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Number Properties IQUANT LIVE SESSION

Session Layout

ConceptConcept

Test 3 GMAT 700+ QEven-Odd

1

ConceptConcept

Test 5 GMAT 700+ QPrimes2

ConceptConcept

Test 2 GMAT 700+ QLCM-GCD

3

Part 1EVEN-ODD NUMBERS


Even Odd Numbers - Properties

Odd * Odd = Odd

Every EVEN number can be represented as 2n, where n is an integer

Every ODD number can be represented as 2n+1, where n is an integer

Basic Properties:

(Even)2 = (2n)2 = 4n2

Derived Properties:

Divisible by 4

Properties2 = 2 × 1 4 = 2 × 2

6 = 2 × 3 8 = 2 × 4

3 = 2 × 1 + 1 5 = 2 × 2 + 1

7 = 2 × 3 + 1 9 = 2 × 4 + 1

Even +/- Even = Even2 + 4 = 68 – 2 = 6

Even +/- Odd = Odd2 + 3 = 52 - 3 = -1

Odd +/- Odd = Even1 + 3 = 41 – 3 = -2

Even * Even = Even

Even * Odd = Even

2 × 4 = 8

2 × 3 = 6

3 × 5 = 15

(Even)n +/- (Even)n = Even +/- Even = Even

(Odd)n +/- (Odd)n = Odd +/- Odd = Even

(Even)n +/- (Odd)n = Even +/- Odd = Odd


Test your Understanding

1. A2 + B2 = (Even, Odd, Cannot determine)

Answer: Consecutive => One Even, the other odd A2 + B2 = Even2 + Odd2

Odd

A, B, C, D are consecutive integers > 1. Then

2. A2 + B2 + C2= (Even, Odd, Cannot determine)

Answer: Consecutive => (Even, Odd, Even) OR (Odd, Even, Odd) A2 + B2 + C2 = (Even2 + Odd2 + Even2) OR (Odd2 + Even2 + Odd2) A2 + B2 + C2 = (Even + Odd + Even) OR (Odd + Even + Odd) A2 + B2 + C2 = Odd OR Even

Cannot Determine

Question = MCQ question, Answer choices: Even, Odd, Cannot be determined

Even Odd+

3. A2 + B3 + C3= (Even, Odd, Cannot determine) Answer: Cannot determine


Test your Understanding

4. B7 - D7 = (Even, Odd, Cannot determine)

Answer: Consecutive => (Even, Odd, Even, Odd)

OR(Odd, Even, Odd, Even)

B7 - D7 = (Odd2 - Odd2) OR (Even2 - Even2) B7 - D7 = (Odd - Odd) OR (Even - Even) B7 - D7 = Even

Even

A, B, C, D are consecutive integers > 1. Then

Question = MCQ question, Answer choices: Even, Odd, Cannot be determined

5. A2 + B2 + C2 + D2 = (Even, Odd, Cannot determine) Answer: Even 2 Odd numbers + 2 Even numbers = Even


Apply in GMAT Context: Question 1

Is an – bn odd, if a, b, and n are positive integers?

1. a and b are squares of consecutive natural numbers

2. a2 + b2 is odd

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.

Data Sufficiency Question Solving Process

Step 1: Understand the question

Step 2: Draw Inferences

Step 3: Analyze Statement 1 independently

Step 4: Analyze Statement 2 independently

Step 5: Analyze both Statements together



2. a2 + b2 is odd


Question 1 – Steps 1 and 2 – Understand question statement and Draw Inferences

Question Statement - Is an – bn odd, if a and b are positive integers?

Given –

1. a, b, n > 0

2. a, b, n are integers

To find –

Is an – bn odd?

One of them is odd and the other is even

an – bn odd Even Odd

an

x

bn

x

a

b

Is an – bn odd? Is only one of a or b odd?

Power does not change the even/odd nature of a number• (Even)n = Even• (Odd)n = Odd


Question 1 – Step 3 – Analyze Statement 1 Independently

Statement 1 - a and b are squares of consecutive natural numbers

Is an – bn odd?

Is only one of a or b odd?

