## Papgust's GMAT MATH FLASHCARDS directory

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### Papgust's GMAT MATH FLASHCARDS directory

by papgust » Thu Jun 03, 2010 6:26 pm
Greetings GMAT Aspirants!

I spoke to Eric whether i could share my flashcards with our community. Eric's instant reply was "Why not? It would be awesome to see the notes that you came up with".

Throughout my GMAT preparation, I have collected key core-concepts and tricks in my flashcards. No doubt, I had to spend atleast an hour to go through my flashcards everyday . Eric and I are confident that these flashcards will help our community in some way or the other. Special thanks to Eric for allowing me to share these flashcards with you I would like to share these flashcards on a daily basis rather as a whole. Very simple reason is members can easily digest each and every concept and discuss on the concept by trying out a GMAT/Non-GMAT problem (It is very important to recognize an underlying concept in a GMAT problem and work through the problem). People tend to overlook some vital points if the flashcards are shared as a bunch.

NOTE: My flashcards are in handwritten form in a notebook. My handwriting will not be readable by everybody here. So, the notebook will not be scanned and shared at one go as it serves no purpose at all. Kindly bear with me and get benefited by learning these points everyday. Will do my best to post as many points as possible whenever i logon to BTG.

Thanks!

Very good luck to ALL!! Happy Learning! Last edited by papgust on Sun Jun 06, 2010 1:55 am, edited 1 time in total.

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by papgust » Thu Jun 03, 2010 6:32 pm
I will post my flashcards based on the core topics that are covered in GMAT,

1. Number Properties
2. Inequalities
3. Averages
4. Ratios
5. Sequences & Progressions
6. Set Theory
7. Co-ordinate Geometry
8. Geometry

The points are taken from multiple sources. But the key resource in my preparation is High School Math book. This book has really helped me in understanding the concepts in depth and apply those concepts in GMAT. I attribute my improvement in quants to this book.

I will start the thread with a very important topic in GMAT - "NUMBER PROPERTIES".

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by papgust » Thu Jun 03, 2010 6:45 pm
How to test whether a number is prime or composite:

Before we start off, what is a prime number and a composite number? (For people who are not sure)
A Prime Number is a positive integer that is divisible by ONLY 2 numbers (1 and itself). Whereas, A composite number is a positive integer which has divisor(s) other than the 2 numbers (1 and itself).
Ok, coming back to the point. I will name the number as n for simplicity. Following are the steps to test whether a number is a prime or composite,

1. Identify the perfect square (P.S) closest to the n.
2. Compute the square root of P.S
3. List all prime numbers upto the computed square root
4. Check if all listed prime numbers divide n equally. If not, then n is a prime. Even if atleast one of the listed prime numbers divide n, then n is a composite.

Example:

Take n as 113. To test whether 113 is a prime,

1. 100 is the closest perfect square to 113 (Remember that you take a closest perfect square that is smaller than n itself!)
2. Square root of 100 ==> 10
3. Prime numbers upto the square root (10) ==> 2,3,5,7.
4. Check whether 2,3,5,7 divides 113. None of the numbers divide 113. So, 113 is a prime.

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by papgust » Fri Jun 04, 2010 7:27 pm
Most of you are comfortable with calculating LCM and HCF of 2 integers/numbers. But are you comfortable for doing the same with fractions? If you are not sure how to do for fractions, here is how you do.

How to calculate LCM and HCF of fractions:
L.C.M of 2 fractions = L.C.M of NUMERATORS / H.C.F of DENOMINATORS

H.C.F of 2 fractions = H.C.F of NUMERATORS / L.C.M of DENOMINATORS

Try out an example if required.
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by papgust » Fri Jun 04, 2010 7:31 pm
There is one interesting cool fact to know. I remember applying this fact in actual GMAT. It's good to learn if you don't know.

Product of any 2 numbers = Product of LCM and HCF of those 2 numbers

Product of any 2 fractions = Product of LCM and HCF of those 2 fractions

I will try to find and post a GMAT problem that uses this concept. Please feel free to post a question if you find it.
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by papgust » Sat Jun 05, 2010 7:40 pm
Warning: Some people may not find this approach comfortable. Some may find it comfortable. Please follow and practice only if you are comfortable with this approach. Otherwise, please ignore it.

