting every other digit.
neat trick, but not the same.
And it doesn't seem to work for me.
Can 22 be divided by 11?
2-2 = 0
Can 121 be divided by 11?
1-2+1 = 0
3,861?
1-6+8-3 = 0
What are you trying to pull here?
Same with 9 my 9ger
>11*0=/=0
(Different user but still)
*0=/=0
I'm more confused than ever.
user () shows adding and subtracting backwards digits in a number adds up to -11 in his example: "therefore big number is divisible by 11".
When I try smaller numbers, trick seems to work except they all add up to zero, not -11.
I have NO idea what you're trying to say with "11 * 0".
You can compute almost any squared root using this little iterative method
[math]x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \in \mathbb{N}[/math]
where [math]x_{0}[/math] is your initial guess, I suggest something like [math]x_{0} = \frac{x_{n}}{2}[/math] in order to achieve a fast convergence in most cases.
Is there any formal approach to this? I want to get my hands on books that teach this.
I've gone through most advanced maths you can get in Engineering but thanks to calculators I'm a disgusting arithmetic idiot. Please help, even some basic stuff like bigger multiplications is hard for me.
newton raphson
Your employer doesn't want you doing headmath.
stop copying my comment
[math]\mathbb{R}^{eeeeeeeeeeeeeeeeeeeeeeeeeeeeeee}[/math]
I didn't even read the thread.
bump
Currently reading through this at work, figured it’s pretty relevant. Enjoy.
1/x = 1/y + 1/z => x = yz/(y+z)