Real Numbers

Given: a and b are two odd positive integers such that a>b.
To prove: out of the numbers one is odd and other is even.
Proof: a and b both are odd positive integers,
Since,

which is an odd number as "a" is given to be a odd number.
Hence one of the number out of must be even and other must be odd because adding two even numbers gives an even number and adding two odd numbers gives an even number.
Here, will give an odd number and will give an even number.

EXAMPLE:

Take a = 7 and b = 3 such that a>b

Now, which is odd

And which is even
Conclusion: Out of out of the numbers ,
is an odd number and is an even number.

Let the numbers are a and a-1

Product of these number: a(a - 1) = a²-a

Case 1: When a is even:

a=2p

then (2p)² - 2p ⇒ 4p² - 2p

2p(2p-1) ………. it is divisible by 2

Case 2: When a is odd:

a = 2p+1

then (2p+1)² - (2p+1) ⇒ 4p² + 4p + 1 - 2p – 1

= 4p² + 2p ⇒ 2p(2p + 1) ………….. it is divisible by 2

Hence, we conclude that product of two consecutive integers is always divisible by 2

Let a be any positive integer and b=6. By division lemma there exists integers q and r such that

a = 6q+r where 0≤r<6

a = 6q , or a = 6q+1 or, a=6q+2 or, a=6q+3

or a=6q+4 or a=6q+5


Let n is any positive integer.

Since any positive integer is of the form 6q or, 6q + 1 or, 6q + 2 or, 6q + 3 or, 6q + 4 or, 6q + 5.

Case 1:

If n = 6q then

n(n + 1)(n + 2) = 6q (6q + 1)(6q + 2), which is divisible by 6.

Case 2:

If n = 6q + 1, then

n(n + 1)(n + 2) = (6q + 1)(6q + 2) (6q + 3)
= (6q + 1)2(3q + 1)3(2q + 1)
= 6(6q + 1)(3q + 1)(2q + 1), which is divisible by 6.

Case 3:

If n = 6q + 2, then

n(n + 1)(n + 2) = (6q + 2) (6q + 3)(6q + 4)
= 2(3q + 1)3(2q + 1)(6q + 4)
= 6(3q + 1)(2q + 1)(6q + 4), which is divisible by 6.

Case 4:

If n = 6q + 3, then

n(n + 1)(n + 2) = (6q + 3) (6q + 4)(6q + 5)
= 3(2q + 1)2(3q + 2)(6q + 5)
= 6(2q + 1)(3q + 2)(6q + 5), which is divisible by 6.

Case 5:

If n = 6q + 4, then

n(n + 1)(n + 2) = (6q + 4) (6q + 5)(6q + 6)
= (6q + 4) (6q + 5)6(q + 1)
= 6(6q + 4) (6q + 5)(q + 1), which is divisible by 6.

Case 6:

If n = 6q + 5, then

n(n + 1)(n + 2) = (6q + 5) (6q + 6)(6q + 7)
= (6q + 5) 6(q + 1)(6q + 7)
= 6(6q + 5) (q + 1)(6q + 7), which is divisible by 6.

n³ – n = n(n² – 1) = n(n – 1)(n + 1)

For a number to be divisible by 6, it should be divisible by 2 and 3 both,

Divisibility by 3:

n – 1, n and n + 1 are three consecutive whole numbers.

By Euclid’s division lemma

n + 1 = 3q + r, for some integer k and r < 3

As, r < 3 possible values of ‘r’ are 0, 1 and 2.

If r = 0

n + 1 = 3q

⇒ n + 1 is divisible by 3

⇒ n(n – 1)(n + 1) is divisible by 3

⇒ (n³ – n) is divisible by 3

If r = 1

⇒ n + 1 = 3q + 1

⇒ n = 3q

⇒ n is divisible by 3

⇒ n(n – 1)(n + 1) is divisible by 3

⇒ (n³ – n) is divisible by 3

If r = 2

⇒ n + 1 = 3q + 2

⇒ n + 1 – 2 = 3q

⇒ n – 1 = 3q

⇒ n - 1 is divisible by 3

⇒ n(n – 1)(n + 1) is divisible by 3

⇒ (n³ – n) is divisible by 3

Divisibility by 2:

If n is even

Clearly, n(n – 1)(n + 1) is divisible by 2

If n is odd

⇒ n + 1 is even

⇒ n + 1 is divisible by 2

⇒ n(n – 1)(n + 1) is divisible by 2

Hence, for any value of n, n³ - n is divisible by 2 and 3 both, therefore n³ – n is divisible by 6.