Sometimes, we get one type of question in GMAT where we need to calculate units digit of integers raised to some power. I found a shortcut where you could save time by remembering some patterns.

How to find unit digit of powers of numbers:

Pattern 1:
Unit's place that has digits - 2/3/7/8

Then, unit's digit repeats every 4th value. Divide the power (or index) by 4.

After dividing,
If remainder is 1, unit digit of number raised to the power 1.
If remainder is 2, unit digit of number raised to the power 2.
If remainder is 3, unit digit of number raised to the power 3.
If remainder is 0, unit digit of number raised to the power 4.

Pattern 2:
Unit's place that has digits - 0/1/5/6

Then, all powers of the number have same digit as unit's place.

For e.g., 6^1 = 6, 6^2 = 36, 6^3 = 216, 6^4 = 1296

Pattern 3:
Unit's place that has digit - 4

Then,
If power is odd --> unit's digit will be '4'
If power is even --> unit's digit will be '6'

Similarly,
Unit's place that has digit - 9

Then,
If power is odd --> unit's digit will be '9
If power is even --> unit's digit will be '1'

Example:
Let's take a long number - 122 ^ 94. Find unit's digit.

Unit's place is 2. So, it repeats every 4th term of the power.
So, divide the power by 4. 94 % 4 ==> 2 (remainder).

Raise the unit digit of the base number to the power (2 - remainder). 2^2 = 4.

Thus, 4 is the unit's digit of 122^94.

I found this approach very easy and comfortable. So, see how comfortable it is for you and apply.

Real GMAT Problem: OG-12 PS #190
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by papgust » Sat Jun 05, 2010 7:54 pm
We are often faced to test the divisibility of some number in the exam. Following points may help you in simplifying the process,

Divisibility Tests:

To check whether a number (say n) is divisible

By 2: unit's place of n must be 0 (OR) unit's place of n must be divisible by 2.

By 3: Sum of the digits of n must be divisible by 3.

By 4: Last 2 digits (Unit's place and ten's place) of n are 0's (OR) Last 2 digits of n must be divisible by 4.

By 5: Unit's digit must be a 5 (OR) a 0.

By 6: n must be divisible by both 2 and 3 (Follow the method used for 2 and 3).

By 8: Last 3 digits (units, tens and hundredth place) of n are 0's (OR) Last 3 digits of n is divisible by 8.

By 9: Sum of the digits of n must be divisible by 9.

By 11: (Sum of the digits of n in odd places) - (Sum of the digits of n in even places) ==> Either 0 (OR) divisible by 11.

By 12: n must be divisible by both 3 and 4 (Follow the method used for 3 and 4).

By 25: Last 2 digits (units and tens place) of n are 0's (OR) Last 2 digits of n must be divisible by 25.

By 75: n must be divisible by both 3 and 25 (Follow the method used for 3 and 25).

By 125: Last 3 digits of n are 0's (OR) are divisible by 125.

Try out examples for each divisibility to grasp better.
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by papgust » Sun Jun 06, 2010 6:24 pm
How to find number of factors for a POSITIVE INTEGER:

There are 2 approaches to find number of factors of an integer.

Approach #1: (Factor Pairs Method)

i. Let's take a non-perfect square number such as 32. Keep picking a number (start from 1) that divides 32 until you reach a number that is smaller than the quotient.

Small Large
1 32
2 16
4 8

Stop! If you take 8, you get 4 as quotient which is smaller than the number (8).
Therefore, there are 3*2 = 6 factor pairs or number of factors of 32.

ii. Let's take a perfect square number such as 36. Keep picking a number (start from 1) that divides 36 until you reach a number that is smaller than the quotient.

Small Large
1 36
2 18
3 12
4 9
6 6

Totally, there are 5*2 = 10 factor pairs or number of factors of 36. But, (6,6) gets repeated twice. So, deduct 1 from factor pairs i.e. 10-1 = 9 factor pairs or number of factors of 36.

Approach #2: (RECOMMENDED)

If N is expresses in terms of its prime factors as a^p * b^q * c^r, where p,q,r are positive integers, then N will have (p+1) * (q+1) * (r+1) positive factors.

