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104 x 104

+4 = 108

4x4 = 16

104 x 104 = 10816

109 x 109

+9 = 118

9x9 = 81

109 x 109 = 11881

A little trick I use.

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I am not sure you have explained it well here, but looking at your first example we have

$104^{2} = (100+4)(100+4) = 1000 + 2 (400) + 4^{2} = 10816$.

This method generalises in many ways.

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Sorry, I didn't explain it well

104 x 104

(104 + 4) = 108

(4x4) = 16

Connect both results: 10816

or

(107 + 7) = 114

(7x7) = 49

Connect both results

11449

I also have a trick to substitute long division.

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221013 / 9 = 24557

(dividing by 9 trick)

Copy 1st digit of the initial number.

2

Add the second digit of the original number, to the first, and make a new digit next to it, as the result.

2 + 2 = 4

So now we have

2 4

4 + 1 = 5

So now we have

2 4 5

Carry on in this manner and you'll get:

245569

However you don't accept the 9, you place it in this manner:

221013 / 9

24556 9

The '9' goes into the original 9, once. So you carry 1 back.

Making it 24557

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3 2 1 4 2 / 9

3 5 6(1) 12

357 12

9 goes into 12 once, with a remainder of 3. So carry 1 back.

3571 r3

() = Carrying back.

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What does carry back one mean?

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Handy exam trick: When you know the answer, but not the correct derivation, derive blindly forward from givens and backward from the answer, and join the chains once the equations start looking similar. Sometimes the graders don't notice the seam.

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Handy exam trick: When you know the answer, but not the correct derivation, derive blindly forward from givens and backward from the answer, and join the chains once the equations start looking similar. Sometimes the graders don't notice the seam.

I see you never lose your wit. =P

Did you like my tricks though? Was given to me by my friends maths tutor.

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What does carry back one mean?

I forget that this is a science forum... Carry 1 back is:

Say if I was adding up these numbers in the same fassion as my division trick.

4956

4

4+9 = 13 (it's a two digit number, carry back 1, keep the 3)

53 or 4(1)3 would be the process of carrying back.

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Very interesting, but this method only work for numbers of the form $10^n + x*10^{n-2}$ (ie. 108, 1017, 1048796...the first digit always has to be one, the second always has to be zero and the third mustn't be a zero). To extend this method to work for (almost) all numbers, we've got to modify the procedure a bit.

Given a number $A = x*10^n + y*10^k$:

(Step zero: check if $n-k = 1$; if yes, this procedure becomes somewhat lengthy and thus ineffective, as explained below. In other words, if the second digit isn't a zero, don't bother.)

Step one:

Add $A + y$.

Step two:

Multiply the number you got in step one by $x$, and write it down.

Step three:

Add (as in concatenate) $n-k-2$ zero's to the end of the number you wrote down in step two. If $n-k-2 = -1$, the procedure requires modifications of an increasingly tedious nature, and quite frankly I think simple multiplication by hand would be faster at this point.

Step four:

Add (as in concatenate) $y^2$ to the end of the number you wrote down in step three.

Example:

$8005^2 = 64080025$

Step one:

$8005 + 5 = 8010$

Step two:

$8*8010 = 64080$

Step three:

Instead of computing the equivalent of $n-k-2$ in the example above, I just look at how many zeroes there are between the digits 8 and 5, subtract one, and that's the number of zeroes I have to add (if we were dealing with the number 8050, I wouldn't add any zeroes, if with 800 005 I would add three zeroes and so on). So we get $640800$.

Step four:

We add (as in concatenate) $5^2 = 25$ to 640800, which gives us $64080025 = 8005^2$.

I didn't spend that much time checking whether this procedure really works for all numbers (except the ones described in step zero), so I apologize if I overlooked something. Nevertheless, I think this is a very interesting method with practical applications, so thank you Klaplunk

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By manipulating well-known algebraic formulas, you can devise your own handy arithmetic rules for mental calculation. For example, suppose you want to calculate 91 × 89 mentally.

91 × 89 = (90 + 1)(90 − 1) = 90
2
− 1
2
= 8099

Similarly, to do 122 × 118 mentally:

122 × 118 = (120 + 2)(120 − 2) = 14400 − 4 = 14396

You can also do this for division. For example, what is 22491 ÷ 153? Using the trick above, you should be able to get the answer 147 without using a calculator or long division.

Now suppose you want to find the prime factors of 359999. If you start by observing that

359999 = 360000 − 1 = 600
2
− 1
2
= (600 + 1)(600 − 1) = 601 x 599

it’s then a simple matter to verify that 601 and 599 are both primes. (This is a problem I once set for some people on another forum.)

Edited by shyvera
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you missed out a zero on the first one, it should read 8099

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I’ve corrected it. By the way, ajb also left out a zero when expanding brackets: (100 + 4)(100 + 4) should be 10000 + 2(400) + 42.

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I’ve corrected it. By the way, ajb also left out a zero when expanding brackets: (100 + 4)(100 + 4) should be 10000 + 2(400) + 42.

Yes, I did spot that after.

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• 2 weeks later...

I read your explanation of that method with no prior experience with 'higher than GCSE' alegebra knowledge. By looking at it and comparing it to my own words I managed to understand the whole thing, so basically I understand that alegebra, which is weird.

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• 3 months later...

I prefer to use Euler's Matrix because you can use it to multiply any two numbers and it's easy to check for mistakes.

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I prefer to use Euler's Matrix because you can use it to multiply any two numbers and it's easy to check for mistakes.

Could you run that method past me - I only know Euler Matrices as the result of a three dimensional rotation. I hope there is some other Euler Matrix - cos long multiplication is easier for two digit numbers than messing around with Euler Angles/matrices

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