How do you do it??
any exercises and techniques used would be greatly appreciated...
Anyone think their brain is on a similar level to a calculator in terms of speed???

Fade Into Black
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 29082004 14:17

Revision help in partnership with Birmingham City University

NDGAARONDI
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 29082004 14:19
Mind mapping might help. Heard of Tony Buzan?

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 29082004 14:22
Nope
Is he a mind mapper or just a great mental maths guy?? 
magiccarpet
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 29082004 14:24
just keep practising...add firstly 2 figure numbers in your head, and you find you get faster. then 3 figures...etc. you may find you can do sums using short cuts, for example by adding 10+10+10 then 3 instead of straight adding 37 to something. breaking division sums into fractions eg instead of doin 345/23, treat it as a top heavy fraction. honestly, the more math you do the faster you get. i was cr*p at mental arith until starting alevel math but the more i did the better i got. also... LEARN THOSE TIMES TABLES!!!

NDGAARONDI
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 29082004 14:25
I believe he's one of these experts in the mental field. I'm not sure if his works can be helpful for maths, but then it might be for one and not another. If I find that other people found it useful for law, I might be tempted to get some of his books to improve my memory. Not sure if this will help for maths though.

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 29082004 14:32
What would be a time faster than average whereby mentally someone could calculate 2 figure times multiplied???
How did you mean by breakin up division??
How much should I practise??Start with the basics again???
Tony Buzan seems to be more beneficial to general subjects, but I suppose anybody can benefit with a better memory regardless of the subject but it is probably more useful for a law student like you said.....
If you find more plz Pm me or keep me posted.... 
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 29082004 14:46
do some practice
eg.
you have two 2digit numbers, "ab" and "cd". they are (10a + b) and (10c + d)
say 84 and 36. 84 = 80 + 3; 36 = 30 + 6
so their product will be: (80+3)(30+6) = (80*30)+(30*3)+(80*6)+(3*6)
this is how I do it and after lots of practice you will be able to do it very fast.
Also you can do:
98*54 = (1002)(50+4) = (100*50)+(100*4)(2*50)(2*4)
that's an example.. but practice is the best way. practice lots and you'll be very good at it. but why do you want to get better? it's not a brilliant skill to have, if you want to improve at maths then arithmetic isn't the best thing to focus on.. try algebra or geometry. 
Fade Into Black
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 29082004 14:55
I just thought that it would be helpful because I consider myslef to be quite quick.....but want to improve
What parts of Algebra and Geometry???
i'm approachin aslevel in further maths if that helps your judgement......
others please contribute suggestions.... 
Fade Into Black
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 29082004 14:58
I think I have quick speeds at calculating....
what is a fast speed to calculate 2 figure numbers multiplied by themselves....
then i will kno how i compare... 
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 29082004 15:05
try using the
(10a + b)^2 = 100a^2 + b^2 + 20ab
identity to calculate squares really fast, it's fun for baffling friends.
"Lex, 73 squared" "uh 5329 if I'm not mistaken!" "=o" 
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 29082004 15:05
When I was about 8 or 9, i remember i used to write about 1015 random 2 digit sums or multiplications on a piece of paper, and then see how fast i could do them all mentally. It might not work for you, but I improved very quickly. Im pretty poor at mental maths now tho.

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 29082004 15:08
Also, look on here and find some quick tricks for doing arithmetic

