GRE: Number System

 

Natural Numbers:- These are the numbers (1,2,3 etc) that are used for counting. In other words, all positive integers are natural numbers.

Prime Numbers:- A natural number larger than unity is a prime number if it does not have other divisors except for itself & unity.

Whole Numbers:- The set of numbers that includes all natural numbers and the number zero are called whole numbers.

Composite Numbers:- It is a natural number that has at least one divisor different from unity and itself.

Real Numbers:- All numbers between negative infinity & positive infinity that can be represented on a number line are called real numbers.

Rational Numbers:- A rational numbers is defined as a number of the form (a/b) where a and b are integers and b ≠ 0.

Rational numbers that are not integers have decimal values, can be of two types:

1.Terminating Decimal Fraction

2.Non- Terminating Decimal Fractions

Terminating decimal fractions :- For example  17/4 = 4.24  ,  21/5 = 4.2

Non-Terminating decimal fractions: Non-terminating decimal fractions are of two types:

Non-Terminating Repeating Fractions: These are a non terminating repeating fraction. For Example 5.333…..,   6.161616…..

Neither terminating nor repeating fractions: For example 5.273612…..

 

Concept of GCD (Greatest Common Divisor or Height Common Factor)

Consider two natural numbers n1 and n2.

If the numbers n1 and n2 are exactly divisible by same number x, then x is a common divisor of n1 and n2.

The highest of all the common divisors of n1 and n2 is called as the GCD or the H.C.F:-

 

Steps for calculating HCF

Suppose we have to find H.C.F of 150,210 and 375.

Step 1:- Writing down the standard form of numbers

150=5×5×3×2

210=5×2×7×3

375=5×5×5×3

Step 2:- Writing prime factors common to all the three numbers are taking their product is 51  ×  31

Step 3:- H.C.F = 5×3= 15

 

 Concept of LCM (Least common multiple)

Let n1 and n2 be two natural numbers distinct from each other. The smallest natural number n that is exactly divisible by n1 and n2 is called the least common multiple (LCM)

Steps for calculation of LCM:-

Suppose we have to find LCM of 150,210,375.

Step 1:- Write down the standard form of numbers

150=5×5×3×2

210=5×2×7×3

375=5×5×5×3

Step 2:-  Write down all the prime factors that appear at least once in any of the numbers: 5,3,2,7.

Step3: -Raise each of the prime factor to their highest available power

The LCM= 2 × 3 × 53 × 7 = 5270

 

HCF OF TWO OR MORE FRACTION IS GIVEN BY :

=HCF of Numerators Divided By LCM Of Denomination

 

LCM OF TWO OR MORE FRACTIONS IS GIVEN BY:      

=LCM of Numerators  Divided By HCF of Denominators

 

Problem 1) Three traffic signals change after 36,42 and 72 seconds respectively.If the lights are first switched on at 9:00 sharp. At What time will they change simultaneously?

Sol. In order to distribute $4000 and 180 pencils evenly, the numbers of employees must be a factor of each of these two numbers. We have to find the greatest number of employees possible therefore, we have to find HCF of $4000 and 180.

4000=2×2×2×5×2×5×2×5 =25  ×  53

180= 2×9×2×5 = 22  ×  32 × 5

HCF = 22 × 5  = 20

Correct Answer = 20

 

Problem 2) The number of students who attended a school could be divided among 10,12 or 16 buses, such that each bus transports an equal number of students that could attend the school?

Sol.  No, of students must be divisible by 10,12 and 16. We have to find a minimum number of students that could attend the school or we have to find a minimum number that is divisible by 10,12 and 16.

LCM of 10,12,16                            10= 2×5

=24 × 3×5                                       12=2×2×3

=16×15                                          16=2×2×2×2

ANS:-  (240)

 

Problem 3) There are 576 boys and 448 girls in a school that are to be divided into sections containing equal no. of students of either boys or girls alone. Find the minimum total number of sections thus formed

a) 24

b) 32

c) 16

d) 20

Sol.  We have to find HCF of 576 and 448

576= 32  × 26

448= 26  × 7

HCF= 26 = 64

Number of sections = 576/64 + 448/64  = (9+7) = \({ 16 }^{ Ans }\)

 

Divisibility:- Any integer I (Commonly called dividend)

When divided by a natural number N (Called Divisor), there exist a unique pair of number Q and R ,Which are called the quotient and remainder respectively.

Dividend=Divisor × Quotient + Remainder

Suppose We divide 42 by 5. The result has a quotient of 8 and remainder of 2.