Consecutive natural numbers P, P+1

P2, (P+1)2{a, b}

If P is even P+1 is odd

Each case is an even-odd pair

Answers the question – YES! Only one of a or b is odd

Statement 1 is sufficient

a, b integers >0

P2 is even (P+1)2 is odd

If P is odd P+1 is even

P2 is odd (P+1)2 is even



Statement 2 - a2 + b2 is odd

(a,b) is an even-odd pair



Correct answer = Choice D = Either statement is sufficient to answer the question

Is an – bn odd?

Is only one of a or b odd?

a, b integers >0

a a2 b b2 a2 + b2

Even Even Even Even Even

Even Even Odd Odd Odd

Odd Odd Even Even Odd

Odd Odd Odd Odd Even



If P = k3 – k, where k and P are positive integers, is P divisible by 4?

1. k = (10x)n + 54 where x and n are positive integers

2. (2n+1)k leaves a remainder when divided by 2; n is a positive integer








Question Statement - If P = k3 – k, where k and P are positive integers, is P divisible by 4?

?Given –

P = k3 – kTo find –

If P is divisible by 4.

= k(k2-1)

= k(k-1)(k+1)

= (k-1) k(k+1) = Product of 3 consecutive integersCase 1

Case 2

(k-1) : even

(k+1) : even

=2m

=2m + 2product has 4 P always divisible by 4

(k-1) : odd

(k+1) : odd

=2m

product has 2

P divisible by 4 if

k is divisible by 4

k: odd

k: evenP may or may not be divisible by 4

P is divisible by 4 if either1. k is odd or2. k is even & k is divisible by 4



Statement 1: k = (10x)n + 54 where x and n are positive integers

Answers the question – P is divisible by 4


P = (k-1) k(k+1) Is P is divisible by 4?P is divisible by 4 if either1. k is odd or2. k is even & k is divisible by 4

k = (10x)n + 54

= 2n (5x)n + 54

even odd+

k is odd



Statement 2: (2n+1)k leaves a remainder when divided by 2



P = (k-1) k(k+1) Is P is divisible by 4?P is divisible by 4 if either1. k is odd or2. k is even & k is divisible by 4

(2n + 1) k is odd

oddodd

k is odd




If P = k3 – k2, where k and P are positive integers, is P divisible by 4?

1. k = (10x)n + 54 where x and n are positive integers and n > 1.








Question 2 vs. 3

If P = k3 – k, where k and P are positive integers, is P divisible by 4?

1. k = (10x)n + 54 where x and n are positive integers


If P = k3 – k2, where k and P are positive integers, is P divisible by 4?

1. k = (10x)n + 54 where x and n are positive integers and n > 1.


Q3

Q4



Question Statement - If P = k3 – k2, where k and P are positive integers, is P divisible by 4?

?Given –

P = k3 – k2

To find –

If P is divisible by 4.

= k2(k-1)

Case 1

Case 2

(k-1) : even =2n

product has 4 P always divisible by 4(k-1) : odd

=2n

product has 2

P divisible by 4 if

(k-1) is divisible by 4

k: odd

k: even

P may or may not be divisible by 4

P is divisible by 4 if either1. k is even or2. k is odd & k-1 is divisible by 4

k2: odd

k2: even =(2n)2



Statement 1: k = (10x)n + 54 where x and n are positive integers and n>1



P = k2(k-1) Is P is divisible by 4?

k = (10x)n + 54

= 2n (5x)n + 54

even odd+

k is odd


Is k – 1 divisible by 4?

k – 1 = (10x)n + 54 - 1

n > 1 n ≥ 2 100xn 625 1+ -=

Tens and units digit = 24

(k-1) is divisible by 4



Statement 2: (2n+1)k leaves a remainder when divided by 2; n is a positive integer.

Does not answer the question – P is divisible by 4

Statement 2 is NOT sufficient

(2n + 1) k is odd

oddodd

k is odd

Correct answer = Choice A = Statement 1 is sufficient but statement 2 is not

Is P is divisible by 4?


Don’t know if k-1 is divisible by 4

P = k2(k-1)

Part 2PRIMES


Prime Number & Factors

1. A prime number is a positive integer that has

exactly two different positive factors, 1 and

itself. Examples: 2, 3, 5, 7 . . .

2. 0 and 1

are neither Prime nor Composite.

