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How do you improve mental maths skills??? watch

1. 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???
2. Mind mapping might help. Heard of Tony Buzan?
3. Nope
Is he a mind mapper or just a great mental maths guy??
4. 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!!!
5. 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.
6. 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??

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....
7. do some practice
eg.

you have two 2-digit 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 = (100-2)(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.
8. 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 as-level in further maths if that helps your judgement......
9. 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...
10. 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"
11. When I was about 8 or 9, i remember i used to write about 10-15 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.
12. Also, look on here and find some quick tricks for doing arithmetic
13. 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...
14. (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 as-level in further maths if that helps your judgement......
Okay... C1 exam = no calculator
C2-C4, FP1-3 = 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^(b-c)
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.
15. 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??
16. 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 pre-A-level 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.
17. (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..
18. (Original post by Fade Into Black)
i'm approachin as-level in further maths if that helps your judgement......
For AS further maths, I would recommend practising your algebra, as you will find the whole syllabus easier if your algebra is quick and neat. (My algebra got gradually better as I went along, but in year 12 I did terribly in further maths because of slow algebra!)

Try things like simplifying expressions, rearranging equations, multiplying out and factorising.
19. So the key is improvin the basics so that I can do it with gd pace
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...
20. (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..
ahh, nice try
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.

edit-i 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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