Chapter-1, REAL NUMBERS Extra Questions

Explanation:

The even integers are 2,4,6…

So, it can be written in the form of 2m.

Where m=integer=Z [since integer is represented by Z]

or m=…,1,0,1,2,3…

Therefore 2m=…,2,0,2,4,6

Alternative Method

Let 'a' be a positive integer. On dividing 'a' by 2, let m be the quotient and r be the remainder. Then, by Euclid's division algorithm, we have

a=2m+r, where 

a≤r≤2 i.e. r=0 and r=1

implies a=2m or a=2m+1

When, a=2m for some integer m, then clearly a is given.

Explanation:

The odd integers are 1,3,5….

So, it can be written in the form of 2q+1.

Where, q=integer=Z

where q=integer=Z

or q=…,-1,0,1,2,3…..

Therefore 2q+1=…,-3,-1,1,3,5…

Alternative method

Let 'a' be given a positive integer. On dividing 'a' by 2, let q be the quotient and r be the remainder. Then by Euclid's division algorithm, we have

a=2q+r, where 0≤r≤2

implies a=2q+r, where 0 or r=1

implies a=2q or 2q+1

When, a=2q+1 for some integer q, then clearly a is odd.

Explanation:

Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p.

 Let n=4p+1,

 (n² –1)=(4p+1)² –1=16p² +8p+1=16p² +8p=8p(2p+1)

Implies (n² –1) is divisible by 8. 

(n² –1)=(4p+3)² –1=16p² +24p+9–1=16p² +24p+8=8(2p² +3p+1)

Implies n² –1 is divisible by 8. 

Therefore, n 2 –1 is divisible by 8 if n is an odd positive integer.

Explanation:

The correct answer is (B) False

By Euclid's division algorithm

b = aq+r, 0≤r implies 117=65×1+52

implies 65=52×1+13

implies 52=13×4+0

Therefore HCF (65, 117) = 13 …..(i)

Also, given that, HCF (65, 117) = 65 m - 117 …..(ii)

From the eq (i) and (ii), we get

65m - 117 = 13

65m = 130

m = 2.

Hence, the given statement is false.

Explanation:

We have from Question that,

a = x³y² = x × x × x × y × y

b = xy³ = x × y × y × y

The HCF (Highest Common Factor) of two or more numbers is the highest number among all the common factors of the given numbers.

As HCF is the product of the smallest power of each common prime factor involved in the numbers.

HCF of a and b = HCF (x³y², xy³)

= x × y × y

= xy²

Therefore, HCF (a, b) is xy², So option  B is the right answer.

Explanation:

The answer  is C)

Given,

p=ab²

=a×b×b

q=a³

b=a×a×a×b

The least common multiple of p,q is a³b².

Explanation:

The product of a non-zero rational and an irrational number is always irrational.

Therefore, the answer is a. always irrational.

Explanation:

The LCM of numbers from 1 to 10 needs to be found out to find the least number We know that 2, 3, 5 and 7 are prime numbers

Prime factorization of

4 = 2 × 2

6 = 2 × 3

8 = 2 × 2 × 2

9 = 3 × 3

10 = 2 × 5

So we get required least number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520

Therefore, the least number that is divisible by all the numbers between 1 and 10 is 2520.

Explanation:

Given, 14587/ 1250

The denominator can be written as

= 14587/2x5^4

= 14587/(10 × 5³) × 2³/2³

By further calculation

= (14587 × 8)/(10 × 1000)

So we get

= 116696/10000

= 11.6696

Therefore, it will terminate after four decimal places.

Explanation:

Let a positive integer 'a' is divided by 3,

a = 3q + r, where q and r both are natural numbers and 0≤r<3

Since r is a natural number and  0≤r<3

Therefore, r can be 0, 1 or 2

Hence, the value of the remainder, r, when a positive integer 'a' is divided by 3, is 0, 1, 2.

Explanation:

Here 6n = (2 × 3)n = 2n × 3n ,

2 and 3 are the only primes in the factorization of 6n, and not 5.

Therefore, it cannot end with the digit 5

Explanation:

All the numbers of form 4q + 2, where 'q' is an integer, are even numbers that are not divisible by '4'.

