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10^2=100
10-9=1
1*100=100
100-100=0
1^1=1
0+1=1
Am I missing something? Why does your first trick work?
The way I've always done multiplication in my head is to break it down into a distributive fashion.
So for 44^2 I would break it down to 40^2+(4+4)40+4*4=N
1600+8(40)+16=N
1600+320+16=N
1936=N
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On February 21 2009 14:18 Azrael1111 wrote:
10^2=100
10-9=1
1*100=100
100-100=0
1^1=1
0+1=1
Am I missing something? Why does your first trick work?
It's simple... he didn't explain it at all -_-;;
50 ^ 2 = 2500
44 ^ 2 = ?
44 = 50 - 6
44 ^ 2 = (50 - 6 )^2 = 50^2 - 6 * 50 * 2 + 6^2 = 2500 - 6*100 + 36
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for 100 you have to use a slightly different formula
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here's a random trick to multiply 2 2digit numbers by each other
no idea why you would ever need this but anyway
say 37 * 84
hard to do your head, unless
8*3=24
7*4=28
3*4+7*8=12+56=68
that takes just a few seconds doesn't it? (well it should anyway)
then we have
24(2)8 with a 68 in between, add em up
gives us 3108!
actually this is the worst example i could've thought of since there's 2 1s to be carried. still, works better than my previous approaches (which would ordinarily be breaking it down to something like 37*80+37*4 = 2400+560 + 120+28 = 2960+148=3108.. which is harder i think)
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United States24671 Posts
I find the long multiplication in the video is usually quicker than the method with the lines (especially if the digits are large numbers like 7-9)
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An even more random trick: squaring any number ending in 5
take all the numbers besides the last one (5) and multiply it by the number above it, then add 25 to the end.
ex: 125^2
12 * 13 = 156
125^2 = 15625
Don't know why this particular one came to mind ~_~.
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this is probably the most useless thing you could ever know, but anyway: how to figure out an integer cube root of a # b/w 1 and 1,000,000 (assuming of course, that it has one)
first you'll have to be familiar with cubes from 1 to 10 if you're not already
1=1
2=8
3=27
4=64
5=125
6=216
7=343
8=512
9=729
10=1000
say someone gives you a number, 474,552. now, you know that the number that was cubed was a 2 digit number.
the first digit is a 7 because 343<474<512
the last digit is an 8 because the 474,552 ends on a 2, and so does 8^3.
hence, cube root of 474,552 is 78!
note of course that this isn't very useful for attempting to take cube roots of numbers that don't have integer cube roots. i guess you could approximate it to the nearest 10 but that's about it.
edit: to be fair most of these tricks are rendered useless by having a calculator. so get one.
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Kentor
United States5784 Posts
can you use mutliply instead of times for like transitive verb thanks
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On February 21 2009 14:27 Hypnosis wrote:
nAX^n-1 use it
uh you can't really use single variable calculus without ... a variable ;o
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this is not a shortcut but a fun fact: every prime number (but 2) belongs at least 1 pythagorean triad as a summand..
May be useful if someday you need to find a rectangle triangle with a given integer side as a cathetus
Example find a pythagorean triad wich has 11 has a summand
since a^2+b^2=c^2 => a^2=c^2-b^2=(c+b)*(c-b) we "demand" c-b=1 a^2=c+b and solve
setting a=11 we have
a^2=121, c=61, b=60 thus
11^2+60^2=61^2
a=5 gives b=12, c=13
a=7 gives b=24, c=25
a=13 gives b=84 c,= 85 and so on.
corollary:
every whole number (but 2 or 1) can be put as a cathetus side in a rectangled triangle with both the remaining sides whole numbers.
NOTE: NOT every pythagorean triad is described by this method.8^2+15^2=17^2 is not from this "family"
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ooh, fun tricks... i used to carry a ti-84 in my pocket so i could tell my math teacher that i did, so i could use it on a test....
but now that i don, these could actaully be helpful 
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Google vedic mathematics and you'll find a lot of tricks.
Another very useful tool is a mnemonic system for converting numbers into words if you want to do larger stuff in your head, remembering what numbers you're going to sum in the end is often harder than the partial multiplications.
Most of these tricks can be derived from the rules of parenthesis multiplication if you split your numbers. For example, ab (a on first spot and b on second spot, not a times b) would become (10*a+b). Alternatively you could split it into ([a+1]*10-[10-b]).
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the only trick i use is x11, which basically only saves me a step.
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I don't use any tricks, just brute force long multiplication all the way!
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A trick:
To test whether or not a number is divisible by 3 you add up the digit of the number and if that number is divisible by 3 then the original number is.
For example:
1893
1+8+9+3 = 21
21 is divisible by 3 therefore 1893 is
suppose it is a huge number:
192345
1+9+2+3+4+5 = 24
2+4 = 6
6 is divisible by 3. Therefore 24 is divisible by 3. Therefore 192345 is divisible by 3.
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On February 21 2009 19:41 [r]h_probe wrote:
A trick:
To test whether or not a number is divisible by 3 you add up the digit of the number and if that number is divisible by 3 then the original number is.
For example:
1893
1+8+9+3 = 21
21 is divisible by 3 therefore 1893 is
suppose it is a huge number:
192345
1+9+2+3+4+5 = 24
2+4 = 6
6 is divisible by 3. Therefore 24 is divisible by 3. Therefore 192345 is divisible by 3.
A trick?
Not to be rude, but I learned that in 5th grade and so did my classmates. The same is valid for 9 btw.
There is a more general rule behind this one, which we didn´t learn back then 
The k-digit-sum:
If the k-digit-sum of a number is divisible by any factor of (10^k - 1), the number is divisible by this factor.
Your example is included, it is the case k=1 (normal digit sum, 10^1-1 = 9 with the factors 9 and 3)
Another example, k=2:
10^2-1 = 99
factors of 99: 3,9,11,33,99
some number: 72072
its 2-digit-sum: 07+20+72 = 99
of course the 2-digit-sum 99 is divisible by all factors of 99, so the number 72072 is divisible by 3,9,11,33 and 99
72069 with its 2-digit-sum 96 is only divisible by 3, and NOT divisible by 9,11,33,99
One could go on with k=3,4 etc
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not sure how useful the trick in that video is, but it's pretty cool nonetheless.
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On February 21 2009 14:36 nevake wrote:
An even more random trick: squaring any number ending in 5
take all the numbers besides the last one (5) and multiply it by the number above it, then add 25 to the end.
ex: 125^2
12 * 13 = 156
125^2 = 15625
Don't know why this particular one came to mind ~_~.
I like to use it often 
It's actually very simple to explain. Every number with 5 in the in the end can be presented as 10n+5
So we go:
(10n+5)^2 = 10n*10n+2*5*10n+25 = 10n*(10n+10)+25 = 100*n*(n+1)+25
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Northern Ireland22208 Posts
These are pretty neat, that verdic maths site is like a treasure trove of tricks
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Not bad but is it a shortcut on bigger numbers?
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EPT