Example:

i. 32 = 2^5.
No. of factors = (5+1) = 6.

ii. 1452 = 2^2 * 3 * 11^2
No. of factors = (2+1) * (1+1) * (2+1) = 18.
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by papgust » Sun Jun 06, 2010 6:26 pm
If N is a perfect square, then the number of factors of N will ALWAYS be an ODD number.

If N is a NON-perfect square, then the number of factors of N will ALWAYS be an EVEN number.
Last edited by papgust on Sun Jun 06, 2010 6:31 pm, edited 1 time in total.
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by papgust » Sun Jun 06, 2010 6:31 pm
How to find Sum of all factors of a POSITIVE integer:

If N is expressed in terms of its prime factors as a^p * b^q * c^r, where p,q,r are positive integers, then the sum of all factors of N is

[ (a^(p+1) - 1) / a-1 ] * [ (b^(q+1) - 1) / b-1 ] * [ (c^(r+1) - 1) / c-1 ]
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by papgust » Sun Jun 06, 2010 6:33 pm
Any number whose prime factorization contains even powers of primes, then the number must be a perfect square.

Any number whose prime factorization contains powers of primes with multiples of 3, then the number must be a perfect cube.
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by papgust » Mon Jun 07, 2010 6:39 pm
Hello folks,

I'm not sure whether you guys are following the thread as there is no response or no interest shown by you.

Please let me know if i need to continue posting. Otherwise, i'll stop right here. It's a sheer waste of time if no one is getting benefited.
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by papgust » Tue Jun 08, 2010 6:44 am
Guys are indeed following the SC thread. I hope people are also following this thread. Let me continue to post flashcards.

REMAINDERS:

(I)

When 2 numbers are divided by same divisor and the remainders obtained are the same,
THEN
DIFFERENCE b/w 2 numbers is also divisible by that divisor.

(II)

When 2 positive numbers 'a' and 'b' are divided by the same divisor 'd' and remainders obtained are 'r1' and 'r2' respectively,
THEN
the remainders obtained when a+b is divided by d will be r1+r2
NOTE: If r1+r2 >= d, compute (r1+r2) - d as the remainder.

(III)

When 2 positive numbers 'a' and 'b' are divided by the same divisor 'd' and the remainders obtained are 'r1' and 'r2' respectively,
THEN
the remainders obtained when a*b is divided by d will be r1*r2
NOTE: If r1*r2 >= d, compute (r1*r2) / d as the remainder.

TAKEAWAY:

A remainder can NEVER be greater than or equal to the divisor.
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by papgust » Tue Jun 08, 2010 6:56 am
How to find REMAINDER for LARGE POWERS of numbers:

There are 2 ways to do so:

1. Pattern Method:

Example:

What is the remainder when 2^56 / 7 ?

Solution:
Remainder when 2^1 is divided by 7 is 2
Remainder when 2^2 is divided by 7 is 4
Remainder when 2^3 is divided by 7 is 1
Remainder when 2^4 is divided by 7 is 2 --> Repeats again.

The remainder repeats after 3 steps i.e. in the 4th step.

Now, Divide the power (or index) by 3 (no of steps after which remainder repeats) and compute a new remainder.

56 % 3 --> 2 (remainder)

Now, raise the base (2) to the power 2 (new remainder). 2^2 % 7 --> 4.

Thus, 4 is the remainder when 2^56 / 7.

2. Remainder Theorem Method: (NOT RECOMMENDED unless clear)

Example:

What is the remainder when 2^51 / 7 ?

Solution:
2^51 can be changed to (2^3)^17.
7 can be changed to (8-1) OR (2^3 - 1)

Substitute 'x' in place of 2^3,

x^17 / (x-1)

Remainder is f(1). Substitute 1 in 'x',

Remainder is 1.

Thus, 1 is the remainder when 2^51 / 7.
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by papgust » Tue Jun 08, 2010 6:59 am
Simple Facts:

a^n - b^n:

1. ALWAYS divisible by a-b
2. If n is even, it is divisible by a+b
3. If n is odd, it is NOT divisible by a+b

a^n + b^n:

1. NEVER divisible by a-b
2. If n is odd, it is divisible by a+b
3. If n is even, it is NOT divisible by a+b
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