--- Do not have TWO different positive factors

3. Every positive integer K can be expressed as K

= P1m × P2

n × P3r ……, where P1, P2, P3 …… are

prime factors and m, n , r are non-negative

integers

Basic Properties2 is the smallest Prime number

2 and 3 are the only consecutive numbers

that are prime

2 is the only even Prime number

4 = 22

6 = 2× 3

8 = 23

10 = 2 × 5

12 = 22 × 3

18= 2 × 32

1000 = 23 × 53

2400 = 25 × 3 × 52

Prime Factorization

V/S

Consecutive prime numbers


Prime Number & Factors

Total number of factors = (m+1) (n +1)(r+1) . . .

K = P1m × P2

n × P3r . . . where P1, P2, P3 …… are prime factors and m, n , r

are non-negative integers

Kn will have the same prime factors as K

PnKm will have the same prime factors as K,

if P is a prime factor of K

Prime Factors

Total Factors

Possible powers of P1 in a factor: P10, P1

1, P12 . . . P1

m


1, P22 . . . P2

n


1, P32 . . . P3

r

(m+1) values

(n+1) values

(r+1) values

Eg: 62 & 6 have the same prime factors: 2 and 3

Eg: 23 * 62 have the same prime factors as 6: 2 and 3


Prime Number & Factors – Test your Understanding

Total number of factors = (m+1) (n +1)(r+1) . . .

K = P1m × P2

n × P3r . . . where P1, P2, P3 …… are prime factors and m, n , r

are non-negative integers

Q: How many prime factors does K have if the total number of factors of K is:

a. 1

b. 2

c. 3

d. 4

e. 5

f. 6

g. 7

K = 1

K = P1

K = P12

K = P1 × P2

K = P14

K = P12 × P2

K = P16

0

1

1

2

1

2

1

A perfect square will

have an odd number

of factors


Factors/Prime Factors- Test your Understanding

If X has 3 prime factors and 8 total factors, then how many prime factors will Xn have? (FIB)

Question 1

If X has 3 prime factors and 8 total factors, then how many factors will Xn

have? (FIB)

Question 3

If K is a factor of positive integer X that has 3 Prime factors and 8 total factors, then how Prime factors does K2 Xn

have?

Question 2

Answer– 3

Answer – 3

Answer – (n+1)3

Kn will have the same

prime factors as K

X = P1 × P2 × P3

X = P1 × P2 × P3

Xn = P1n × P2

n × P3n

If K is a factor of positive integer X that has 8 total factors, then how Prime factors does K2 Xn have?

Question 4

X = P1 × P2 × P3 X = P17 X = P1

3 × P2 or or

Answer – Cannot Determine




1. a and b are squares of consecutive prime numbers

2. a2 + b2 is odd







2. a2 + b2 is odd

Even- Odd Question 1


From Even- Odd Question 1


Statement 1 - a and b are squares of consecutive prime numbers

Is an – bn odd? Is only one of a or b odd?

Consecutive natural numbers P, P+1

{a, b} {P2, (P+1)2}

Each case is an even-odd pair



Consecutive prime numbers 2, 3, 5, 7…

Squares 4, 9, 25, 49…

{a, b} {4, 9} {9, 25} {25, 49}odd-oddeven-odd

Two possible scenarios as shown

Cannot answer the question – is only one of a or b odd?


Understanding Information Given in the question is very critical.

Choice B Choice D

If P is even P+1 is odd

P2 is even (P+1)2 is odd

If P is odd P+1 is even

P2 is odd (P+1)2 is even



Is an – bn + cn + dn odd, if a, b, c, and d are positive integers >1?

1. a, b, c, and d are squares of consecutive prime numbers

2. a4 when divided by 200 has the quotient 1








Question Statement - Is an – bn + cn + dn odd, if a, b, c, and d are positive integers >1?

Given –

1. a, b, c, and d > 1

2. a, b, c, and d are integers

To find –

Is an – bn + cn + dn odd?

Is 1 or are 3 of the 4 numbers odd?

Power does not change the even/odd nature of a number• (Even)n = Even• (Odd)n = Odd

• a ± b ± c ± d = e or o?

• o ± o ± o ± o = e

• e ± o ± o ± o = o

• e ± e ± e ± o = o

• e ± e ± e ± e = e

• e ± e ± o ± o = e

An expression with sum or difference of integers is odd if odd number of terms are odd.



