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1

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) 

Quote:
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.


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. 


Quote:
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.


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


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.



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.

2

Prime numbers

Prime Numbers

A prime number can be divided, without a remainder, only by itself and by 1. For example, 17 can be divided only by 17 and by 1.



Some facts:

The only even prime number is 2. All other even numbers can be divided by 2.
If the sum of a number's digits is a multiple of 3, that number can be divided by 3.
No prime number greater than 5 ends in a 5. Any number greater than 5 that ends in a 5 can be divided by 5.
Zero and 1 are not considered prime numbers.
Except for 0 and 1, a number is either a prime number or a composite number. A composite number is defined as any number, greater than 1, that is not prime.
To prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can't be a prime number. If you don't get a whole number, next try dividing it by prime numbers: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below).

3

The Sieve of Eratosthenes

The Sieve of Eratosthenes

Eratosthenes (275-194 B.C., Greece) devised a 'sieve' to discover prime numbers. A sieve is like a strainer that you use to drain spaghetti when it is done cooking. The water drains out, leaving your spaghetti behind. Eratosthenes's sieve drains out composite numbers and leaves prime numbers behind.
To use the sieve of Eratosthenes to find the prime numbers up to 100, make a chart of the first one hundred positive integers (1-100):

1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100

Cross out 1, because it is not prime.

Circle 2, because it is the smallest positive even prime. Now cross out every multiple of 2; in other words, cross out every second number.

Circle 3, the next prime. Then cross out all of the multiples of 3; in other words, every third number. Some, like 6, may have already been crossed out because they are multiples of 2.

Circle the next open number, 5. Now cross out all of the multiples of 5, or every 5th number.
Continue doing this until all the numbers through 100 have either been circled or crossed out. You have just circled all the prime numbers from 1 to 100!

4

How to test whether a number is 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.

5

How to calculate LCM and HCF of fractions:



Quote:
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

6

Product of two numbers
Product of two fractions

Quote:
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.

7

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.

8

Area of triangle
Area of rectangle
Area of trapezoid
Area of elipse

Triangle
Area = ½ × b × h
b = base
h = vertical height

Rectangle
Area = w × h
w = width
h = height

Trapezoid (US)
Trapezium (UK)
Area = ½(a+b) × h
h = vertical height

Ellipse
Area = πab

9

Area of square
Area of parallelogram
Area of circle
Area of sector

Square
Area = a2
a = length of side

Parallelogram
Area = b × h
b = base
h = vertical

Circle
Area = π × r2
Circumference = 2 × π × r
r = radius

Sector
Area = ½ × r2 × θ
r = radius
θ = angle in radians

10

Recognize multiples of
2
3
4
5
6
9

2. Last digit is even
3. Sum of digits is a multiple of 3
4. Last two digits are multiples of 4
5. Last digit is 5 or 0
6. Sum of digits is a multiple of 3 and the last digit is even
9. Sum of digits is a multiple of 9
10. Last digit is 0
12. Sum of digits is a multiple of 3 and the last two digits are a multiple of 4

11

Isosceles triangle
Equilateral triangle
Pythagorean theorem
30-60-90 triangle
45-45-90 triangle

Isosceles triangle - two equal sides and two equal angles
Equilateral triangle- all sides equal and all equal angles
Pythagorean theorem- a^2 + b^2 = c^2
30-60-90 triangle - 1 / root 3 / 2
45-45-90 triangle - 1 / 1 / root 2

12

Slope
Permutation
Combination

Slope = change in y / change in x
Permutation - n! / (n-k)!
Combination- n!/[k!(n-k)!]

13

Sum of all angles of a regular polygon
Area of sector
Volume of cylinder
Volume of sphere

Sum of angles = (n-2)*180
area of Sector - r/360 * Pi * r ^2
Volume of cylinder - Pi * r^2 * h
Volume of sphere- 4/3 * Pi * r^3

14

Squares of 2 till 10

2-4
3-9
4-32
5-25
6-36
7-49
8-64
9-81
10-100

15

Squares of 11 to 15

11-121
12-144
13-169
14-212
15-225

16

Squares of 16 to 20

16-256
17-289
18-324
19-361
20-400

17

Squares of 21 to 25

21-441
22-484
23-529
24-592
25-625

18

Cubes of 1 to 5

1-1
2-8
3-27
4-64
5-125

19

Cubes of 6 to 10

6-216
7-343
8-512
9-729
10-1000

20

Cubes of 11 to 15

11-1331
12-1728
13-2197
14-2744
15-3375

21

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.

22

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.

23

Number of factors

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.

_________________
Download GMAT Math and CR questions with Solutions from Instructors and High-scorers:
http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html

-----------

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Daily Quote:
“Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”

24

How to find Sum of all factors of a POSITIVE integer

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 ]

25

Factorization rule

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

26

Remainders

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

Quote:
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

Quote:
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.

27

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.

28

Simple Facts:

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

29

Multiples of N

Playing with Multiples of N:


(I)
If you add/subtract multiples of number 'N', the result is also a multiple of 'N'.

Examples:
35+21 = 56 [Multiple of 7]
20-15 = 5 [Multiple of 5]


TAKEAWAY:
In general, if N is a divisor of both x and y, then N is a divisor of both x+y and x-y.


(II)
If you add/subtract a multiple of N to/from a non-multiple of N, the result is a non-multiple of N.

Example:
9-5 = 4 [(Multiple of 3) - (Non-Multiple of 3) = (Non-multiple of 3)]



(III)
If you add/subtract 2 non-multiples of N, the result could either be a multiple or a non-multiple of N.

Examples:
19+13 = 32 [(Non-Multiple of 3) - (Non-Multiple of 3) = (Non-multiple of 3)]
19+14 = 33 [(Non-Multiple of 3) - (Non-Multiple of 3) = (Multiple of 3)]

EXCEPTION:
When N = 2, two odds always sum to an even number (Multiple of 2).

30

GCF Facts

GCF Facts:


1. GCF of integers 'm' and 'n' CANNOT be larger than the difference between 'm' and 'n'.

Assume that GCF of m and n is 12. m and n are both multiples of 12. Consecutive multiples of 12 are 12 units apart from each other on the number line. Therefore, m and n CANNOT be less than 12 units apart.


2. Consecutive multiples of n have a GCF of n.

4 and 8 are multiples of 4. Thus 4 is a common factor of both the numbers. 4 and 8 are exactly 4 units apart from each other on the number line. Thus, 4 is the greatest common factor (GCF) of 4 and 8. That is why GCF of any 2 consecutive numbers is ALWAYS 1 as both are multiples of 1.

_________________
Download GMAT Math and CR questions with Solutions from Instructors and High-scorers:
http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html

-----------

GO GREEN..! GO VEG..!

Daily Quote:
“Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”