Let N = 5p + 1. Then,

According to the condition:

N² = 25p² + 10p + 1 ⇒ 5(5p² + 2p) + 1 ⇒ 5A+1

Where A = 5p² + 2p

Therefore N² is of the form 5m + 1.

we know, that any positive integer N is of the form 3q, 3q + 1 or, 3q + 2.

When N = 3q, then

N² = 9q² = 3(3q)² = 3m where m = 3q²

When N = 3q + 1,

then N² = (3q + 1)² = 9q² + 6q + 1 ⇒ 3q(3q + 2) + 1 = 3m + 1 where m = q(3q + 2)
And, as q is an integer, m = q(3q + 2) is also an integer.

When N = 3q + 2, then N² = (3q + 2)² = 9q² + 12q + 4 ⇒ 3(3q² + 4q + 1) + 1

3m + 1 where, m = 3q² + 4q + 1

Therefore,

N² is of the form 3m, 3m + 1 but not of the form 3m + 2

Since any positive integer n is of the form 2p or, 2p + 1

When n = 2p, then n² = 4p² = 4a where a = p²

When n = 2p + 1, then n² = (2p + 1)² = 4p² + 4p + 1 ⇒ 4p(p + 1) + 1

⇒ 4m + 1 where m = p(p + 1)

Therefore square of any positive integer is of the form 4q or 4q + 1 for some integer q

Since any positive integer n is of the form 5p or 5p + 1, or 5p + 2 or 5m + 3 or 5p +4.

When n = 5p, then

n² = (5p)² = 25p²⇒ 5 (5p²) = 5a, where a =5p²

When n = 5p + 1,

then n² = (5p + 1)² = 25p² + 10p + 1

⇒ 5p(5p + 2) + 1 ⇒ 5a + 1 Where a = p(5p + 2)

When n = 5p + 2,

then n² = (5p + 2)²⇒ 25p² + 20p + 4 ⇒ 5p(5p + 4) + 4

⇒ 5a + 4 where a = p(5p + 4)

When n = 5p + 3, then n² = 25p² + 30p + 9

⇒ 5(5p² + 6p + 1) + 4 ⇒ 5a + 4 where a = 5p² + 6m + 1

When n = 5p + 4, then n² = (5p + 4)² = 25p² + 40p + 16 ⇒ 5(5p² + 8m + 3) + 1 = 5a + 1 where a = 5p² + 8m + 3

Therefore from above results we got that n2 is of the form 5q or, 5q + 1 or, 5q + 4.

To show: the square of an odd positive integer is of the form 8q + 1, for some integer q.

Solution:

Let a be any positive integer and b=4.

Applying the Euclid's division lemma with a and b=4 we have

a = 4p + r where 0≤r<4 and p is some integer,

⇒ r can be 0,1,2,3

⇒ a = 4p + 0 , a = 4p + 1 , a = 4p+ 2 , a = 4p + 3 ,

Since a is odd integer,

So a = 4p + 1 or a = 4p + 3

So any odd integer is of the form a = 4p+ 1 or a = 4p+ 3 .

Since any odd positive integer n is of the form 4p + 1 or 4p + 3.

When n = 4p + 1,

then n² = (4p + 1)²

Apply the formula (a + b)² = a² + b² + 2ab


⇒ (4p + 1)² = 16p² + 8p + 1 ..... (1)


Take 8p common out of 16p² + 8p


⇒ 8p(2p+ 1) + 1 = 8q + 1 where q = p(2p + 1)

If n = 4p + 3, then n² = (4p + 3)²

Apply the formula (a + b)² = a² + b² + 2ab

⇒ (4p + 3)²= 16p² + 24p + 9

Now 16p² + 24p + 9 can be written as 16p² + 24p + 8 + 1


⇒ (4p + 3)² = 8(2p² + 3p + 1) + 1 = 8q + 1 where q = 2p² + 3p + 1

From above results we got that n² is of the form 8q + 1.

Note: To show that the square of an odd positive integer is of the form 8q + 1 We have started the question from taking b=4 initially because when
we take square of any form of 4 such as 4p+1 we end up having the values which are the multiple of 8 as in (1).

