Real Numbers

We know that an integer is said to be even when it is divisible by 2.

Let m be a integer i.e. m = ..., -3, -2, -1, 0, 1, 2, 3, …

Clearly, 'm' may or may not be even, therefore, 'm' can't be an answer.

m + 1 = ..., -2, -1, 0, 1, 2, 3, 4...

Here also, 'm + 1' may or may not be even, therefore, 'm + 1' can't be an answer.

Now, consider 2m i.e. 2m = ...,-6, -4, -2, 0, 2, 4, 6,...

Here, for some integer m, 2m is always integer,

Now, consider 2m + 1, i.e. 2m + 1 = ..., -5, -3, -1, 1, 3, 5, 7,...

Clearly, 2m + 1 is an odd integer, therefore it can't be an answer


Hence, 2m is only answer.

We know that an integer is said to be odd if it is not divisible by 2 .

Let m be a integer i.e. q = -1, 0, 1, 2, 3, 4,…..

Multiplying both the sides by 2

⇒ 2q = -2, 0, 2, 4, 6, 8,.……

Adding 1 on both the sides

⇒ 2q + 1 = -2 + 1, 0 + 1, 2 + 1, 4 + 1, 6 + 1, 8 + 1,…..

⇒ 2q + 1 = -1, 1, 3, 5, 7, 9,….

Hence, for some integer q, every even integer is of the form 2q + 1.

Let a be of the form n²– 1, n can be odd or even.

Let us consider two cases.

Case I If n is even i.e. n = 2k, where k is an integer.

⇒ a = (2k)² -1

⇒ a = 4k² -1

At k = 1, = 4 (1)² -1 = 4 -1 = 3, which is not divisible by 8.

At k = 2, a = 4 (2)² -1 = 16-1 = 15 which is not divisible by 8.

Case II If n is odd i.e. n = 2k + 1, where k is an integer.

⇒ a = (2k + 1)² -1

⇒ a = 4k² + 1 + 4k-1

⇒ a = 4k² + 4k = 4k(k + 1)

At k = 1, a = 4(1)(1 + 1) = 4 × 2 = 8, which is divisible by 8.

At k = 2, a = 4(2)(1 + 1) = 8 × 2 = 16 which is divisible by 8.

Hence, we can conclude from above two cases that if n is odd, then n² -1 is divisible by 8

According to Euclid’s division algorithm,

b = a × q + r, 0 ≤ r < a [using, dividend = divisor × quotient + remainder]

⇒ 117 = 65 × 1 + 52

⇒ 65 = 52 × 1 + 13

⇒ 52 = 13 × 4 + 0

∴ HCF (65, 117) = 13 (i)

Also given that, HCF (65, 117) = 65 m – 117

⇒ 65 m – 117 = 13 [from (i)]

⇒ 65 m = 130

⇒ m = 2

Since, it is given that 5 and 8 are the remainders of 70 and 125 respectively. On subtracting these remainders from the numbers we get 65 = (70-5) and 117 = (125-8), which is divisible by the required number.

Now, required number = HCF (65,117) [for the largest number]

According to Euclid’s division algorithm,

b = a × q + r, 0 ≤ r < a [∴dividend = divisor × quotient + remainder]

⇒ 117 = 65 × 1 + 52

⇒ 65 = 52 × 1 + 13

⇒ 52 = 13 × 4 + 0

⇒ HCF = 13

Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8

Let a = x³y² = x × x × x × y × y

And b = xy³ = x × y × y × y

⇒ HCF of a and b = HCF (x³y², xy³) = x × y × y = xy²

[Since, HCF is the product of the smallest power of each common prime factor involved

in the numbers]

Let p = ab² = a × b × b

And q = a³b = a × a × a × b

⇒ LCM of p and q = LCM (ab², a³b) = a × b × b × a × a = a³b²

[Since, LCM is the product of the greatest power of each prime factor involved in the number]

Product of a non-zero rational and an irrational number is always irrational

For example, this is irrational number.

