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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.

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.

Re: What arithmetic should I memorize? [#permalink]

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31 Oct 2009, 09:56

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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.

Re: What arithmetic should I memorize? [#permalink]

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31 Oct 2009, 20:21

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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

Re: What arithmetic should I memorize? [#permalink]

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11 Feb 2010, 22:41

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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!
_________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

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.

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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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04 Jan 2012, 16:59

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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.

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.

Re: What arithmetic should I memorize? [#permalink]

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14 Feb 2010, 06:12

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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!!
_________________

"Great people don't do different things, they do things differently"

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14 Feb 2010, 06:21

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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.
_________________

"Great people don't do different things, they do things differently"

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26 Dec 2012, 18:19

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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.

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09 Jan 2010, 20:05

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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.

Re: What arithmetic should I memorize? [#permalink]

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29 Nov 2012, 10:10

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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.

Re: What arithmetic should I memorize? [#permalink]

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05 Jan 2014, 23:40

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Dividing 1 with any two digit no ends with 9 eg. 1/19, 1/29, 1/39... Let’s take 1/19 1.Divide 1 by 2(1+1). It will give quotient 0 and reminder 1. Here we will set 0 as our first decimal value and 1 as prefix. So it will be .0 and keep 1 in mind which will be used for our next dividend. 2.Now divide 10(1= reminder and 0=quotient from above point) by 2. It will give quotient=5 and reminder=0. So our next digit is 5. Now it becomes 0.05. 3.Now divide 05(0=reminder and 1=quotient from point no2) by 2. It will give quotient=2 and reminder=1. Now ans is 0.052 You may continue if you want more decimal value. Now let’s test it for 1/89. 1.Divide 1/9(8+1). We will get 0 as quotient and 1 as reminder. So it becomes 0.0 2. Divide 10 by 9. Q=1, R=1. Second digit is 1 i.e 0.01 3. Divide 11 by 9. Q=1 R=2. Third digit is 1 i.e 0.011 4. Divide 21 by 9. Q=2 R=2. Ans is 0.0112

gmatclubot

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