While attempting these types of questions always remember to start with the value which would end up giving the value the question demands.
and follow the above steps.

To prove: any positive odd integer is of the form 6q + 1 or, 6q + 3 or, 6q + 5, where q is some integer.
Solution:

Let ‘a’ be any odd positive integer we need to prove that a is of the form 6q+1, or 6q+3, or 6q+5, where q is some integer.

Since a is an integer consider b = 6 another integer
applying Euclid's division lemma there exist integers q and r such that we get,

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

Therefore according to question:

a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

However since a is odd so ‘a’ cannot take the values 6q, 6q+2 and 6q+4

(since all these are even integers, hence divisible by 2)

Definition of HCF (Highest Common Factor): The largest positive integer which divides two or more integers without any remainder is called Highest Common Factor (HCF) or Greatest Common Divisor or Greatest Common Factor (GCF).

(i) Prime factorization of 32 and 54 are:

32 = 2 × 2 × 2 × 2 × 2

54 = 2 × 3 × 3 × 3

From above prime factorization we got that the highest common factor of 32 and 54 is 2

(ii) Prime factorization of 18 and 24 are:

18 = 2 × 3 × 3

24 = 2 × 2 × 2 × 3

From above prime factorization we got that the highest common factor of 18 and 24 is 3×2 ⇒ 6

(iii) Prime factorization of 30 and 70 are:

30 = 2 × 3 × 5

70 = 2 × 5 × 7

From above prime factorization we got that the highest common factor of 30 and 70 is 2×5 ⇒ 10

(iv) Prime factorization of 56 and 88 are:

56 = 2 × 2 × 2 × 7

88 = 2 × 2 × 2 × 11

From above prime factorization we got that the highest common factor of 56 and 88 is 2×2×2 ⇒ 8

(v) Prime factorization of 475 and 495 are:

475 = 5 × 5 × 19

495 = 3 × 3 × 5 × 11

From above prime factorization we got that the highest common factor of 475 and 495 is 5

(vi) Prime factorization of 75 and 243 are:

75 = 3 × 5 × 5

243 = 3 × 3 x 3 × 3 × 3

From above prime factorization we got that the highest common factor of 75 and 243 is 3

(vii) Prime factorization of 75 and 243 are:

240 = 2 × 2 ×2 × 2 × 3 × 5

6552 = 2 × 2 × 2 × 3 x 3 × 7 × 13

From above prime factorization we got that the highest common factor of 240 and 6552 is 2×2×2×3 ⇒ 24

(viii) Prime factorization of 155 and 1385 are:

155 = 5 × 31

1385 = 5 × 277

From above prime factorization we got that the highest common factor of 155 and 1385 is 5

(ix) Prime factorization of 100 and 190 are:

100 = 2 × 2 × 5 × 5

190 = 2 × 5 ×19

From above prime factorization we got that the highest common factor of 155 and 1385 is 2×5 ⇒ 10

(x) Prime factorization of 105 and 120 are:

105 = 3 × 5 × 7

120 = 2 × 2 × 2 × 3 × 5

From above prime factorization we got that the highest common factor of 105 and 120 is 3×5 ⇒ 15

(i)

Concept used :

To obtain the HCF of two positive integers, say c and d, with c > d,we follow
the steps below:
Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.
Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
Now,
We know that,

= 225>135

Applying Euclid’s division algorithm:
(Dividend = Divisor × Quotient + Remainder)

225 = 135 ×1+90

Here remainder = 90,

So, Again Applying Euclid’s division algorithm

135 = 90×1+45

Here remainder = 45,

So, Again Applying Euclid’s division algorithm

90 = 45×2+0

Remainder = 0,

Hence,

HCF of (135, 225) = 45

(ii)
Concept used :
To obtain the HCF of two positive integers, say c and d, with c > d,we follow
the steps below:
Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.
Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
Now,
We know that,

38220>196

So, Applying Euclid’s division algorithm

38220 = 196×195+0 (Dividend = Divisor × Quotient + Remainder)

Remainder = 0

Hence,

HCF of (196, 38220) = 196

(iii)

Concept used :
To obtain the HCF of two positive integers, say c and d, with c > d,we follow
the steps below:
Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.
Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
Now,
We know that,