For eg:

When q=1,

4q+2 = 4(1) + 2= 6.

When q=2,

4q+2 = 4(2) + 2= 10

When q=0,

4q+2 = 4(0) + 2= 2 and so on.

So, any number which is of form 4q+2 will give only even numbers that are not multiples of 4.

Hence, every positive integer cannot be written in the form 4q+2

Explanation:

Let the two consecutive positive integers = a, a + 1

According to Euclid’s division lemma,

We have below equation,

a = bq + r, where 0 ≤ r < b

For b = 2, we have a = 2q + r, where 0 ≤ r < 2 ... (i)

Substituting r = 0 in equation (i),

Here we will have,

a = 2q, is divisible by 2.

a + 1 = 2q + 1, is not divisible by 2.

Substituting r = 1 in equation (i),

We get,

a = 2q + 1, is not divisible by 2.

a + 1 = 2q + 1+1 = 2q + 2, is divisible by 2.

Thus, we can conclude that, for 0 ≤ r < 2, one out of every two consecutive integers is divisible by 2. So, the product of the two consecutive positive numbers will also be even.

Hence, the statement “product of two consecutive positive integers is divisible by 2” is true.

Explanation:

let the integer be=x,x+1,x+2

x(x+1)(x+2)=6

x(x+1)(x+2)=1×2×3

compare x=1,x+1=2

so numbers are 1,2,3

Hence given statement is true.

Explanation:

No, the square of any positive integer cannot be written in the form 3m + 2 where m is a natural number.

From Euclid’s division lemma, we know that

A positive integer ‘a’ can be written in the form bq + r

a = bq + r,

where b, q, and r are any integers,

For b = 3

a = 3(q) + r,

where r can be an integer,

For r = 0, 1, 2, 3……….

3q + 0, 3q + 1, 3q + 2, 3q + 3……. are positive integers,

(3q)² = 9q²

= 3(3q²)

= 3m (where 3q² = m)

(3q + 1)² = (3q + 1)²

Using the algebraic identity (a + b)² = a² + b² + 2ab– (a)

= 9q² + 1 + 6q

Taking out 3 as common

= 3(3q² + 2q) + 1

= 3m + 1 (Where, m = 3q² + 2q)

(3q + 2)² = (3q + 2)²

Using the algebraic identity –   equation (a)

= 9q² + 4 + 12q

Taking out 3 as common

= 3(3q² + 4q) + 4

= 3m + 4 (Where, m = 3q² + 2q)

(3q + 3)² = (3q + 3)²

Using the algebraic identity –   equation (a)

= 9q2 + 9 + 18q

Taking out 3 as common

= 3(3q² + 6q) + 9

= 3m + 9 (Where, m = 3q² + 2q)

Therefore, the square of any positive integer cannot be of the form 3m + 2.

Explanation:

By Euclid's division algorithm, a=bq+r where a,b,q,r are non-negative integers and 0≤r<b.

On putting b=3 and r=1 we get 

a=3q+1 

Squaring both sides

⇒a^2 =(3q+1)^2

⇒a^2 =(3q)^2 +(1)^2 +2(3q) 

⇒a^2 =3(3q^2 +2q)+1 

⇒a^2 =3m+1 , where $$ m = 4q^2 + 2q $ is any integer.

Hence, the square of a positive integer of form 3q+1 can not be written in any form other than 3m+1.

Explanation:

Since, the common factors of 525 and 3000 are 3, 5, 15, 25, 75.

So, HCF (525 and 3000) = 75

Finding HCF by Euclid's Lemma, 

3000=525×5+375

[∵ Dividend=Divisor×Quotient+Remainder]

525=375×1+150

375=150×2+75

150=75×2+0

⇒ HCF (525 and 3000) = 75

The numbers 3, 5, 15, 25, and 75 divide the numbers 525 and 3000. It means these terms are common in both 525 and 3000.

So, the highest common factor among these is 75.

Explanation:

3×5×7+7=105+7=112. 112 is an even number and is therefore a composite number. Also, 112 is divisible by other numbers such as 2,4,7,8,14,16,28, and 56 apart from itself 112 and 1

Hence 3×5×7+7 is a composite number.

