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What arithmetic should I memorize?

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I find myself constantly using long division and multiplication for simple things like 13 x 11.

What are some good arithmetic calcs to memorize? I created a 20x20 multiplication table but this seems like a bit much right?

Perfect squares up to 100? \(\sqrt{3}\)? \(\sqrt{5}\)?

I am not really talking about formulas to memorize since you should definitely memorize things like nCr, nPr, sum of all #'s in an evenly spaced set, etc.

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Fantastic question :!:

This file should give you an idea where you are lacking.
I think you should definitely know the squares from 0-10 and preferrably from 10 to 20 as well.
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As far as common squares are concerned, I only remember the ones with 0 or 5 in the units digit. For the latter category, I use the following process:

For example: 65*65

1) Always write down 25 as this is always the last two digits of the result:
...25

2) Multiply (non-units digits) times (non-units digits + 1)
6 * (6+1) = 42

2c) Combine:
4225

This way I always have the important squares handy... very useful for estimations!

Any other math shortcuts? Anyone :)

Steve

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I'm new here, and try to go through all the topics - this site is a treasure! :)

Maybe not the right place to share my experience with "multiplication for simple things like 13 x 11", but anyway...if you need to multiply any two-digits number by 11, just sum those digits and put the result in between. For example,
13x11 -> 1+3=4 -> 143 is the result.
Or, 36x11 -> 3+6=9 -> the result is 36x11=396.

It really saves time.

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Thanks for sharing the Doc bb...@Shelen & @jeckll...Thanks for the tip but I would like to add some more.

What if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below

For 3 digits
133x11 --> 1 1+3 3+3 3=1463
And for 4 digits
1243x11 --> 1 1+2 2+4 4+3 3=13673

And for 5 digits
15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983
and so on.
you can plug other numbers in and check it out.

Hope it helps! :)
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Shelen wrote:
I'm new here, and try to go through all the topics - this site is a treasure! :)

Maybe not the right place to share my experience with "multiplication for simple things like 13 x 11", but anyway...if you need to multiply any two-digits number by 11, just sum those digits and put the result in between. For example,
13x11 -> 1+3=4 -> 143 is the result.
Or, 36x11 -> 3+6=9 -> the result is 36x11=396.

It really saves time.

And if it equals more than 10, add a 1 to the first digit

EG

68*11 -> 6+8=14 -> 6 14 8 -> 748

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I would like to add one more trick or shortcut to find the square of a number... :roll:

Suppose you know square of 13 but you want to know whats the square of 14, then follow this..

13 square = 169

For 14 square, 13 square + 13 - { present number } + 14 - { the next number }

i.e 169 + 13 + 14 = 196 { which is 14 square }

Note - you may calculate 14 square by knowing 15 square also! :lol:

14 square = 15 square - 15 {present number} - 14 {previous number}

14 square = 225 - 15 -14

14 square = 196 :!:

Additionally if you club this technique with "finding squares of numbers ending with 5" then finding the square of big numbers id just a matter of few seconds.

This technique is very helpful when finding squares of big numbers, it will be mere addition or subtraction.

Consider Kudos if you like this post. :o :-D

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Guys! I just found another way of checking whether a number is divisible by 8 or not ( the rule is same but another approach or route ). It's for a number with more than two digits.

Let's take 1936
1) First of all check whether last two digits of the number are divisible by 4 or not.
For 1936, we do this way 36/4=9

2) If it is divisible by 4 then add the quotient to the 3rd last digit of the number and if the sum of them is divisible by 2 then the whole number is divisible by 8.

--> 9 (quotient)+ 9 ( 3rd digit from right)= 18, and -->18/2=9
So the whole number is divisible by 8.

Once you understand it and do a little practice, you'll find it easy and fast.
**You can try other numbers to see whether it is true or not
Hope it helps!
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RE:
What if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below

For 3 digits
133x11 --> 1 1+3 3+3 3=1463
And for 4 digits
1243x11 --> 1 1+2 2+4 4+3 3=13673

And for 5 digits
15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983
and so on.





Hi guys,
When multiplying by 11 I find it much easier to multiply the # by 10 (or add 0) and then add the original # to it.
For example, 1234x11 = 12340 + 1234

When dividing by 11 and the following condition is satisfied one could factor the # as follows:
671/11 = 61x11/11 = 61 because in 671, 6+1=7 which is the # in the middle of 671. This works for 3digit #'s

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These are a few math tricks my brother used when studying for the MCAT. Hope they help!
(attached a word document)

1. A nice math trick is multiplying two integers that have multiple digits relatively quickly. It does not apply to all integers the following has to be met:
TENS DIGIT in both integers HAS TO BE SAME
ONES DIGIT in each integer HAS TO ADD UP TO 10
Also, if the ones digits are 1 and 9 you just write 09.

