What arithmetic should I memorize?

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.

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!

I found this post a little too late but agree with bb that there are set number of things that everyone must know, out cold, by test day. This can save valuable time and give you opportunity to spend more time analyzing and solving questions rather than doing mundane computations.

Till 2 weeks back, I only knew tables up to to 10. The day and age we live in, I never felt the need. I either have excel or a calculator handy on my smart phone. These are habits one has to develop so I started putting together a Brain Dump sheet. I have roughly 30 print-outs on my desk, and first thing I do when I get to work is take 5 minutes to fill them out. Initially it will be difficult, but start by breaking them down into smaller sections till you have average proficiency.

You might want to add first 100 prime numbers to the list attached.

2 3 5 7
11 13 17 19
23 29
31 37
41 43 47
53 59
61 67
71 73 79
83 89
97

Learning aid:
2x, 5x and 8x are similar.
3x, 6x are similar
4x and 7x are similar.

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

Added TTP's 15-page PDF with required formulas and illustrations.

https://blog.targettestprep.com/gmat-equation-guide/

Included in our Best Downloads of GMAT Club

I'd just caution test takers that the GMAT is not a test of how many formulas you can memorize, and whether you can plug numbers into them. Especially at the higher difficulty levels, it's a test of how well you think about mathematical concepts. I could write all the GMAT Quant facts I ever use on one sheet of paper (if I draw the Geometry diagrams small enough ). Most prep company materials I've seen advise memorizing far too many things, things that won't ever help on the actual test.

Hi,

I believe this to be one of the most important hacks to learn by heart: Pythagorean triple.
A Pythagorean triple is a triple of positive integers a, b, and c such that a right triangle exists with legs a,b and hypotenuse c. By the Pythagorean theorem, this is equivalent to finding positive integers a, b, and c satisfying
a^2+b^2=c^2. Please find below the ones that are most popular to memorize:

3, 4, 5
5, 12, 13
7, 24, 25
8, 15, 17
12, 35, 37

Remember that this only works for right triangles. Good luck

Hello everyone,
I'm a newcomer. Just decided to learn GMAT for a week. Thank you so much for the tip. I try to solve the logic.
Let's take an interger n = Abcd (A can be any number) Abcd = 1000A + 100b + 10c + d.
If last two digit is divisible by 4: 10c + d = 4x. Then we have: Abcd = 125*8A + 100b + 4x
If sum of the third digit and x is divisible by 2: b+x=2y or x = 2y-b
Abcd = 125*8A + 100b + 4(2y-b) = 125*8A + 96b + 8y = (125A +12b+y)*8

So n is divisible by 8.
Actually simply just put if the last 3 digits is divisible by 8 then the whole number is, too.

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