Explanation:

HCF of the two numbers is 18

LCM of the two numbers is 380

We know that the LCM of two numbers is always divisible by their HCF

Here, 380 is not divisible 18.

So, the LCM of the two numbers is not divisible by their HCF.

Thus, the same two numbers cannot have LCM as 380 and HCF as 18.

Explanation:

If a fraction, in it's lowest terms, has no other

Prime factors except 2 and 5 can be expressed

as a terminating decimal.

first, write the fraction in its lowest

terms for that

Write the numbers into a product of prime

and find the HCF of them

987 = 3 × 7 × 47

10500 = 2 × 2 × 3 × 5 × 5 × 5 × 7

HCF of 987 and 10500 = 3 × 7 = 21

Therefore,

Least form of the fraction = 987 / 10500

Divide the numerator and denominator with HCF,

we get,

= ( 987 / 21 ) / ( 10500 / 21 )

= 47 / 500

987 / 10500 = 47 / 500

Denominator = 500

= 2^2 × 5^2

We have only 2 and 5 as factors .

Therefore, 987 / 10500 is a terminating decimal.

Explanation:

We know that the rational number having denominators 3,7,9,11,13,17,23,27... and multiple of these numbers, will have a non-terminating decimal expansion.

Here the decimal expansion is 327.7081, which is terminating.

Then, the prime factors of q, when this number is expressed in the form  p/q  will be 2 or 5 or both.

Explanation:

Let us find the HCF of each pair of numbers. 

 (a) 

396 = 231 × 1 + 165  

231 = 165 × 1 + 66  

165 = 66 × 2 + 33  

66 = 33 × 2 + 0  

Therefore, HCF = 33. 

Hence, numbers are not co-prime. 

(b) 

2160 = 847 × 2 + 466  

847 = 466 × 1 + 381  

466 = 381 × 1 + 85  

381 = 85 × 4 + 41  

85 = 41 × 2 + 3 

 41 = 3 × 13 + 2  

3 = 2 × 1 + 1 

 2 = 1 × 2 + 0  

Therefore, the HCF = 1. Hence, the numbers are co-prime.

Explanation:

We know that any positive odd integer is of the form 4q+1 or 4q+3.

let a be any positive integer, then:

case 1: a=4q+1

a^2= (4q+1)^2 = 16q^2+8q+1

= 8(2q^2+q)+1

= 8m+1 , m=(2q^2+q)

case 2: a=4q+3

a^2=(4q+3)^2 =16q^2+24q+9

= 8(2q^2+3q+1)+1

=8m+1 ,m=(2q^2+3q+1)

hence, the square of any positive odd integer is of the form 8m+1

Explanation:

Let us assume that √2 + √3 is a rational number.

So it can be written in the form a/b

√2 + √3 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving

√2 + √3 = a/b

√2 = a/b – √3

On squaring both sides we get,

=> (√2)2 = (a/b – √3)2

We know that

(a – b)2 = a2 + b2 – 2ab

So the equation (a/b – √3)2 can be written as

(a/b – √3)2 = a2/b2 + 3 – 2 (a/b)√3

Substitute in the equation we get

2 = a2/b2 + 3 – 2 × √3 (a/b)

Rearrange the equation we get

a2/b2 + 3 – 2 = 2 × √3 (a/b)

a2/b2 + 1 = 2 × √3 (a/b)

(a2 + b2)/b2 × b/2a = √3

(a2 + b2)/2ab = √3

Since, a, and b are integers, (a2 + b2)/2ab is a rational number.

√3 is a rational number.

It contradicts our assumption that √3 is irrational.

∴ our assumption is wrong

Thus √2 + √3 is irrational.

Explanation:

Let 'a' be any positive integer.

b=4

by Euclid's division lemma,

a=bq+r

a²=(bq+r)²-------1.

r=0,1,2,3

from 1.

for r=0,

a²=(4q+0)²

a²=16q²

a²=4(4q²)

=4q, where q=4q².

for r=1,

a²=(4q+1)²

a²=16q²+1+8q

a²=4(4q²+2q)+1

a²=4q+1, where q=4q²+2q.

Explanation:

Let a be the positive integer and b = 4. 