Example 34x36:
Step 1: Add one to tens place then multiply (1+3)x3 = 12
Step 2: Ones place 4x6 = 24
Step 3: Place Steps 1 & 2 = 1224

Example 1 & 9 in ones place: 21x29
Step 1: Add one to tens place then multiply (1+2)x2 = 6
Step 2: Ones place 1x9= 09
Step 3: Place Steps 1 & 2 together = 609

2. This one is for people having difficulty memorizing a few square root numbers!
sqrt 1 = 1 (as in 1/1 = New Year's Day)
sqrt 2 = 1.4 (as in 2/14 = Valentine's Day)
sqrt 3 = 1.7 (as in 3/17 = St. Patrick's Day)

3. The next math trick is the Babylonian Method it can be useful when estimating square roots.
First guess roughly what you think it would be

Step 1: For a number less than 1 guess bigger.
For a number greater than 1 guess smaller.
Step 2: Divide your guess into the square root number.
Step 3: Take your answer (from step 2) and add it to your guess
Step 4: Divide by 2

Example sqrt(.78):
Step 1: sqrt of .78 < 1 so guess = .85
Step 2. .78/.85 = ~.9
Step 3: (.85+.9) = 1.75
Step 4: 1.75/2 = ~.88 And this should be your answer (or close enough)

If you guess really wildly just use the answer from your first guess and run through the process again. You can do it in seconds once you get good at it.

Wild Guess Example: sqrt 70 = ?
Step 1 sqrt of 70>1 so guess 10
Step 2: 70/10= 7
Step 3: 7+10= 17
Step 4: 17/2 = 8.5
*8.5x8.5 = 72.25 Still off (10 kind of a wild guess, so repeat process with the new answer from step 4)
Step 1: guess = 8.5
Step 2: 70/8.5= 8.2
Step 3: 8.2+8.5= 16.7
Step 4: 16.7/2 = ~8.4
8.4*8.4 = 70.6

4. If two numbers (both even or odd) are close together and their average is an integer, then this method can be used.
Need to recognize that:
x^2 - y^2 = (x + y)(x - y) and vice-versa (x + y)(x - y) = x^2 - y^2
Example 1
48*52 = (50-2)(50+2) = 50^2 - 2^2 = 2496.

Example 2
3^2 - 2^2 = 3 + 2
4^2 - 3^2 = 4 + 3
5^2 - 4^2 = 5 + 4

5. When x and y are consecutive integers, then (x - y) = 1.

It's useful for calculating large squares:

Example: 71^2 = ?
71^2 - 70^2 = 71 + 70
so 71^2 = 70^2 + (141)
= 4900 + 141 = 5041

Example: 53^2 = ?
53^2 – 52^2 = (53 + 52) + (52 + 51) + (51 + 50)
53^2 = 50^2 + (105) + (103) + (101)
= 2500 + 309 = 2809

0.79^2
80^2 - 79^2 = 80 + 79
so 80^2 - (80 + 79) = 79^2
6400 - 159 = 6241 so 0.79^2 = 0.6241
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Hi,

I've created an excel file I use to test myself on a daily basis to hopefully strengthen my basics while studying for the exam. It tests 20x20 multiplication and some of the the common powers, roots, fractions etc. that bb included in his doc.

Just thought I'd share incase there are others that might find it useful.

Cheers
Rupert
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AtifS wrote:
Thanks for sharing the Doc bb...@Shelen & @jeckll...Thanks for the tip but I would like to add some more.

What if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below

For 3 digits
133x11 --> 1 1+3 3+3 3=1463
And for 4 digits
1243x11 --> 1 1+2 2+4 4+3 3=13673

And for 5 digits
15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983
and so on.
you can plug other numbers in and check it out.

Hope it helps! :)








welldone dude,
just a small addition which i tried and will help to avoid confusion:

start writing the answer from right hand side in case if the addition of two no. exceeds 10, and add it to consecutive no. on left hand side. try out!!
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AtifS wrote:
Guys! I just found another way of checking whether a number is divisible by 8 or not ( the rule is same but another approach or route ). It's for a number with more than two digits.