Factors of 1 to 10 numbers

1 = 1

2 = 1 × 2

3 = 1 × 3

4 = 1 × 2 × 2

5 = 1 × 5

6 = 1 × 2 × 3

7 = 1 × 7

8 = 2 × 2 × 2

9 = 1 × 3 × 3

⇒ LCM of number 1 to 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

= 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520

Simplifying the given fraction

Hence, given rational number will terminate after four decimal places.

No

By Euclid’s Lemma,

b = a × q + r, 0 ≤ r < a [Using dividend = divisor × quotient + remainder]

Here, b is any positive integer.

According to the question, a = 4

⇒ b = 4q + r where 0 ≤ r < 4

⇒ r = 0, 1, 2, 3

So, this must be in the form 4q, 4q + 1, 4q + 2 or 4q + 3.

Hence, every positive integer cannot be of the form 4q + 2.

True

Let the consecutive positive integers be n and (n + 1).

Since, the numbers are consecutive one must be even between these. The product of these is n(n + 1) which will also be divisible by 2

True

Let the consecutive positive integers be n, (n + 1) and (n + 2).

One number of these three must be even i.e. divisible by 2. Hence, product of numbers is divisible by 2 and another one must be divisible by 3. Hence, product of numbers is divisible by 6

False

By Euclid’s lemma, b = a × q + r, 0 ≤ r ≤ a Here, b is any positive integer.

According to the question, a = 3 and b = 3q + r for 0 ≤ r ≤ 2

So any positive integer is of the form 3k, 3k + 1 or 3k + 2.

Squaring each of these terms,

(3k)² = 9k² = 3m [where, m = 3k²]

and (3k + 1)² = 9k² + 6k + 1 [Using (a + b)² = a² + 2ab + b²]

⇒ (3k + 1)² = 3(3k² + 2k) + 1 = 3m + 1 [where, m = 3k² + 2 k]

Also (3k + 2)² = 9k² + 12 k + 4 [Using (a + b)² = a² + 2ab + b²]

⇒ (3k + 1)² = 9k² + 12k + 3 + 1

⇒ (3k + 1)² = 3(3k² + 4k + 1) + 1

⇒ (3k + 1)² = 3m + 1 [where m = 3k² + 4k + 1]

Hence, the square of any positive integer cannot be of the form 3m + 2, where m is natural number.

No

By Euclid’s Lemma, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer

According to the question, a = 3 ⇒ b = 3q + r for 0 r < 3

So, this must be in the form 3 q, 3q + 1 or 3 q + 2.

and (3 q + 1)² = 9q² + 6q + 1

⇒ (3 q + 1)² = 3 (3q² + 2q) + 1

⇒ (3 q + 1)² = 3m + 1

[where m = 3q² + 2q ]

Hence, square of a positive integer of the form 3q + 1 is always in the form 3m + 1 for some integer m.

We will find the HCF of 525 and 3000

By Euclid’s Lemma, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer

3000 = 525 × 5 + 375 [Using dividend = divisor × quotient + remainder]

525 = 375 × 1 + 150

375 = 150 × 2 + 75

150 = 75 × 2 + 0

So, the highest common factor among 3, 5, 15, 25 and 75 of 525 and 3000 is 75.

According to the definition of Composite numbers, any number with more than 2 factors is said to be composite.

Here, 3 × 5 × 7 + 7 = 105 + 7 = 112

The factors of 112 are : 2 × 2 × 2 × 2 × 7 = 2⁴ × 7.

Since it has more than 2 factors, Hence, itis a composite number.

HCF stands for Highest Common Factor while LCM stands for Least Common Multiple.

⇒ HCF of two numbers must be a factor of LCM of those numbers

According to the question, HCF = 18 and LCM = 380

But 380 is not divisible by 18. So, two numbers cannot have 18 and 380 as their HCF and LCM respectively.

Simplifying the given fraction

and we know,,
if is a rational number, such that the prime factorisation of q is of the form 2ⁿ5^m where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Hence, it terminates.