Statement 1 - a, b, c, and d are squares of consecutive prime numbers

Consecutive prime numbers 2, 3, 5, 7, 11, 13…

Squares 4, 9, 25, 49, 121, 169…

{a, b, c, d} {4, 9, 25, 49}

e, o, o, o

Does not answer the question – the expression can be either even or odd

Statement 1 is not sufficient

a, b, c, and d > 1Is an – bn + cn + dn odd?


{9, 25, 49, 121} {25, 49, 121, 169}

o, o, o, o o, o, o, o



Statement 2 – a4 when divided by 200 has the quotient 1

Does not answer the question – the expression can be either even or odd




a4/200 has quotient 1

200 < a4 < 400

a = 2

a = 3

a4 = 16

a4 = 81

a = 4 But we don’t know if b, c, and d are odd or even

a ≠ 1

a = 4 a4 = 256

a = 5 a4 = 625


Question 2 – Step 5 – Analyze Statements 1 & 2 Together

Statements 1& 2

Answers the question

Both together are sufficient



a = 4

Statement 1 {a, b, c, d} {4, 9, 25, 49}

e, o, o, o

{9, 25, 49, 121} {25, 49, 121, 169}

o, o, o, o o, o, o, o

Statement 2

{a, b, c, d} = {4, 9, 25, 49}

e, o, o, o

3 of the numbers are odd

Correct answer = Choice C = Both together are sufficient



What is the remainder when b is divided by a, if a and b are consecutive perfect squares and b is greater than a?

1. Both a and b are squares of prime numbers.

2. Both a and b have 3 positive factors.








Question Statement - What is the remainder when b is divided by a, if a and b are consecutive perfect squares and b is greater than a?

?Given –

1. b > a

2. a and b are consecutive perfect squares

To find –

Remainder of b/a = ?

Need to know values of b and a

Consecutive numbers 1, 2, 3, 4, 5…

Squares 1, 4, 9, 16, 25…

{a, b} {1, 4} {4, 9} {9, 16} {16, 25} …


Question 3– Step 3 – Analyze Statement 1 Independently

Statement 1 - Both a and b are squares of prime numbers.


Need to know values of a and b

Values of a, b known

Answers the question – remainder can be calculated


Per question {a, b} squares of consecutive numbers

Prime numbers 2, 3, 5, 7…

Squares of prime numbers that are consecutive

{a, b} = {1, 4} {4, 9} {9, 16} {16, 25}…

a, b are consecutive squares

{2,3}



Statement 2 - Both a and b have 3 positive factors.

Values of a, b known

Answers the question – remainder can be calculated


a = P12m × P2

2n × …

{a, b} = {4, 9}


Need to know values of a and b{a, b} = {1, 4} {4, 9} {9, 16} {16, 25}…

b = P’12r + P’2

2s +…

a, b are consecutive squares

Total number of factors of a = (2m+1)(2n+1) × …

= (2m+1)(2n+1) × …3

= (2m+1)(2n+1) × …(2x1+1)(2x0+1)

a = P12

a is square of a prime number

Similarly b is square of a prime number

{2, 3} are the only consecutive numbers that are prime


Square of prime number has 3 factors

m = 1; n . . . = 0



How many distinct prime factors does √Q have, if Q is a perfect square of a positive integer?1. Q is odd and 8Q8 has four distinct prime factors2. 8Q and Q2 do not have the same set of prime factors








Question Statement - How many distinct prime factors does √Q have, if Q is a perfect square of a positive integer?

? Given –

Q is a perfect square of a positive integer

To find –

Number of prime factors of √Q

Number of Prime Factors of Q

NOT Total number of factors!

√Q is a positive integer

Q = (√Q)2


(√Q)2 will have the same prime factors as √Q



Statement 1: Q is odd and 8Q8 has four distinct prime factors

Answers the question – Q has 3 distinct prime factors


√Q is a positive integer # of Prime factors of √Q # of Prime factors of Q

8Q8 = 23Q8

Prime factors of 8Q8 = 2, Prime factors of Q8

Q is odd

Prime factors of Q

Odd

4 1

32

If Q were even

2 is a prime factor of Q

Q has 4 prime factors




Statement 2: 8Q and Q2 do not have the same set of prime factors


√Q is a positive integer # of Prime factors of √Q # of Prime factors of Q

8Q = 23Q

Prime factors of 8Q = 2, Prime factors of Q

Prime factors of Q2

Xn has the same prime factors as X

Q is oddOdd

How Many?

