The Quick Math Tip

T H E  Q U I C K  M A T H  T I P

Now you can blow up the GRE Math

SQUARE A 2 DIGIT NUMBER :

The handy in GRE

Square a 2 Digit Number, for this example 37:

•  Look for the nearest 10 boundary
  
•  In this case up 3 from 37 to 40.
  
•  Since you went UP 3 to 40 go DOWN 3 from 37 to 34.
  
•  Its like [37-3] * [37+3]
  
•  Now mentally multiply 34x40
  
•  This can be split as 34*10*4
  
•  Now 34x10=340;
  
•  (Multiply 340 * 4)Double it mentally to 680 & Double it again mentally to 1360
  
•  This 1360 is the FIRST interim answer.
  
•  Now,as we added
  
•  3x3=9 which is the SECOND interim answer.
  
•  Add the two interim answers to get the final answer.
  
•  Answer: 1360 + 9 = 1369
  

An alternative to 2digit no ending with 5 alone

SQUARE A 2 DIGIT NUMBER ENDING IN 5 :

For this example we will use 25

• Take the "tens" part of the number (the 2 and add 1)=3
  
•  Multiply the original "tens" part of the number by the new number (2x3)
  
•  Take the result (2x3=6) and put 25 behind it. Result the answer 625.
  

Try a few more 75 squared ... = 7x8=56 ... put 25 behind it is 5625.
55 squared = 5x6=30 ... put 25 behind it ... is 3025. Another easy one! Practice it on paper first

The genral one : (a+b)^2 = (a^2 + 2ab + b^2)

THE 11 RULE :

 You likely all know the 10 rule (to multiply by 10, just add a 0 behind the number) but do you know the 11 rule? It is as easy! You should be able to do this one in you head for any two digit number. Practice it on paper first!

To multiply any two digit number by 11:

•  For this example we will use 54.
  
•  Separate the two digits in you mind (5__4).
  
•  Notice the hole between them!
  
•  Add the 5 and the 4 together (5+4=9)
  
•  Put the resulting 9 in the hole 594. That's it! 11 x 54=594
  

The only thing tricky to remember is that if the result of the addition is greater than 9, you only put the "ones" digit in the hole and carry the "tens" digit from the addition. For example 11 x 57 ... 5__7 ... 5+7=12 ... put the 2 in the hole and add the 1 from the 12 to the 5 in to get 6 for a result of 627 ... 11 x 57 = 627
Practice it on paper first!

MULTIPLY UP TO 20X20 IN YOUR HEAD :

In just FIVE minutes you should learn to quickly multiply up to 20x20 in your head. With this trick, you will be able to multiply any two numbers from 11 to 19 in your head quickly, without the use of a calculator.
I will assume that you know your multiplication table reasonably well up to 10x10.

Try this:

• Take 15 x 13 for an example.
  
• Always place the larger number of the two on top in your mind.
  
• Add the ones digit of the smaller number with the large number (15 + 3 = 18)
  
• Add a zero behind it (multiply by 10) to get 180.
  
• Multiply both the ones digit with each other (3x5= 15)
  
• Add 180 + 15 = 195.
  

That is It! Wasn't that easy? Practice it on paper first!

 THE PERCENT :

Find 7 % of 300. Sound Difficult?

*This useful technique is contributed by yanni in a forum
Percents: First of all you need to understand the word “Percent.” The first part is PER , as in 10 tricks per listverse page. PER = FOR EACH. The second part of the word is CENT, as in 100. Like Century = 100 years. 100 CENTS in 1 dollar… etc. Ok… so PERCENT = For Each 100.

So, it follows that 7 PERCENT of 100, is 7. (7 for each hundred, of only 1 hundred).
8 % of 100 = 8. 35.73% of 100 = 35.73
But how is that useful??

Back to the 7% of 300 question. 7% of the first hundred is 7. 7% of 2nd hundred is also 7, and yep, 7% of the 3rd hundred is also 7. So 7+7+7 = 21.

If 8 % of 100 is 8, it follows that 8% of 50 is half of 8 , or 4.

Break down every number that’s asked into questions of 100, if the number is less then 100, then move the decimal point accordingly.

EXAMPLES :
8%200 = ? 8 + 8 = 16.
8%250 = ? 8 + 8 + 4 = 20.
8%25 = 2.0 (Moving the decimal back).
15%300 = 15+15+15 =45.
15%350 = 15+15+15+7.5 = 52.5

Also it’s usefull to know that you can always flip percents, like 3% of 100 is the same as 100% of 3.

35% of 8 is the same as 8% of 35.

THE MASTERS MASTER-PIECE
A must for GRE 

“is x divisable by y?”

2:A number is divisable by 2 if it’s an even number.
3: The sum of the individual digits is divisable by 3, e.g. 174 => 1+7+4 = 12 (=> 1+2 = 3)
4: The last two digits is divisable by 4
5: The last digit is 5 or 0
6: The number is divisable by 2 and 3
7: Double the last digit and subtract it from the rest. If the result is divisable by 7, the original number also is. E.g. 483 => 48-(3*2) = 42, divisable by 7
8: The last three digits is divisable by 8
9: The sum of digits is 9
10: The last digit is 0
11: Subtract the sum of “even” digits from the sum of “odd” digits - if divisable by 11 (incl 0), the original number is divisable by 11: E.g. 7084 => (7+8)-(0+4)=11. 7084 is divisable by 11.

