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Old 04-18-2008, 01:34 PM
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Default Mental calculations tips..

Hello,

1.*1.

This has nothing to do with extraordinary technique, but you gotta think about it (I'm confident that some of you know about it)..

(10+a)(10+b)=10(10+a)+b(10+a)=10(10+a)+10b+ab=10(10+a+b)+ab

So multiplying two such numbers is equivalent to :

- sum one of the numbers with the unit digit of the other one
- multiply it by 10
- add the product of the two unit digits.

19 \times 17= ?

- 17+9=26

- 26 \times 10=260

- 7 \times 9=63

- 19 \times 17=260+63=323


I hope this would help at least one person in the future

Enjoy calculations !
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\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{      and       } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}

Last edited by Moo; 08-21-2008 at 09:02 AM. Reason: calculation, not calculus...
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Old 04-18-2008, 01:58 PM
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Hello

I didn't know about this hint. Usually, I simply develop (10+a)(10+b)=100+10(a+b)+ab.

In the case of 19 \times 17, where 9 and 7 are close to 10, I'd rather do 19\times17=(20-1)(20-3)=400-20(1+3)+3 which avoids manipulating too large numbers. (yes, I consider 63 as large )
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Old 04-18-2008, 02:07 PM
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Well, this goes too for 12x15 for example, which involves smaller numbers
I only gave an example with random numbers... Sometimes, there are easier ways of calculating
When I'm talking about mental calculations, I consider it as being without any paper & pen... Unfortunately, (20-1)(20-3) for some people needs some writing :-)

Another thing :

The square of a number whose unit digit is 5.

If you're looking for the square of a number N=X5 (with X representating all the digits of the number, meaning that actually, N=10X+5), don't worry, this can be shortened ^^

First of all, calculate X(X+1)=Y

N^2 will simply be Y25 \ (100Y+25)

For example :

35^2 = ?

X=3
X(X+1)=3 \times 4={\color{red}12}

Hence 35^2={\color{red}12}25

Another example :

115^2= ?

X=11
X(X+1)=11 \times 12={\color{red}132}

Hence 115^2={\color{red}132}25
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\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{      and       } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}

Last edited by Moo; 04-18-2008 at 02:52 PM. Reason: calculation, not calculus...
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Old 04-18-2008, 02:18 PM
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Good hint, but it hardly deals with Calculus concepts... more with arithmetic and algebra!
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Old 04-18-2008, 02:25 PM
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I'm sorry, I don't know that much about the different categories :s
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shinhidora production

\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{      and       } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}
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Old 04-18-2008, 02:37 PM
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Using binomial to avoid long calculations...

If you're asked to multiply two numbers, you can rapidly go through this checking.
See if the numbers are symmetric with respect to a number N.
If so, the numbers a and b can be written as : a=N-x and b=N+x
The product ab will be :

ab=(N-x)(N+x)

By using a binomial formula, you get :

ab=N^2-x^2

Which can sometimes be more sympathical to calculate..


For example :

58 \times 42 = ?

Notice that these numbers are symmetric with respect to 50.

58=50+8
42=50-8

58 \times 42 =(50+8)(50-8)=50^2-8^2=2500-64=2436




I know that these methods look elementary, but it sometimes helps when you are in a hurry and not allowed to use any calculators
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shinhidora production

\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{      and       } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}

Last edited by Moo; 04-18-2008 at 02:52 PM. Reason: claculation not calculus...
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Old 04-18-2008, 02:51 PM
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Moo, I believe the word you are looking for is calculations, not calculus.

Other than that, very nice, good to see someone besides Colby and TPH post some interesting tutorials on things.
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Old 04-18-2008, 02:54 PM
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Quote:
Originally Posted by janvdl View Post
Moo, I believe the word you are looking for is calculations, not calculus.

Other than that, very nice, good to see someone besides Colby and TPH post some interesting tutorials on things.
Thanks, I've corrected the mistakes... And I've learnt one more thing !
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shinhidora production

\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{      and       } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}
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Old 04-20-2008, 05:49 AM
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Check for the unit digits (and cubic roots of numbers)

In general cases, when you gotta check rapidly if your calculation seems correct, look at the digit number.

Let X=x*10+a and Y=y*10+b

XY=(ay+x+bx+y)*10+ab

Hence the unit digit of XY will be the unit digit of ab, which is easier to get.

For example :

46 \times 73

Here, a=6 and b=3. So ab=1{\color{blue} 8}
Hence the unit digit of 46 \times 73 is {\color{blue} 8}
If you find a result whose digit number is not 8, you can directly calculate again.




Here is only a beginning for cubic roots... And shall be known

1^3={\color{red}1}
2^3={\color{red}8}
3^3=2{\color{red}7}
4^3=6{\color{red}4}
5^3=12{\color{red}5}
6^3=21{\color{red}6}
7^3=34{\color{red}3}
8^3=51{\color{red}2}
9^3=72{\color{red}9}
10^3=100{\color{red}0}

Notice that each number appears once in the unit digit.

This means that if, for example, you're looking for the cubic root of a number whose unit digit is 6, its cubic root necessarily ends with a 6.
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shinhidora production

\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{      and       } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}
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Old 04-20-2008, 08:14 AM
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Quote:
Originally Posted by Moo View Post
Hello,

1.*1.

This has nothing to do with extraordinary technique, but you gotta think about it (I'm confident that some of you know about it)..

(10+a)(10+b)=10(10+a)+b(10+a)=10(10+a)+10b+ab=10(10+a+b)+ab

So multiplying two such numbers is equivalent to :

- sum one of the numbers with the unit digit of the other one
- multiply it by 10
- add the product of the two unit digits.

19 \times 17= ?

- 17+9=26

- 26 \times 10=260

- 7 \times 9=63

- 19 \times 17=260+63=323


I hope this would help at least one person in the future

Enjoy calculations !
Hi Moo,

I was jus wondering if you do this....say I solve an equation and get x=\frac{288}{128}...my mind works like this x=\frac{200+88}{100+28}=\frac{4(50+22)}{4(25+7)}=\frac{77}{37} that is how my mind simplifies things...its much quicker in my mind though...is that how you process things?
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