867>255

So, Applying Euclid’s division algorithm

867 = 255×3+102 (Dividend = Divisor × Quotient + Remainder)

Remainder = 102

So, Again Applying Euclid’s division algorithm

255 = 102×2+51

Remainder = 51

So, Again Applying Euclid’s division algorithm

102 = 51×2+0

Remainder = 0

Hence,

(HCF 0f 867 and 255) = 51

(i) Using Euclid’s Division Lemma

a = bq + r, (o ≤r<b)

963 = 657×1 + 306

657 = 306×2 + 45

306 = 45×6+36

45 = 36×1+9

36 = 9×4+0

∴ HCF (657, 963) = 9

(ii) Using Euclid’s Division Lemma

a = bq + r, (o ≤r<b)

592 = 252×2+88

252 = 88×2+76

88 = 76×1+12

76 = 12×6+4

12 = 4×3+0

∴ HCF (592, 252) = 4

(iii) Using Euclid’s Division Lemma

a = bq + r, (o ≤r<b)

1155 = 506×2+143

506 = 143×3+77

143 = 77×1+66

77 = 66×1+11

66 = 11×6+0

∴ HCF (506, 1155) = 11

(iv) Using Euclid’s Division Lemma

a = bq + r, (o ≤r<b)

1288 = 575×2+138

575 = 138×4+23

138 = 23×6+0

∴ HCF (1288, 575) = 23

HCF of 468 and 222 is found by division mehtod:

Therefore, HCF (468,222) = 6

Now, we need to express the HCF of 468 and 222 as 468x + 222y where x and y are any two integers.

Now, HCF i.e. 6 can be written as,

HCF = 222 - 216 = 222 - (24 × 9)

Writing 468 = 222 × 2 + 24, we get,

⇒ HCF = 222 - {(468 – 222 x 2) × 9}

⇒ HCF = 222 - {(468 ×9) – (222 × 2 × 9)}

⇒ HCF = 222 - (468 × 9) + (222 × 18)

⇒ HCF = 222 + (222 × 18) - (468 × 9)

Taking 222 common from the first two terms, we get,

⇒ HCF = 222[1 + 18] – 468 × 9

⇒ HCF = 222 × 19 – 468 × 9

⇒ HCF = 468 ×(-9) + 222 ×(19)

Let, say, x = -9 and y =19

Then, HCF = 468 ×(x) + 222 ×(y)

Therefore the HCF of 468 and 222 is written in the form of 468x + 222y where, -9 and 19 are the two integers.

Prime factors of 408: 2 × 2 × 2 x 3 × 17

Prime factors of 1032: 2 × 2 ×2 × 3 × 43

HCF of 408 and 1032 = 2 × 2 × 2 × 3 = 24

According to question:

24 = 1032 m – 408 × 5

24 = 1032 m – 2040

2064 = 1032 m

∴ m = ⇒ 2

Therefore m = 2

Prime factors of 657: 3 × 3 × 73

Prime factors of 963: 3 × 3 × 107

HCF of 657 and 963 = 3 × 3 = 9

According to question:

the HCF of 657 and 963 is expressible in the form 657 x + 963 × -15

⇒ 9 = 657x + 963 × -15

⇒ 9 = 657x - 14445

⇒ 9 + 14445 = 657x

∴ x = 14454/657 = 22

Therefore x = 22

To find: the largest number which divides 615 and 963 leaving remainder 6 in each case.

Solution:

Let the HCF of 615 and 963 be x.

Since it is given remainder is 6 in each case,

Therefore for the numbers to be completely divisible, 6 should be subtracted from both the numbers.

Therefore new numbers are:

615-6 = 609

963-6 = 957

Prime factors of 609 = 3 × 3 × 29

Prime factors of 957= 3 × 11 × 29

Therefore HCF of 609 and 957 is: 3 × 29 = 87

Hence the largest number which divides 615 and 963 leaving remainder 6 in each case is 87.