Correct answer = Choice A = Statement 1 is sufficient but statement 2 is not


IF STATEMENT II IS NOT DECOUPLED FROM STATEMENT 1

Question 4 – ERROR ALERT!!!


Prime factors of Q

Odd

4 1

32

8Q = 23Q


Prime factors of Q2

Q is oddOdd

How Many?

Statement 1 Statement 2

You may assume that there are 4 factors of 8Q

And you may consider Statement 2 to be sufficient as well

Answer = Choice D = Either statement is sufficient to answer the question



How many distinct factors does √Q have, if Q is a perfect square of a positive integer?1. Q is odd and 8Q8 has four distinct prime factors2. 8Q and Q2 do not have the same set of prime factors








Question Statement - How many distinct factors does √Q have, if Q is a perfect square of a positive integer?

Given –

Q is a perfect square of a positive integerTo find –

Number of distinct factors of √Q

Each prime factor of Q has even power

Q = P12m × P2

2n × P32q × . . .

Where P1 , P2 , P3 . . . are prime numbers

m, n, q . . . are positive integers

Total number of factors of Q = (2m+1)(2n+1)(2q+1) …

√Q = P1m × P2

n × P3q × . . .

(m+1)(n+1)(q+1) …



Statement 1: Q is odd and 8Q8 has four distinct prime factors


Q = P12m × P2

2n × P32q × . . . # of factors of √Q (m+1)(n+1)(q+1) . . .

8Q8 = 23Q8


Q is odd

Prime factors of Q


Odd

4 1

32

Q = P12m × P2

2n × P32q m, n, q



Statement 2: 8Q and Q2 do not have the same set of prime factors


8Q = 23Q


Prime factors of Q2


Q is oddOdd

How Many?

Q = P12m × P2


What powers?


Question 5 – Step 5 – Analyze Statements 1 & 2 Together

Statements 1& 2

Statement 1

Statement 2

Correct answer = Choice E = Both together are not sufficient

Q = P12m × P2


Q = P12m × P2

2n × P32q

P1 ≠ P2 ≠ P3 ≠ 2

Q is oddStatement 1 + 2 are NOT sufficient

m, n, q

3 prime factors, all odd

Part 3LCM AND GCD


Concept Slide on GCD and LCM

GREATEST COMMON DENOMINATOR LEAST COMMON MULTIPLE

40

1. Find the prime factors of the given numbers

40 = 2 × 2 × 2 × 5 98 = 2 × 7 × 7

2. Write the prime factors in exponential form

98 = 2 × 7240 = 23 × 5

3. Pick the SMALLEST power of each prime factor

4. Multiply the numbers from 3.

GCD

2 5 71 0 0

21 × 50 × 70 = 2

1. Find the prime factors of the given numbers

40 = 2 × 2 × 2 × 5 98 = 2 × 7 × 7

2. Write the prime factors in exponential form

98 = 2 × 7240 = 23 × 5

3. Pick the GREATEST power of each prime factor

4. Multiply the numbers from 3.

LCM

2 5 73 1 2

23 × 51 × 72

98× = GCD × LCM


GCD - Test your Understanding

If GCD of two numbers (both integers, greater than 1) is 1, then which of the following can be true?

1. They are prime.2. They are consecutive.3. They do not have a common prime factor4. They do not have a common factor other than 1

I. Only 1II. Only 2III. Only 3 and 4IV. Only 2 and 3V. 1, 2, 3 and 4

Answer V – 1,2,3,4

If GCD of two numbers (both integers, greater than 1) is 1, then which of the following must be true?

1. They are prime.2. They are consecutive.3. They do not have a common prime factor4. They do not have a common factor other than 1

I. Only 1II. Only 2III. Only 3 and 4IV. Only 1 and 4V. 1, 2, 3 and 4

Answer III – Only 3 and 4

Question 1 Question 2

Question = MCQ question, Answer choices: A, B, C, D, E


LCM - Test your Understanding

Answer: Only #3 and #5 Answer: 1, 2, 4, 5

If the LCM of two integers a, b (where b> a and a>1) is b, then which of the following can be true?

1. Both a and b can be Prime Numbers.2. Both a and b can be consecutive integers.3. All prime factors of a must be prime

factors of b.4. All prime factors of b must be prime

factors of a.5. b must be a multiple of a.