Expressing the given decimal as fraction

⇒ prime factors of q = 2 and 5

By Euclid’s Lemma, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer.

Let a be an arbitrary positive integer. Then corresponding to the positive integers a and 4, there exist non-negative integers m and r, such that

a = 4m + r, where 0 ≤ r < 4

Squaring both the sides using (a + b)² = a² + 2ab + b²

⇒ a² = (4m + r)² = 16m² + 8mr + r²

Case I When r = 0 we get

a² = 16m² which is an integer.

Case II When r = 1 we get

a² = 16m² + 1 + 8m

⇒ a² = 4(4m² + 2m) + 1 = 4q + 1

where q = (4m² + 2m) in an integer.

Case III When r = 2 we get

a² = 16m² + 4 + 16m

⇒ a² = 4 (4m² + 4m + 1) = 4q

where q = (4m² + 4m + 1) is an integer.

Case IV When r = 3 we get

a² = 16m² + 9 + 24m = 16m² + 24m + 8 + 1

⇒ a² = 4 (4m² + 6m + 2) + 1 = 4q + 1

where q = (4m² + 6m + 2) is an integer.

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

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer.

Let a be an arbitrary positive integer. Then corresponding to the positive integers a and 4, there exist non-negative integers q and r such that

a = 4q + r, where 0 ≤ r < 4

Cubing both the sides using (a + b)³ = a³ + b³ + 3ab² + 3a²b

⇒ a³ = (4q + r)³ = 64q³ + r³ + 12qr² + 48q²r

⇒ a³ = (64q³ + 48q²r + 12qr²) + r³

where 0 ≤ r < 4

Case I When r = 0.

a³ = 64q³ = 4(16q³)

⇒ a³ = 4m where m = 16q³ is an integer.

Case II When r = 1 we get

a³ = 64q³ + 48q² + 12q + 1

⇒ a³ = 4 (16q³ + 12q² + 3q) + 1

⇒ a³ = 4m + 1

where m = (16q² + 12q² + 3q) is an integer.

Case III When r = 2 we get

a³ = (64q³ + 96q² + 48q) + 8

⇒ a³ = 4 (16q³ + 24q² + 12q + 2)

⇒ a³ = 4m

where, m = (16q³ + 24q² + 12q + 2) is an integer.

Case IV When r = 3 we get

a³ = (64q³ + 144q² + 108q) + 27

⇒ a³ = (64q³ + 144q² + 108q) + 24 + 3

⇒ a³ = 4 (16q³ + 36q² + 27q + 8) + 3

⇒ a³ = 4m + 3

where m = (16q³ + 36q² + 27q + 8) is an integer.

Hence, of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.

By Euclid’s division algorithm, By Euclid’s Lemma, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer.

Let a be an arbitrary positive integer. Then corresponding to the positive integers a and 5, there exist non-negative integers m and r such that

a = 5m + r, where 0 ≤ r < 5

Squaring both the sides using (a + b)² = a² + 2ab + b²

⇒ a² = (25m² + r² + 10mr)

⇒ a² = 5(5m² + 2mr) + r²

Where 0 ≤ r < 5

Case I When r = 0 we get

a² = 5(5m²)

⇒ a² = 5q

where q = 5m² is an integer.

Case II When r = 1 we get

a² = 5(5m² + 2m) + 1

⇒ a² = 5q + 1

where q = (5m² + 2m) is an integer.

Case III When r = 2 we get

⇒ a² = 5(5m² + 4m) + 4

⇒ a² = 5q + 4

Where q = (5m² + 4m) is an integer.

Case IV When r = 3 we get

⇒ a² = 5(5m² + 6m) + 9 = 5 (5m² + 6m) + 5 + 4

⇒ a² = 5(5m² + 6m + 1) + 4 = 5q + 4

where, q = (5m² + 6m + 1) is an integer.

