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Re: What arithmetic should I memorize? [#permalink]
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14 Feb 2010, 11:28
target2011 wrote: 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. Thank you for explaining, you are totally right but I am a bit lazy & I like to do mental math so instead of dividing all 3 digits by 8 in my mind was not faster than the method I described above ( it is really easy & fast to divide 2 digits than dividing by 3 digits). Your method is right & is the one described in "Number Theory", but I think I would like to split the problem into easy steps & then solve it easily and fast. And the method I described is easy & fast if u do it after a little practice as it is broken down into two easy 2 digits numbers. Maybe the way I explained my method was not clear otherwise the method is really easy & fast once you understand as if I take my previously described number 1936 here we have to take out 936 & then dividing it by 8 mentally would take time and it won't save the time as much as we can save with the method I described just plug numbers and check it. Take some numbers with last 3 digits of high value & try. Hope it helps!
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17 Feb 2010, 19:53
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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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Re: What arithmetic should I memorize? [#permalink]
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08 Mar 2010, 01:21
Thanks. Guys I was searching for quant document for tables,squares and it was not there. If anyone had compiled such documents having: Tables, squares,cubes, arithmatic formulae, Pythagoras triplets etc. Please share.
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Re: What arithmetic should I memorize? [#permalink]
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08 Mar 2010, 02:28
bb wrote: Fantastic question This file should give you an idea where you are lacking. I think you should definitely know the squares from 010 and preferrably from 10 to 20 as well. thanks for the doc BB kudos

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08 Mar 2010, 12:05
gurpreetsingh wrote: Thanks.
Guys I was searching for quant document for tables,squares and it was not there. If anyone had compiled such documents having: Tables, squares,cubes, arithmatic formulae, Pythagoras triplets etc.
Please share. There is plenty of these things around: newtothemathforumpleasereadthisfirst77764.htmlalluneedforgmatquant90489.html#p689013gmatmathbook87417.html
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Re: What arithmetic should I memorize? [#permalink]
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09 Mar 2010, 05:34
gurpreetsingh wrote: Thanks.
Guys I was searching for quant document for tables,squares and it was not there. If anyone had compiled such documents having: Tables, squares,cubes, arithmatic formulae, Pythagoras triplets etc.
Please share. I have compiled some notes and also GMAT Math Book, will need official permission to share, when got permission I'll share them in a separate new Post.
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25 Jul 2010, 21:52
very good compilation. It is indeed very helpful. Thanks

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Re: What arithmetic should I memorize? [#permalink]
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25 Jul 2010, 22:42
Hi,
Concerning the divisibility of numbers by 8
A number is divisible by if it is divisible by2 three times....
Or
If the last three digits are divisible by 8
e.g 18248 is divisible by 8 because 248 in divisible by 8.
I hope this helps.
Regards Farukh

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25 Jul 2010, 22:44
Also,
Thanks BB for your great compilation!
Regards Farukh

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Re: What arithmetic should I memorize? [#permalink]
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17 Aug 2010, 07:50
Thanks BB. Makes my calculations much simpler.

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Re: What arithmetic should I memorize? [#permalink]
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12 Nov 2010, 18:50
Here is an easy way to multiply by 11.
For instance 13x11, just add 13 to 13 but move the second # one space to the left, then add down.
.....1st no 13 .2nd no + 13 ......Add  Total .... 143
(Sorry about the formatting. Only way I knew how to show it.) So far it has worked for every 2 and 3 digit number that I have tried.
Good luck.

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29 Dec 2010, 13:04
jeckll wrote: 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 twodigits 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 nice trick. had forgotten about it
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20 Apr 2011, 03:13
target2011 wrote: 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!! Well, the whole method explained here is nothing but the conventional way of multiplying the numbers. e.g. 15453 X 11  + 1 5 4 5 3 1 5 4 5 3  1 1+5 5+4 4+5 5+3 3 which just a single line representation of the two lines resulting from conventional method. Hope that clarifies any doubt over the method, if any.
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Re: What arithmetic should I memorize? [#permalink]
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21 Sep 2011, 23:29
Regarding Multiplication by 11, an easy way I use is to Multiply the number by 10 and then add the number to it.
i.e. A*11 = A*10 + A e.g: 13*11 = 13*10 + 13 = 143 42*11 = 42*10 + 42 = 420 + 42 = 462
This is a generic method that can be used for many different multiplications:
A*21 = A*20 + A A*19 = A*20  A A*110 = A*100 + A*10
Hope this helps somebody!

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22 Sep 2011, 00:44
Nice nifty trick enfinity wrote: 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 (nonunits digits) times (nonunits 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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22 Sep 2011, 00:52
gottabwise wrote: 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 twodigits 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 How about 46 X 11 = ? ; Just found out we cannot apply this rule to this so wondering
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31 Oct 2011, 07:44
just add the digits and insert the result in between the number. If there is a carry over add it to the ten's digit of the number
So in this case 46*11 ; 4+6 = 10. Insert 0 in between 4 and 6, carry forward the 1, that gives us 506

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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 viceversa (x + y)(x  y) = x^2  y^2 Example 1 48*52 = (502)(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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19 Jan 2012, 20:58
thanks for posting this download

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Re: What arithmetic should I memorize? [#permalink]
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08 Mar 2012, 00:46
Thank you for the document! Have already found it useful.

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