\(Let\quad N={ p }_{ 1 }^{ { \alpha }_{ 1 } }.{ p }_{ 2 }^{ { \alpha }_{ 2 } }.{ p }_{ 3 }^{ { \alpha }_{ 3 } }—–{ p }_{ k }^{ { \alpha }_{ k } }\quad Where\quad { p }_{ 1 },{ p }_{ 2 },{ p }_{ 3 }—–{ p }_{ k }\\ are\quad different\quad primes\quad and\quad { \alpha }_{ 1 },{ \alpha }_{ 2 },{ \alpha }_{ 3 }——{ \alpha }_{ k }\quad are\quad natural\quad numbers.\\ Total\quad number\quad of\quad divisors\quad of\quad N\quad =\\ =\quad (1+{ \alpha }_{ 1 })(1+{ \alpha }_{ 2 })(1+{ \alpha }_{ 3 })——-(1+{ \alpha }_{ k })\\ Sum\quad of\quad these\quad divisors\quad =\\ =\quad ({ p }_{ 1 }^{ 0 }+{ p }_{ 1 }^{ 1 }+{ p }_{ 1 }^{ 2 }—–{ p }_{ 1 }^{ { \alpha }_{ 1 } })({ p }_{ 2 }^{ 0 }+{ p }_{ 2 }^{ 1 }+{ p }_{ 2 }^{ 2 }——{ p }_{ 2 }^{ { \alpha }_{ 2 } })——-({ p }_{ k }^{ 0 }+{ p }_{ k }^{ 1 }—-{ p }_{ k }^{ { \alpha }_{ k } })\\\)

 

Problem 4) Find the sum of factors and the number of factor of 240?

Sol.  240 = 24 × 41 × 51

Total Number of divisor = (1+4) (1+1) (1+1)

=5×2×2= 20

Sum of these divisors

= (20 + 21 + 22 + 24 ) (30 + 31) (S0 + S1)

= (1+2+4+8+16) (4) (6)

=24 (31) = (744)

 

Problem 5)

Quantity A  The number of positive factor of 20.      

Quantity B  The number of positive factor of 32.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  32 = 25                                        ,                               20= 22  51

Number of distinct positive factors of 32 = (6)

Number of distinct positive factors of 20 = 3×2 = 6

The two quantities are equal.

 

The highest power of a prime number p, which divides n! exactly is given by

= [n/p] + [n/p2] +[n/p3]……..

Where [x] denotes the greatest integer less than or equal to x

 

Remainder Theorem:-

The remainder of the expression

[A1 × B1 × C1 ………………+ A2 × B2 × C2] / m

will be the same as the remainder of the expression

[A1R × B1R × C1R ………………+ A2R × B2R × C2R] / m

Where A1R  is the remainder when A1 is divided by M. B1R is the remainder when B1 is divided by M and so on.

 

Problem 6) Find the remainder when 17 ×23×126×38 divided by 8

Sol.     ( 17 ×23×126×38 / 8 )

              = (1×7×6×6 / 8 ) = ( 42×6 / 8) = (2*6 / 8 )     =    ( 12 / 8 )

 Remainder=   ( 4 )

 

Problem 7) If a=16 b and b is a prime number greater than 2, How many positive distinct factors does a have?

Sol.   a =16b = 24. b1

No. of positive distinct factors = ( 4 + 1 ) ( 1 + 1 )    = (10)

 

Problem 8) If x is a positive integer, which one of the following could be the remainder. When 73x  is divisible by

a) 0

b) 1

c) 2

d) 3

e) 4

f) 5

g) 6

h) 7

i) 8

j) 9

Indicate all such remainders

 Sol.  We have to find the unit digit of 73x

As with multiplication, when an integer is raised to a power, the unit digit is determined solely by the product of units digits.

Those products will form a repeating pattern.

31= 3. 32 =9. 33 =27. 34 = 81 and 35 =243

Here  the pattern returns to its original value of 3 and we have got a repeating pattern

3,9,7,1,3,7,1,3 —–

Thus the unit digit of  73x  must be 1,3,7,9

 

Divisibility Rules:

Divisibility by 2-A number is divisible by 2 if the last digit is divisible by 2.

Divisibility by 5 – A number is divisible by 5 if the last digit is 5 or 0

Divisibility by 4 – A number is divisible by 4 if the last  two digit  are divisible by 4

Divisibility by 8 – A number is divisible by 8 if the last  3 digit  are divisible by 8

Divisibility by 11 – A number is divisible by 11 if the difference of the sum of the digits in the odd places and the sum of the digits in the even places is zero or divisible by 11.

 

Problem 9) w, x, y and z are consecutive odd integers such that \(w<x<y<z.\) which of the following statements must be true?

Indicate all such statements.

a) wxyz is odd.

b) w+x+y+z is odd

c) w+z=x+y

Sol.  Odd×odd = odd

wx is odd. yz is odd and wxyz is also odd.

Odd+odd = even

Even+even=even

w+x=even, y+z=even

Therefore,w+x+y+z is also even.

w,x,y and z are consecutive odd integers, all can be defined in terms of w.

w=w, y=w+4

x=w+z, z=w+6

w+z=2w+6, x+y=2w+6

Therefore, w+z=x+y

 

Problem 10)

Quantity A Remainder when \({ (67) }^{ 99 }\) is divided by 7.