Then, by Euclid’s algorithm, a = 4m + r for 

some integer m ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4. 

So, a = 4m or 4m + 1 or 4m + 2 or 4m + 3. 

So,  (4m)^2 = 16m^2 

= 4(4m^2) = 4q, where q is some integer. (4m + 1)^2 

= 16m^2 + 8m + 1 

= 4(4m^2 + 2m) + 1

 = 4q + 1, where q is some integer. 

(4m + 2)^2 = 16m^2 + 16m + 4

 = 4(4m^2 + 4m + 1) 

= 4q, where q is some integer. (4m + 3)^2

 = 16m^2 + 24m + 9 

= 4(4m^2 + 6m + 2) + 1

 = 4q + 1, where q is some integer. 

Hence, The square of any positive integer is either of the form 4q or 4q + 1, where q is some integer.

Explanation:

let a be any positive integer. by Euclid's Division lemma,

a = bm + r, where b= 5 and r can be 0, 1, 2,3,4,5.

Now,

Point: 1

where r = 0

a= 5m

a² = ( 5m )² = 25m²

a² = 5( 5m )²

a² = 5q

here ,q = 5m²

Point: 2

a= 5m + 1

a² = ( 5m + 1 )²

= 25m² + 10m + 1

= 5( 5m² + 2m ) +1

= 5q + 1

here, q= 5m²+ 2m

point : 3

a = 5m + 2

a² = ( 5m + 2 ) ²

= 25m² + 20m + 4

= 5 ( 5m²+ 4m ) +4

= 5q + 4

here, q = 5m² + 4m

Point: 4

a = 5m + 3

a² = ( 5m + 3 ) ²

= 25m² + 30m + 9

= 5 ( 5m²+ 6m ) + 9

= 5q + 9

here, q= 5m² + 6m

point: 5

a= 5m+ 4

a² = ( 5m+ 4 )²

= 25m²+ 40m+ 16

= 5(5m² + 8m) +16

= 5q + 16

here, q = 5m²+ 8m

As we can see from all the above cases the square of any positive ( +ve ) integer cannot be of the form 5q + 2, or 5q + 3. For any integer q.

Explanation:

Let a be the positive integer and b = 6. 

Then, by Euclid’s algorithm,

 a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 5. 

So, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5. (6q)2 

= 36q2 = 6(6q2) 

= 6m, where m is any integer. (6q + 1)2

= 36q2 + 12q + 1 

= 6(6q2 + 2q) + 1 = 6m + 1, 

where m is any integer. (6q + 2)2 

= 36q2 + 24q + 4 = 6(6q2 + 4q) + 4 

= 6m + 4, 

where m is any integer. (6q + 3)2

 = 36q2 + 36q + 9 

= 6(6q2 + 6q + 1) + 3 

= 6m + 3, 

where m is any integer. (6q + 4)2 

= 36q2 + 48q + 16 

= 6(6q2 + 7q + 2) + 4

 = 6m + 4, 

where m is any integer. (6q + 5)2

 = 36q2 + 60q + 25 

= 6(6q2 + 10q + 4) + 1

 = 6m + 1, 

where m is any integer. Hence, The square of any positive integer is of the form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Explanation:

I am taking q as some integer.

Let positive integer a be any positive integer.

Then, b = 4.

By division algorithm, we know here

0 ≤ r < 4 , So r = 0, 1, 2, 3.

When r = 0

a = 4m

Squaring both sides, we get

a² = ( 4m )²

a² = 4 ( 4m​²)

a² = 4q , where q = 4m²

When r = 1

a = 4m + 1

squaring both sides, we get

a² = ( 4m + 1)²

a² = 16m² + 1 + 8m

a² = 4 ( 4m² + 2m ) + 1

a² = 4q + 1 , where q = 4m² + 2m

When r = 2

a = 4m + 2

Squaring both hand sides, we get

a² = ​( 4m + 2 )²

a² = 16m² + 4 + 16m

a² = 4 ( 4m² + 4m + 1 )

a² = 4q , Where q = ​ 4m² + 4m + 1

When r = 3

a = 4m + 3

Squaring both hand sides, we get

a² = ​( 4m + 3)²

a² = 16m² + 9 + 24m

a² = 16m² + 24m ​ + 8 + 1

a² = 4 ( 4m² + 6m + 2) + 1

a² = 4q + 1 , where q = 4m² + 6m + 2

Hence, the Square of any positive integer is in the form of 4q or 4q + 1, where q is any integer.