Fade Into Black
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 29082004 15:11
ok thanx how often should I practice these sums...
how quick should i be??
thats a start but i suppose it may vary with age as to whether the technique stil works for those older as effeciently... 
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 29082004 15:33
(Original post by Fade Into Black)
I just thought that it would be helpful because I consider myslef to be quite quick.....but want to improve
What parts of Algebra and Geometry???
i'm approachin aslevel in further maths if that helps your judgement......
others please contribute suggestions....
C2C4, FP13 = calculator
so if you want to prepare, don't prepare that kind of stuff.
Make sure you know algebra well/
eg. laws of indices:
(a^b)(a^c) = a^(b+c)
(a^b)/(a^c) = a^(bc)
a^0 = 1
a^(1/b) = b root a, so:
a^(b/c) = c root (a^b) = (c root a)^b
erm.. also do some logs for practice,
log(a) b basically is equal to the number of times a needs to be multiplied by itself to make b.
eg.
if: log(4) 5 = x
then 4^x = 5
and if log(m) a = p,
then m^p = a
erm.. geometry is easy at a level, just learn all the rules..
differentiation and integration.. well you'll be taught that properly
proof.. this is easy so you might as well read this:
you can prove things in general 3 ways:
first is the simplest.. eg. prove that (a+b)^3 = a^3 + 3ba^2 + 3ab^2 + b^3
start with the left hand side, and finish with the right through valid steps.
LHS = (a+b)^3
= (a+b)(a+b)(a+b)
= (a^2 + 2ab + b^2)(a+b)
= a^3 + 3ba^2 + 3ab^2 + b^3
= RHS
second is by contradiction. basically you get the claim, say that it is wrong and show that this leads to an error, therefore the claim MUST be right. eg.
prove that sqrt2 is irrational (irrational means cannot be written as a/b, where a and b are integers)
say that is isn't irrational. therefore:
sqrt2 = a/b , where a and b are relatively prime (meaning they cannot be cancalled)
2 = a^2/b^2 (squaring both sides)
2b^2 = a^2 (multiplying both sides by b^2)
therefore a^2 is even, as b^2 is a positive integer multiplied by 2.
this means that a is even! (even number squared is also even)
So if a is even, then we can write it as a = 2m, where m is an integer (as 2m must also be even)
going back to the start,
sqrt2 = a/b = 2m/b
2 = 4m^2/b^2 (squaring both sides again)
2b^2 = 4m^2 (multiplying both sides by b^2)
b^2 = 2m^2 (dividing both sides by 2)
and from this we can tell that b^2 is also even, as 2m^2 is even.
this tells us that b is even too.
but here's the contradiction... if a and b are both even, then the fraction is reducible  they are not relatively prime! we can then say that
sqrt2 = a/b = 2p/2q = p/q, and we have reduced the fraction. applying the proof again to p/q, we find that the fraction is ALWAYS reducible, which cannot happen. therefore the original statement that sqrt2 is rational is wrong, and so sqrt 2 is irrational.
last type of proof: induction.
this is perhaps the hardest. say, for example, you wanted to prove that (4^n)1 is always a multiple of 3, for all positive integers n. (it is!)
we first prove that is is true when n is 1. we then ASSUME it to be true for all values of n, and use this fact to prove that it is true for all values (n+1). then simple repetition can show that if it is true for n, it is true for n+1, if it's true for n+1 it's true for n+2, and onto infinity, starting at 1.
is (4^n)1 is a multiple of 3, then 4^n  1 = 3a, for some integer a.
4^n  1, when n = 1:
4^1  1 = 3 = 3a. a = 1, 1 being an integer. therefore 4^1  1 is a multiple of 3.
now assume it is true for all n:
4^n  1 = 3a. (this means that 4^n = 3a + 1) (remember this, we will use it later)
now to prove it is true for all (n+1) using this fact:
4^(n+1)  1 (think about this... the first part means the number 4 times itself n times, then once more) we can write 4^(n+1) = (4^n)*4.
4^(n+1)  1 = 4*4^n  1 (like we just said)
substituting (4^n = 3a + 1 into this,)
4^(n+1)  1 = 4(3a + 1)  1
= 12a + 4  1
= 12a + 3
= 3(4a + 1)
as a is an integer, (4a+1) is an integer, and 3(4a+1) is always a multiple of 3.
This is the inductive step, and so by putting n to 1, 2, 3,... n we can prove it is true for 2, 3, 4... n+1, and therefore ALL positive integers.
I'm having a break. hope this helps. 
Fade Into Black
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 29082004 15:51
Damnnnnnnnnn.......
THANXXXXX
BIG TIME
that must have taken you ages
a big thanx
i think you deserve a break...
my onli problem is that i havent learnt half of the stuff and its slightly confusin on a Pc but a big thanx
If you could teach me a few more of each or tell me where the questions are that i cud practice..
should i recap on basic algebra?? 
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 16
 29082004 15:57
yes, make sure you know what you can and can't do
eg.
you can't go from:
a + b = c
to
a^2 + b^2 = c^2
anything you do to both sides you must square all the sides:
a + b = c
(a + b) = (c)
(a + b)^2 = (c)^2
a^2 + 2ab + b^2 = c^2
see you were missing a 2ab on the LHS if you'd have squared the terms.
try this:
2^(16^x) = 16^(2^x)
find x. if you get stuck, think in context what powers mean. a^b means a*a*a*a*...*a, where there are b amount of as. Or a^b = a times itself b times.
this is a tricky one for preAlevel so give it a lot of thought, and don't do trial and error.. it's messy and doesn't work most of the time. 
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 29082004 16:19
(16^x)^2 = (2^x)^16
256^x = 65536x
divide both sides by 256
^x = 256x
Not too sure tbh neva encountered a question lik this before.. 
figgetyfig
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 29082004 16:34
(Original post by Fade Into Black)
i'm approachin aslevel in further maths if that helps your judgement......
Try things like simplifying expressions, rearranging equations, multiplying out and factorising. 
Fade Into Black
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 29082004 16:39
So the key is improvin the basics so that I can do it with gd pace
thanx for the advice...
I would offer my advice but u dnt seem to need it..
By the way do u kno the answer to the question above as mik1a isnt online at the mo... 
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 29082004 16:59
(Original post by Fade Into Black)
(16^x)^2 = (2^x)^16
256^x = 65536x
divide both sides by 256
^x = 256x
Not too sure tbh neva encountered a question lik this before..
the first step is not allowed.
a^(b^c) does not equal (a^c)^b
eg. 3^(2^4) = 3^(2*2*2*2) = 3^16 = (3*3*3*3)^4 = 81^4
but! (3^4)^2 = 81^2
and of course 81^2 =/= 81^4.. the RHS is a factor of 81^2 out.
editi also mention that ^x is meaningless. you must always have a base... as it is, what you've said is nothing raides to the power of x... or "undefined multiplied by itself x times". every term has a base and a power:
5 ; the base is 5, the power is 1 (5 = 5^1)
pi ; the base is pi, the power is 1 (pi^1 = pi)
4^2 ; the base is 4, power is 2
1 ; the base is 1, power 1. Or you could say the base is x, the power is 0, where x is any real number (as x^0 = 1)

2^(16^x) = 16^(2^x)
16 = 2^4
so if you change all the "16"s in the equation to "2^4"s, you'll make some progress... havign said this, it's way too hard for GCSE level! sorry lol
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