Question 1 (MAQ)

If the LCM of two integers a, b (where b> a and a>1) is a*b, then which of the following can be true?

1. Both a and b can be Prime Numbers.2. Both a and b can be consecutive integers.3. All Prime factors of a must be Prime

factors of b.4. a and b do not share any Prime factors.5. a and b do not have a common factor

Question 2 (MAQ)


LCM- Test your Understanding

How prime factors does K2 Xn have, if K is Prime, X has 3 Prime factors and the LCM of K and X is KX?

Question 4

Answer = 3+ 1 = 4

Question 3

If the LCM of two integers a, b where b> a and a>1 is a*b/5, then what is the GCD of a & b?

Answer: 5, property used a*b = LCM * GCD

K and X have NO prime factors in common

1 Prime factor

3 Prime factors



Does P have a factor X where1<X<P and X and P are positive integers?

1. GCD (P2, k) = k, where k is a prime number

2. 36*20 + 2 < P < 36*20+6








Question Statement - Does P have a factor X where1<X<P and X and P are positive integers?

Given –

X, P are positive integers

To find –

Does P have a factor X between 1 and P?

Is P Prime?

1 < X < P

P > 1

A prime number has only two factors: 1 and the number itself

A composite number will have at least one factor between 1 and the number itself

Think:

4: {1, 2, 4}

12: {1, 2, 3, 4, 6, 12}



Statement 1: GCD (P2, k) = k, where k is a prime number


X, P are positive integers Does P have a factor X between 1 and P?

Is P prime?

P2 is divisible by k

k is a prime number

k is a prime factor of P2 k is a prime factor of P


P = nk

If n = 1P is Prime

If n ≠ 1P is NOT Prime

Think:Let k = 5 P = 5n

(n is an integer)



Statement 2: 36*20 + 2 < P < 36*20+6

X, P are positive integers Does P have a factor X between 1 and P?

Is P prime?

Possible values of P:

36*20 + 3 36*20 + 4 36*20 + 5

3(12*20 + 1) 4(8*20 + 1) 5(36*4 + 1)

Divisible by 3 Divisible by 4 Divisible by 5

NOT Prime NOT Prime NOT Prime


Correct answer = Choice B = Statement 2 is sufficient but statement 1 is not

You need to be smart about simplifying information in order to arrive at the answer.



If P and Q are positive integers and Q = 10 + 4P, find the GCD of P and Q

1. Q = 10 x, where x is a positive integer

2. P = 10 y, where y is a positive integer








Question Statement - If P and Q are positive integers and Q = 10 + 4P, find the GCD of P and Q

Given –

P, Q are positive integers

To find –

GCD of P and Q

Q = 10 + 4P

P = (Q- 10)/4



Statement 1: Q = 10 x, where x is a positive integer


Q = 10 + 4P P = (Q- 10)/4GCD of P and Q = ?P, Q are integers

P = (Q- 10)/4

P = (10x- 10)/4

P = 5(x- 1)/2P is an integer

x – 1 is even

ODD

x P Q GCD (P,Q)

3 5 30 5

5 10 50 10

7 15 70 5

GCD may be 5 or 10



Statement 2: P = 10 y, where y is a positive integer


Q = 10 + 4P P = (Q- 10)/4GCD of P and Q = ?P, Q are integers

Q = 10 + 4P

Q = 10 + 4(10y)

Q = 10(4y + 1)

y P Q GCD (P,Q)

1 10 50 10

2 20 90 10

3 30 130 10

GCD is always10

Correct answer = Choice B = Statement 2 is sufficient but statement 1 is not

Final Words


Key Takeaways

Know your concepts well.

Go through the concept files in detail and retain the basic and derived properties.

Follow the process with due diligence

Steps 1 & 2 – Understand the question and draw inferences

Step 3 – Analyze Statement 1

Step 4 – Analyze Statement 2

Practice how to simplify information in order to answer the question at hand

thoroughly completely

Decouple from statement 1

Will come with practice


Next Steps

Analyze your performance in the Live

Session

Work on Weak Areas

◦ In Quant Online

◦ Number Properties block

Attempt Advanced Quiz

Prepare for next Session

◦ In Quant Online

◦ Divisibility and Remainders, Rounding, Statistics

e gmat number properties

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