Explanation:

we know,

odd number in the form of (2P +1) where P is a natural number,

so, n² -1 = (2P + 1)² -1

= 4P² + 4P + 1 -1

= 4P² + 4P

now, checking :

P = 1 then,

4P² + 4P = 4(1)² + 4(1) = 4 + 4 = 8 , it is divisible by 8.

P =2 then,

4P² + 4P = 4(2)² + 4(2) =16 + 8 = 24, it is also divisible by 8 .

P =3 then,

4P² + 4P = 4(3)² + 4(3) = 36 + 12 = 48 , divisible by 8

hence, we conclude that 4P² + 4P is divisible by 8 for all natural numbers.

hence, n² -1 is divisible by 8 for all odd values of n.

Explanation:

Let the two odd numbers be

(2a+1) & (2b+1) because if we add 1 to any even no. it will be odd.

x²+y²

=>(2a+1)²+(2b+1)²

=>(4a²+4a+1)+(4b²+4b+1)

=>4(a²+b²+a+b)+2

4 Is not a multiple of 2 it means clearly that 4 is not a multiple of x²+y², so x²+y² is even but not divisible by 4.

Hence proved.

Explanation:

a=bq+r

First, find the HCF of 693 and 567,

693=567(1)+126

567=126(4)+63

126=63(2)+0

HCF of 693 and 567 is 63.

Now find the HCF of 63 and 441,

441=63(7)+0

The HCF of 63 and 441 is 63.

Therefore, the HCF of 441,567 and 693 is 63.

Explanation:

1251 - 1 = 1250

9377 - 2 = 9375

15628 - 3 = 15625

[ by Euclid's division lemma ]

a = bq + r

9375 = 1250 × 7 + 625

1250 = 625 × 2 + 0

HCF of 9375 & 1250 is 625

Now,

15625 = 625 × 25 + 0

So,

HCF is 625.

Explanation:

Nos. are

Assume that the total of √3 +√ 5 is a rational number.

Now, it can be written in the form a/b:

a/b = √3 + √5 

a and b are coprime numbers

and b ≠ 0

Solving

√3 + √5 = a/b

On squaring both sides, we get,

(√3 + √5)^2 = (a/b)^2

(√3)^2 + (√5)^2+ 2(√5)(√3) = a^2/b^2

3 + 5 + 2√15 = a^2/b^2

8 + 2√15 = a^2/b^2

2√15 = a^2/b^2 – 8

√15 = (a^2– 8b^2)/2b

Here a and b are integers, then (a^2-8b^2)/2b is a rational number.

Then √15 is also a rational number.

However, this is incompatible because 15 is an irrational number.

Our assumption is incorrect.

√3 + √5 is an irrational number.

Hence, proved.

Explanation:

clearly

prime factorization of 12 = 3×2×2

hence

prime factorization of 12n =(3×2×2)n

since its prime factorization didn't contain 5

hence it cannot end with 0 or 5.

Explanation:

In this problem, we have to find the Least Common Multiple also known as the LCM in order to find the minimum required distance and we use LCM to find the minimum distance which is asked in the question.

Finding LCM:

40 = 2 * 2 * 2 * 5

42 = 2 * 7 * 3

45 = 3 * 3 * 5

LCM = 2 * 2 * 2 * 5 * 3 * 3 * 7 = 40 * 63= 2520 cm

Therefore, the minimum distance which each person has to walk or take is 2520 cm to cover the same distance.

The Distance covered by each person individually:

Distance covered by First person = 2520 / 40 = 63 steps

Distance covered by the Second person = 2520 / 42 = 60 steps

Distance covered by the Third person = 2520 / 45 = 56 steps

Explanation:

The denominator of the rational number 257/ 5000 is 5000.

Now, 5000 = 2 × 2 × 2 × 5 × 5 × 5 × 5

                   = (2)^ 3 × (5)^ 4 , which is of the type 2m × 5n , (where m = 3 and n = 4 are non-negative integers.)