Let's take 1936
1) First of all check whether last two digits of the number are divisible by 4 or not.
For 1936, we do this way 36/4=9

2) If it is divisible by 4 then add the quotient to the 3rd last digit of the number and if the sum of them is divisible by 2 then the whole number is divisible by 8.

--> 9 (quotient)+ 9 ( 3rd digit from right)= 18, and -->18/2=9
So the whole number is divisible by 8.

Once you understand it and do a little practice, you'll find it easy and fast.
**You can try other numbers to see whether it is true or not
Hope it helps!




hi,
i feel the process is bit complicated as it doesn't give the value of quotient, it just tells you whether no. is divisible by 8.
here is another tric, if the no. formed by last three digits is divisible by 8 then the whole no. is divisible by 8.

953360 is divisible by 8 since 360 is divisible by 8,
529418: not divisible as 418 is not divisible by 8.

plug in different values and try.
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I found a trick or a shortcut to find the sum of the first half of consecutive integers( STRAIGHT,EVEN AND ODD) given the sum of one half of the set. This works only when the total number of elements in the set is even. but definitely saves a few seconds.

Let me illustrate with examples:

1. CONSECUTIVE INTEGERS : 6 TO 15
The Sum of the greater 5 numbers in a set of 10 consecutive integers is 65. Find the sum of the first 5 numbers.

Short cut:
Step 1: Multiply the no.of elements in each half : in this case 5 each and the spacing between each number in the set. in this case 1. ie., 5*5*1 = 25
Step 2: If the given sum is that of the greater numbers in the set, then subtract '25' to get the sum of the lower 5 numbers, i.e, 65-25 = 40 is the Answer
or if the given sum is that of the lower 5 consecutive numbers, then add '25' to get the sum of the greater 5 numbers i.e., 40+25 = 65.

2.Lets try this with 6 consecutive EVEN integers:

Find the sum of lower half of the numbers in a set of 6 consecutive even integers if the sum of the latter half is 30.

Step 1: 3*3*2 ( remember each half has 3 elements and the spacing between the elements is 2 as they are even) = 18
step 2: the given sum - step 1= 30 - 18 = 12 ( the sum of the even integers 2,4 and 6) is the answer.


3.Lets try for the 16 consecutive integers from 8 to 23. given sum of the greater 8 numbers in the set = 156

step 1: 8*8*1=64
step 2: given sum - step 1 = 156-64= 92 (which is the sum of numbers starting 8 thru 15)

4. Now lets try 24 consecutive ODD integers:

Find the sum of the second half of the elements of a set when the first half sums up to 168. The set contains Consecutive Odd integers.

Step 1: 12*12*2 = 288 ( Odd numbers are spaced evenly)
Step 2: given sum + 168 = 288+168 = 456

Check it out : the numbers are 3 to 49 inclusive.


try more examples. But remember it works only on CONSECUTIVE INTEGERS WITH EVEN NUMBER OF ELEMENTS. And when the sum of lower half is given, you need to ADD the given sum to step 1 and when the sum of greater half is given u need to SUBTRACT step 1 from the given sum.

Hope this helps! :)

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Very gud doc....it will be helful for anyone who's taking GMAT...

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@bb: Thanks
@enfinity: good technique but I guess one need time to get familiar and comfortable with it. Up to then the conventional way is better.

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Good document. Should bump this up for ppl to download and memorize

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Also, see this post for some good theory: math-number-theory-88376.html
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Shelen wrote:
I'm new here, and try to go through all the topics - this site is a treasure! :)

Maybe not the right place to share my experience with "multiplication for simple things like 13 x 11", but anyway...if you need to multiply any two-digits number by 11, just sum those digits and put the result in between. For example,
13x11 -> 1+3=4 -> 143 is the result.
Or, 36x11 -> 3+6=9 -> the result is 36x11=396.

Thanks for the tips. I recently downloaded some math apps to learn shortcuts. I plan to use them minimially though given the geometry and other formulas I need to memorize.
It really saves time.


Posted from my mobile device
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One useful tip (from vedic maths) -
multiplying numbers close to 100 or 1000 or so on
e.g. multiply 97 * 98
Take the base as 100 as both numbers are close to 100

Step 1) First, we multiply the offsets 2 and 3
We get 6.
Since our base is 100, which has 2 zeros, the product of offsets must also have 2 digits. Hence we write 6 as 06 and these are our last 2 digits.

Step 2) Now substract one of the integers ( 97 or 98) from the peer number's offset i.e. 97 - 2 or 98 -3. Either will give you 95.
These are our first 2 digits.

So the answer is 9506.
Hope this helps.

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