Case V when r = 4 we get

⇒ a² = 5(5m² + 8m) + 16 = 5 (5m² + 8m) + 15 + 1

⇒ a² = 5(5m² + 8m + 3) + 1 = 5q + 1

where, q = (5m² + 8m + 3) is an integer.

Hence, the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

Let a be an arbitrary positive integer, then corresponding to the positive integers a and 6, there exist, non-negative integers q and r such that

a = 6q + r, where 0 ≤ r < 6

Squaring both the sides using (a + b)² = a² + 2ab + b²

⇒ a² = (6q + r)² = 36q² + r² + 12qr

⇒ a² = 6(6q² + 2qr) + r²

where,0 ≤ r < 6

Case I When r = 0 we get

⇒ a² = 6(6q²) = 6m

where, m = 6q² is an integer.

Case II when r = 1 we get

⇒ a² = 6(6q² + 2q) + 1 = 6m + 1

where, m = (6q² + 2q) is an integer.

Case III When r = 2 we get

⇒ a² = 6(6q² + 4q) + 4 = 6m + 4

where, m = (6q² + 4q) is an integer.

Case IV When r = 3 we get

⇒ a² = 6(6q² + 6q) + 9

⇒ a² = 6(6q² + 6a) + 6 + 3

⇒ a² = 6(6q² + 6q + 1) + 3 = 6m + 3

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

Case V when r = 4 we get

⇒ a² = 6(6q² + 8q) + 16

⇒ a² = 6(6q² + 10q) + 24 + 1

⇒ a² = 6(6q² + 8q + 2) + 4 = 6m + 4

where, m = (6q² + 8q + 2) is an integer

Case VI When r = 5 we get

⇒ a² = 6(6q² + 10q) + 25

⇒ a² = 6(6q² + 10q) + 24 + 1

⇒ a² = 6(6q² + 10q + 4) + 1 = 6m + 1

where, m = (6q² + 10q + 1) + 1 is an integer.

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

On putting b = 4 we get

a = 4q + r, where 0 ≤ r < 4 (i)

If r = 0

⇒ a = 4q, 4q is divisible by 2

⇒ 4q is even.

If r = 1

⇒ a = 4q + 1, (4q + 1) is not divisible by 2.

If r = 2

⇒ a = 4q + 2,2(2q + 1) is divisible by 2

⇒ 2(2q + 1) is even.

If r = 3

⇒ a = 4q + 3, (4q + 3) is not divisible by 2.

So, for any positive integer (4q + 1) and (4q + 3) are odd integers.

Case I For (4q + 1)

Squaring both the sides using (a + b)² = a² + 2ab + b²

a² = 16q² + 1 + 8q

⇒ a² = 4(4q² + 2q) + 1 = 4m + 1

Where m = (4q² + 2q) in an integer.

Case II For (4q + 3)

Squaring both the sides using (a + b)² = a² + 2ab + b²

a² = 16q² + 9 + 24q = 16q² + 24q + 8 + 1

⇒ a² = 4 (4q² + 6q + 2) + 1 = 4m + 1

where m = (4q² + 6q + 2) is an integer.

Hence, for some integer m, the square of any odd integer is of the form 4m + 1.

Let a be of the form n²– 1, where n is odd

i.e. n = 2k + 1, where k is an integer.

⇒ a = (2k + 1)² -1

⇒ a = 4k² + 1 + 4k-1

⇒ a = 4k² + 4k = 4k(k + 1)

At k = 1, a = 4(1)(1 + 1) = 4 × 2 = 8, which is divisible by 8.

At k = 2, a = 4(2)(1 + 1) = 8 × 2 = 16 which is divisible by 8.

And so on.

Hence, we can conclude from above two cases that if n is odd, then n² -1 is divisible by 8.