∴ Rational number = 257/5000

                               = 257/2^3×5^4 × 2/2  

                               = 514/2^3×5^4

                               = 514/(10)^4

                               = 514/10000

                               = 0.0514   

So, 0.0514 is the required decimal expansion of the rational 257/5000 and it is also a terminating decimal number.

Explanation:

let's assume that √p is rational,

⇒ √p = a/b ( where 'a' and 'b' are co-primes, meaning they don't have any common factors except for 1)

From squaring both sides,

p = a²/b²

⇒pb² = a²

⇒ b² = a²/p

Since 'p' divides a², it also divides 'a' meaning 'a' has a factor of p

Let 'a' = pm (where m is a positive integer) ⇒ a² = p²m²

Now, pb² = a²

pb² = p²m²

pb²/p²= m²

b²/p =m²

∴ 'p' divides 'b' ⇒ 'b' also has a factor 'p'

∴ 'a' and 'b' are not co primes and our assumption was wrong

⇒ √p is irrational

Similarly, √q is irrational

∴⇒ √p + √q is irrational.

Explanation:

let a be any positive integer

then

b=5,0≤r<b,0≤r<6

r=0,1,2, 3,4, 5

let

r=0,a=bq+r,6q+0

(6q)^2

36q^2

6(6q^2)

let 6q^2 be m

=6m

let.

r=1

a=bq+r

(6q+1)^2

(6q^2)+2*6q*1+1^2

36q^2+12q+1

6(6q^2+2q)+1

let 6q^2+2q be m

= 6m+1

let

r=2

(6q+2)^2

36q^2+24q+4

6(6q^2+4q)+4

let 6q^2+4q be m

= 6m+4

let

r=3

(6q+3)^2

36q^2+36q+9

36q^2+36q+6+3

6(6q^2+6q+1)+3

let the 6q^2+6q+1 be m

= 6m+3

Let

r=4

(6q+4)^2

36q^2+48q+16

36q^2+48q+12+4

6(6q^2+8q+2)+4

let 6q^2+8q+4 be m

6m+4

Let

r=5

(6q+5)^2

36q^2+60q+24+1

6(6q^2+10q+4)+1

let 6q^2+10q+4 

= 6m+1

note= I have taken m instead of q

from above it is proved.

Explanation:

Consider a positive integer = 6q + r

where q is an integer and

r = 0, 1, 2, 3, 4, 5.

We know that positive integers are of the form 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4 and 6q + 5.

By taking the cube on both sides

For 6q,

(6q)^3 = 216q^3

= 6(36q)^3 + 0

= 6m + 0, (where m is an integer = (36q)^3).

For 6q + 1,

(6q + 1)^3 = 216q^3 + 108q^2 + 18q + 1

= 6(36q^3 + 18q^2 + 3q) + 1

= 6m + 1, (where m is an integer = 36q^3 + 18q^2 + 3q).

For 6q + 2,

(6q + 2)^3 = 216q^3 + 216q^2 + 72q + 8

= 6(36q^3 + 36q^2 + 12q + 1) +2

= 6m + 2, (where m is an integer = 36q^3 + 36q^2 + 12q + 1).

For 6q + 3,

(6q + 3)^3 = 216q^3 + 324q^2 + 162q + 27

= 6(36q^3 + 54q^2 + 27q + 4) + 3

= 6m + 3, (where m is an integer = 36q^3 + 54q2 + 27q + 4).

For 6q + 4,

(6q + 4)^3 = 216q^3 + 432q^2 + 288q + 64

= 6(36q^3 + 72q2 + 48q + 10) + 4

= 6m + 4, (where m is an integer = 36q^3 + 72q^2 + 48q + 10).

For 6q + 5,

(6q + 5)^3 = 216q^3 + 540q^2 + 450q + 125

= 6(36q^3 + 90q^2 + 75q + 20) + 5

= 6m + 5, (where m is an integer = 36q^3 + 90q^2 + 75q + 20).

Therefore, the cube of a positive integer of the form 6q + r, q is an integer, and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Explanation:

Euclid's division Lemma any natural number can be written as: .

where r = 0, 1, 2,. and q is the quotient.