Quantity B Remainder when \({ (75) }^{ 80 }\) is divided by 7.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  Using remainder theorem

\((\frac { { 67 }^{ 99 } }{ 7 } )\rightarrow (\frac { { 4 }^{ 99 } }{ 7 } )=\frac { { { (4 }^{ 3 }) }^{ 33 } }{ 7 } =\frac { { 64 }^{ 33 } }{ 7 } \\ Remainder\quad =\quad 1\\ \frac { { 75 }^{ 80 } }{ 7 } \rightarrow \frac { { 5 }^{ 80 } }{ 7 } \rightarrow \frac { { ({ 5 }^{ 6 }) }^{ 13 }\times { 5 }^{ 2 } }{ 7 } \rightarrow \frac { { 1 }^{ 13 }\times 25 }{ 7 } \\ Remainder\quad =\quad 4\)

Quantity B is greater.

 

Problem 11) If set S consists of all positive integers that are multiples of both 2 and 7, how many numbers in set S are between 140 and 240, inclusive?

Sol.  Correct answer is 8.

A positive integer that is a multiple of both 2 and 7 is a multiple of 14. Since 140 is a multiple of 14, start listing there and count the terms in the range:

140, 154, 168, 182, 196, 210, 224, 238

 

Problem 12) If then which of the following must be true?

a) \({ a }^{ 2 }>a>b>{ b }^{ 2 }\)

b) \(b>a>{ a }^{ 2 }>{ b }^{ 2 }\)

c) \({ b }^{ 2 }>a>{ a }^{ 2 }>b\)

d) \({ b }^{ 2 }>{ a }^{ 2 }>b>a\)

e) \({ b }^{ 2 }>b>a>{ a }^{ 2 }\)

Sol.  The correct answer is \(({ b }^{ 2 }>b>a>{ a }^{ 2 })\)

The goal in this question is to order \(a,{ a }^{ 2 },b\quad and\quad { b }^{ 2 }\) by magnitude. Based on the original inequality

\(0<a<\frac { 1 }{ b } <1\), several things are true.

\(\frac { 1 }{ b } <1\quad \& \quad b>1\)

Therefore, \({ b }^{ 2 }>b\)

Second, note that \(a<1\) in the given inequality.

a is a positive number less than 1,

\({ a }^{ 2 }<a<1\\ { b }^{ 2 }>b>1\\ { ({ b }^{ 2 }>b>a>{ a }^{ 2 }) }^{ Ans }\)

 

Problem 13) On a number line, the distance from A to B is 4 and the distance from B to C is 5.

Quantity A  The distance from A to C

Quantity B   9

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  If the points A, B and C are in alphabetical order from left to right, then the distance from A to C will be 9. However, alphabetical order is not required. If the points are in the order C, A and B from left to right,then the distance from A to C is 5-4=1. Therefore, the relationship cannot be determined.

 

Problem 14) The integer a is even and the integer b is odd. Which of the following statements must be true?

Indicate all such statements

a) a+2b is even

b) 2a+b is even

c) ab is even

d) \({ (a) }^{ b }\) is even

e) \({ (a+b) }^{ 2 }\) is even

f) \({ a }^{ 2 }-{ b }^{ 2 }\) is odd

Sol.  If b is odd, then 2b must be even.

even+even = even

(a+2b) must be even

even+odd = odd

2a+b must be odd.

even×odd = even

ab must be even.

\({ (even) }^{ odd }\) must be even.

\({ (a) }^{ b }\) must be even

\({ (odd) }^{ even }\) must be odd

\({ (odd) }^{ odd }\) must be odd

(a+b) is odd

Therefore, \({ (a+b) }^{ 2 }\) must be odd.

\({ (a) }^{ 2 }\) must be even and \({ (b) }^{ 2 }\)  must be odd.

\(({ a }^{ 2 }-{ b }^{ 2 })\) must be odd.

 

Problem 15) Find the number of zeros in 137!?

Sol.  The highest power of a prime number p, which divides n! exactly is given by

\(\left[ \frac { n }{ p } \right] +\left[ \frac { n }{ { p }^{ 2 } } \right] +\left[ \frac { n }{ { p }^{ 3 } } \right] ——-\)

Where \(\left[ x \right]\) denotes the greatest integer less than or equal to x.

\(\left[ \frac { 137 }{ 5 } \right] +\left[ \frac { 137 }{ 25 } \right] +\left[ \frac { 137 }{ 125 } \right] \\ =1+5+27=\quad 33\quad zeros\)

Since the restriction on the number of zeroes is due to the number of fives.

Therefore, 137! Has 33 zeros.

 

Problem 16) Find the number of numbers from 1 to 100 which are not divisible by any one of 2 & 3.

a) 16

b) 17

c) 18

d) 33

Sol.  Number of numbers not divisible by 2 & 3 = Total no. of numbers – number of numbers divisible by either 2 or 3.

Number of numbers divisible by 2 = 50

Numbers divisible by 3 (but not by 2, as it has already been counted) are 3,9, 15, 21, 27, 33—–93, 99.

Number of numbers divisible by 3 (but not by 2)

\(=\frac { (99-3) }{ 6 } +1=17\)

Hence, number of numbers divisible by either 2 or 3

= 50+17 = 67

Number of numbers not divisible by any of 2 & 3 =

Total number of numbers – no. of numbers divisible by either 2 or 3

= 100 – 67

= \({ 33 }^{ Ans }\)

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