PRIME FACTORIZATION 

Prime factorization also requires us to tell how many times each of those prime factors go into a number, along with the other prime factors.
Take the prime factors, and multiply them by themselves
2 x 3 x 5 = 30. So you only need one of each to have the prime factorization of 30.

But if I had the number, say, 40, we'd have the prime factors:
2, and 5. (all other factors of 40 are not prime).
But 2 x 5 is not 40.
So how can we get 40 with just twos and fives?
Well, we'd need 2 x 2 x 2 x 5 to make 40. (That's the same as 8 x 5).
A simpler way to write 2 x 2 x 2 x 5 is to write 2 ^3 x 5 (which is read, "Two to the third power times 5.")
So the prime factors of 40 are 2 and 5, but the prime factorization of 30 is 2^3 x 5.

Are we good so far?

To lock it in, let's take your first number:
36.
The factors are 2, 3, 4, 6, 9, 12, and 18.
Not all of those numbers are prime. Which are?
2 and 3 are the only prime factors of 36.

How many of each do you need to make 36?
2 x 2 x 3 x 3 will give you 36.
That is 2^2 x 3^2, and that's your answer.

An easy recipe to prime factorize a number is this:
List the prime factors. To do this, start with 2, see if that will go into the number. If it does go in, divide the number by 2.
Now you have a new number. Repeat the above step until 2 will no longer go into the number evenly.
For the answer so far, write the number 2 (if it actually does go into the original number at all) and after the caret (the ^) write the amount of times it went in.
Then use the next highest prime number, which is 3, and repeat the above steps with the number 3.

After that use the next highest prime number and repeat the above. (The next highest prime number is not 4!)

Let's try the whole thing with the number 293:
Does 2 go into 293 evenly? No. You can tell this immediately because 293 is not an even number. So 2 is not a prime factor of 293.
Try 3. Does 3 go into 293?
The best way to tell is if you learned the ways to test for divisibility. If you haven't, or you don't know what that means, you will just have to divide 293 by 3 and see if it comes up with no remainder. (If you can't divide, give up on factorization for now, and go get good at division )

You'll see that 3 is not a factor of 293.
Let's go to the next prime number, which is 5. You can tell if a number is divisible by 5 if it ends in a five or a zero. This number doesn't, so 5 is not a factor.
So far, you have nothing to write in the answer.



What's the next prime? Is it 6? No, because 6 has, besides itself and one, 2 and 3 as factors. (Remember, a prime has only 1 and itself as factors).
So the next prime number is 7. Divide 293 by 7, and you'll get 21, with a remainder of 6, so it doesn't go in evenly, and is therefore not a factor of 293.

8, 9 and 10 all have factors other than 1 and themselves, so they are not prime.

The next highest prime number is 11. Divide 293 by 11 and get 26, with a remainder of 7, so that won't work.

12 is not prime. (By the way, an number that isn't prime is known as a "composite" number.)

13 is prime. Divide 293 by 13 and get 22 with a remainder of 7, so 13 is NOT a factor.

14, 15, and 16 are composite numbers.

The next highest prime is 17. Divide 293 by 17 and get 17 with a remainder of 4. Wow, this is getting to be a pain.

18 is composite.

19 is prime. Divide 293 by 19 and still get a remainder. (Now I'm getting mad!)

20, 21 and 22 are composite.

Divide 293 by 23 and get 12 with a remainder. (GRRRRR!)

So it follows till 146

Any number higher than 146 is not going to be a factor of 193. Can you see why?
It is because Any number higher than half of the original number cannot possibly be a factor of the number because it cannot possibly divide into it evenly.

That means that 293 is the only prime factor of 293. Which means that 293 is a prime number.

Here's a tip : whenever you come to a large prime factor of a number, make a list of all the multiples of that number up to half of the original number.
For example, when we get to 13, make a list of its multiples:

13
26
39
52
65
78
91
104
117
130
143
156

So now you know, whenever you come to any of these numbers as you do the above method, you'll know that they are composite numbers, and you don't have to divide to test them. It will save you some work. Especially with numbers like 91, 117 and 143 which are not otherwise easily recognized as composites.

but this is too costly as it eats our time., so we better skip this and we follow nice simple and a cool thing to memorize all the primes up to 100.

They are:
1 (which doesn't count in prime factorization), 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.

Let me explain ,

Any number higher than 146 is not going to be a factor of 193. Can you see
why?
"It is because Any number higher than half of the original number cannot
possibly be a factor of the number because it cannot possibly divide into it
evenly."

It turns out that we could have stopped at 17.
Once you reach a factor that is the square root (or larger) of the original number, you need to look no further. This is why:
The square root of 293 is around 17. (17 squared is 289.) We have already tested 17 and found that it is not a factor.
Consider this now - If a number higher than 17 was a factor of a 293, the other number would have to be lower than 17. We know this because if we multiply 17 by 17 we already get 289, so if we multiply 17 (or any number larger than 17) by anything higher than 17 we'd get something larger than 293.
Therefore, we'd have to multilpy 17 (or any number larger than 17) by some number that is smaller than 17. But we know that no prime numbers that are smaller than 17 are factors of 293, therefore there is no chance that a larger number could be multiplied by one of them to give you 293.
So you can stop at 17.



If you have more short cuts on related topics ., kindly drop them here.