Let us consider two odd positive integers x = 2m + 1 and y = 2n + 1

Then, x² + y² = (2m + 1)² + (2n + 1)²

Squaring these terms using (a + b)² = a² + 2ab + b²

⇒ x² + y² = 4m² + 1 + 4m + 4n² + 1 + 4n

⇒ x² + y² = 4(m + n)² + 4(m + n) + 2
which is even, as each term is divisible by 2

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

First we take b = 693 and a = 567 and get the required HCF.

⇒ 693 = 567 × 1 + 126

⇒ 567 = 126 × 4 + 63

⇒ 126 = 63 × 2 + 0

So, HCF(693,567) = 63

Now, take b = 441 and a = 63 and get the required HCF.

⇒ 441 = 63 × 7 + 0

So, HCF (441, 63) = 63

Hence, the HCF (441, 567, 693) = 63

Since, 1, 2 and 3 are the remainders of 1251, 9377 and 15628 respectively. Thus after subtracting these remainders from the numbers , we get

1251 – 1 = 1250, 9377 − 2 = 9375 and 15628 − 3 = 15625 which is divisible by the required number.

Now, required number = HCF (1250, 9375, 15625)

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

Firstly put b = 15625 and a = 9375

⇒ 15625 = 9375 × 1 + 6250

⇒ 9375 = 6250 × 1 + 3125

⇒ 6250 = 3125 × 2 + 0

So, HCF (9375, 15625) = 3125

Now, put a = 1250 and b = 3125

⇒ 3125 = 1250 × 2 + 625

⇒ 1250 = 625 × 2 + 0

So, HCF (1250, 3125) = 625

Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3, respectively.

Let us suppose that √3 + √5 is rational. Let √3 = √5 − a, were a is rational.

On squaring both sides, we get

(√3)² = (√5 – a )²

⇒ 3 = 5 + a² − 2a√5 [ Using (a-b)² = a² + b² + 2ab]

⇒ 2a√5 = 2 + a²

This is not possible because right hand side is rational while left hand side i.e. √5 is irrational.

So, our assumption is wrong. √3 + √5 is irrational.

If any number ends with the digit 0 or 5, it is always divisible by 5. If 12n ends with the digit zero it must be divisible by 5. This is possible only if prime factorisation of 12n contains the prime number 5.

12 = 2 × 2 × 3 = 2² × 3

⇒ 12n = (2² × 3)n = 2²n × 3n

Since its prime factorisation does not contain 5.

Hence, 12n cannot end with the digit 0 or 5 for any natural number n.

We have to find LCM of 40, 42 and 45 to get the required minimum distance.

For this we find prime factorisations,

40 = 2 × 2 × 2 × 5

42 = 2 × 3 × 7

45 = 3 × 3 × 5

LCM (40, 42, 45) = 2 × 3 × 5 × 2 × 2 × 3 × 7 = 2520

Hence, Minimum distance each should walk 2520 cm. So that each can cover the same distance in complete steps.

Simplifying the given fraction

Hence, given rational number will terminate after four decimal places.

Let us suppose that √p + √q is rational. Let √p = √q − a, were a is rational.

On squaring both sides, we get

(√p)² = (√q – a )²

⇒ p² = q² + a² − 2a√q [ Using (a-b)² = a² + b² + 2ab]

⇒ 2a√q = p² + q² + a²

This is not possible because right hand side is rational while left hand side i.e. √q is irrational.

So, our assumption is wrong. √p + √q is irrational.

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

Let a be an arbitrary positive integer. Then corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that

a = 6q + r where 0 ≤ r < 6

Cubing both the sides using (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ a³ = (6q + r)³ = 216q³ + r³ + 3 × 6q × r(6q + r)

⇒ a³ = (216q³ + 108q²r + 18qr²) + r³

where 0 ≤ r < 6

Case I When r = 0 we get a³ = 216q³

⇒ a³ = 6(36q³) ⇒ a³ = 6m + 0 where m = 36q³ is an integer

Case II When r = 1 we get

a³ = (216q³ + 108q² + 18q) + 1

⇒ a³ = 6(36q³ + 18q² + 3q) + 1

⇒ a³ = 6m + 1, where m = (36q³ + 18q² + 3q) is an integer.