∵ Thus any number is in the form of 3q, 3q+1, or 3q+2.

→ Case a: if n =3q

⇒n = 3q = 3(q) is divisible by 3,

⇒ n + 2 = 3q + 2 is not divisible by 3.

⇒ n + 4 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.

→ Case b: if n =3q + 1

⇒ n = 3q + 1 is not divisible by 3.

⇒ n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3.

⇒ n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 is not divisible by 3.

→ Case c: if n = 3q + 2

⇒ n =3q + 2 is not divisible by 3.

⇒ n + 2 = 3q + 2 + 2 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.

⇒ n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) is divisible by 3.

Thus one and only one out of n, n+2, n+4 is divisible by 3.

Hence, Proved.

Explanation:

Let three consecutive positive integers be n, n + 1, and n + 2.

As per Euclid division lemma whenever a number is divided by 3, 

the remainder obtained is either 0 or 1 or 2.

∴ n = 3p or 3p + 1 or 3p + 2, 

where p is some integer.

Case a:

If n = 3p,

 then n is divisible by 3 but n+1 and n+2 are not divisible by 3.

Case b:

If n = 3p + 1

Then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3

 but n+1 and n+2 are not divisible by 3.

Case c:

If n = 3p + 2

Then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So, we can say that one of the numbers among n, n + 1, and n + 2 is always divisible by 3.

Explanation:

n³ - n = n(n²-1) = n(n -1)(n + 1) is divided by 3 then the possible reminder is 0, 1, and 2 [ ∵ if P = ab + r, then 0 ≤ r < a by Euclid lemma ]

∴ Let n = 3r , 3r +1 , 3r + 2 , where r is an integer

Case 1 :- when n = 3r

Then, n³ - n is divisible by 3 [∵n³ - n = n(n-1)(n+1) = 3r(3r-1)(3r+1) , clearly shown it is divisible by 3 ]

Case2 :- when n = 3r + 1

e.g., n - 1 = 3r +1 - 1 = 3r

Then, n³ - n = (3r + 1)(3r)(3r + 2) , it is divisible by 3

Case 3:- when n = 3r - 1

e.g., n + 1 = 3r - 1 + 1 = 3r

Then, n³ - n = (3r -1)(3r -2)(3r) , it is divisible by 3

From the above explanation we observed n³ - n is divisible by 3, where n is any positive integers

Explanation:

we know that any positive number will be in the form of  5k, 5k+1, 5k+2,5k+3 

5k+4 

if n = 5 k  n is perfectly divisible by 5  now sir, n = 5k we will add 4 on both sides 

we get 4 +5k = n +4  and if we divide the number it will leave a remainder 4 

when we divide this number by 5 now sir , 

n = 5k we will add 8 both side we get 

n+8 = 5k +8 which means = >  5k +5 +3 which means = 5(k+1) + 3 

which means it will give a remainder 3 when we divide it by 5 

if sir n = > 5 k  when we add 12 on both sides we get: - ) 

n+12 = 5k + 12 which means = > 5k +10 +2 => 5(k+2) +2 so it means when we divide this quantity  by 5 we get the remainder 2 

5k = n we will ad 16 on both sides we get 

5k +16 = n +16 which means 5 (k+3) +1 so it will give a remainder of 1 

case 2  = > 

where n = 5k +1 the number leaves a remainder when it is divisible by 5 

n  =  5k +1 adding both sides 2 we get

n+3 = 5k +3 which means it will leave a remainder of 3 

n  = 5k + 1

adding 4 on both sides which means 

n+4 = 5(k+1) which means it will leave no remainder  and perfectly divisible by 5 

n = 5k +1 adding both sides 8 we get 

n+8 = 5(k+1) +4 which means it will leave 4 as a remainder when we divide it by 5 

n = 5 k +1 adding both sides 12 we get 

n+12 = 5(k+2) + 3 so sir it will leave a remainder 3 when it is divisible by 5 

n = 5k +1 adding both sides 16 we get 

n+16 => 5(k+3) +2 which means it will leave a remainder of 2 when it is divisible by 5 

so it means only in 1 case 5 is divisible by the number