Case III When r = 2 we get

⇒ a³ = 6(36q³ + 36q² + 12q + 1) + 2

⇒ a³ = 6m + 2

where m = (36q³ + 36q² + 12q + 1) is an integer.

Case IV When r = 3 we get

a³ = (216q³ + 324q² + 162q) + 27

⇒ a³ = (216q³ + 324q² + 162q + 24) + 3

⇒ a³ = 6(36q³ + 54q² + 27q + 4) + 3

⇒ a³ = 6m + 3

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

Case V When r = 4 we get

a³ = (216q² + 432q² + 288q) + 64

⇒ a³ = 6(36q³ + 72q² + 48q) + 60 + 4

⇒ a³ = 6(36q³ + 72q² + 48q + 10) + 4

⇒ a³ = 6m + 4

where m = (36q³ + 72q² + 48q + 10) is an integer.

Case VI When r = 5 we get

a³ = (216q³ + 540q² + 450q) + 120 + 5

⇒ a³ = 6(36q³ + 90q² + 75q + 20) + 5

⇒ a³ = 6m + 5

where m = (36q³ + 90q² + 75q + 20) + 5

Hence, 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 forms 6m, 6m + 1, 6m + 2, 6m + 3,6m + 4and 6m + 5i.e., 6m + r.

Let a = n, b = n + 2 and c = n + 4

Let 3 consecutive positive integers be p, p + 1 and p + 2

Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2.

:

Therefore:

p = 3q or 3q + 1 or 3q + 2, where q is some integer

If p = 3q, then n is divisible by 3

If p = 3q + 1, then n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3

If p = 3q + 2, then n + 1 = 3q + 2 + 1 = 3q + 3 = 3(q + 1) is divisible by 3

Thus, we can state that one of the numbers among p, p + 1 and p + 2 is always divisible by 3

Let a = n³ – n

⇒ a = n × (n² -1)

⇒ a = n × (n-1) × (n + 1) [Using (a² –b²) = (a−b) × (a + b)]

⇒ a = (n-1) × n × (n + 1)

We know that

I If a number is completely divisible by 2 and 3, then it is also divisible by 6.

II If the sum of digits of any number is divisible by 3, then it is also divisible by 3.

III. If one of the factors of any number is an even number, then it is also divisible by 2.

Since, a = (n-1) × n × (n + 1)

Sum of the digits = n−1 + n + n + 1 = 3n which is a multiple of 3, where n is any positive integer.

And (n-1) × n × (n + 1) will always be even, as one out of (n-1) or n or (n + 1) must be even.

Hence, by condition I the number n³-n is always divisible by 6, where n is any

positive integer.

Given numbers are n, (n + 4), (n + 8), (n + 12) and (n + 16) where n is any positive integer.

Then let n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4 for q є N

Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

Then n = 5q + r, where 0 ≤ r < 5.

⇒ n = 5q or 5q + 1 or 5q + 2 or 5q + 3 or 5q + 4.

Case I If n = 5q, then n is only divisible by 5.

So, in this case, n is only divisible by 5.

Case II If n = 5q + 1, then n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1) which is only divisible by 5.

So, in this case, (n + 4) is only divisible by 5.

Case III If n = 5q + 2, then n + 8 = 5q + 2 + 8 = 5q + 10 = 5(q + 2) which is only divisible by 5.

So, in this case, (n + 8) is only divisible by 5.

Case IV If n = 5q + 3, then n + 12 = 5q + 3 + 12 = 5q + 15 = 5(q + 3) which is only divisible by 5

So, in this case, (n + 12) is only divisible by 5.

Case V If n = 5q + 4, then n + 16 = 5q + 20 = 5(q + 4), which is divisible by 5.

So, in this case, (n + 16) is only divisible by 5.

Hence, one and only one of n, n + 4, n + 8 n + 8, n + 12 and n + 16 is divisible by 5 where any positive integer is an integer.