The new numbers after subtracting remainders are:

285-9 = 276

1249-7 = 1242

Prime factors of 276 = 23 × 3 × 2

Prime factors of 1242 = 2 × 3 × 3 × 3 × 23

Therefore HCF of 276 and 1242 is: 2 × 3 × 23 = 138

Hence the greatest number which divides 285 and 1249 leaving remainder 9 and 7 respectively is138

The new numbers after subtracting remainders are:

280-4 = 276

1245-3 = 1242

Prime factors of 276 = 23 × 3 × 2

Prime factors of 1242 = 2 × 3 × 3 × 3 × 23

Therefore HCF of 276 and 1242 is: 2 × 3 × 23 = 138

Hence the greatest number which divides 280 and 1245 leaving remainder 4 and 3 respectively is138

The new numbers after subtracting remainders are:

626-1 = 625

3127-2 = 3125

15628-3 = 15625

Prime factors of 625 = 5 × 5 × 5 × 5

Prime factors of 3125 = 5 × 5 × 5 × 5 × 5

Prime factors of 15625 = 5 × 5 × 5 × 5 × 5 ×x 5

Therefore HCF of 625, 3125 and 15625 is: 5 × 5 × 5 × 5= 625

Hence the largest number which divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively is 625

The new numbers after subtracting remainders are:

445-4 = 441

572-5 = 567

699-6 = 693

Prime factors of 441= 3 × 3 × 7 × 7

Prime factors of 567 = 3 × 3 × 3 × 3 × 7

Prime factors of 693 = 3 × 3 × 7 ×11

Therefore HCF of 441, 567 and 693 is: 3 x 3 x7 = 63

Hence the greatest number that will divide 445, 572 and 699 leaving remainder 4, 5 and 6 respectively is 63

The new numbers after subtracting remainders are:

2011-9 = 2002

2623-5 = 2618

Prime factors of 2002= 2 × 7 × 11 × 13

Prime factors of 2618 = 2 × 7 × 11 × 17

Therefore HCF of 2002 and 2618 is: 2 × 7 × 11 = 154

Hence the greatest number which divides 2011 and 2623 leaving remainder 9 and 5 respectively is 154

To find maximum number of columns we should find HCF of 616 and 32

Using Euclid’s algorithms:

Let a = 616 and b = 32

a = bq + r, (o ≤r<b)

616 = 32×19+8

32 = 8×4+0

∴ HCF of 616 and 32 is 8

Therefore the maximum number of columns in which army contingent to march is 8

To find greatest capacity of tin we should find HCF of 120 and 180 and 240

Prime factors of 120= 2 × 2 × 2 × 3 × 5

Prime factors of 180 = 2 × 2 × 3 × 3 × 5

Prime factors of 240 = 2 × 2 × 2 × 2 × 3 × 5

Therefore HCF of 120, 180 and 240 is: 2 × 2 × 3 × 5 = 60

Therefore the greatest capacity of a tin is 60 Liters

For having the full packs of both and the same number of pencils and crayons, we need to find LCM of 24 and 32

Prime factors of 24 = 2 × 2 × 2 × 3

Prime factors of 32 = 2 × 2 × 2 × 2 × 2

Therefore LCM of 24 and 32 is: 2 × 2 × 2 × 2 × 2 × 3 = 96

Number of colour pencil packs = = 4

Number of crayons packs = = 3

Hence 4 packets of colour pencils and 3 packets of crayons would be bought.

To find greatest number of cartons each stack would have, we should find HCF of 144 and 90

Prime factors of 144 = 2 × 2 × 2 × 2 × 3 × 3

Prime factors of 90 = 2 × 3 × 3 × 5

Therefore HCF of 144 and 90 is: 2 × 3 × 3 = 18

Therefore the greatest number of cartons each stack would have is: 18

Given : Two brands of chocolates are available in packs of 24 and 15 respectively.
To find : the least number of boxes of each kind.
Solution:

Prime factors of 24 = 2 × 2 × 2 × 3

Prime factors of 15 = 3 × 5

Therefore LCM of 24 and 15 is: 2 × 2 × 2 × 3 × 5 = 120

Number of boxes for first chocolate kind = = 5

Number of boxes for second chocolate kind = = 8

Hence 5 boxes of first kind and 8 boxes of second kind needed to buy.

Given: The size of the bathroom is 10 ft. by 8 ft.

To find: the size in inches of the tile required that has to be cut and how many such tiles are required?

Solution:

To find the largest size of tile, we should find HCF of 10 and 8

Prime factors of 10 = 2 × 5

Prime factors of 8 = 2 × 2 × 2

Therefore HCF of 10 and 8 is: 2 ft

Since i ft = 12 inches

Therefore the largest size of tile is: 2 × 12 inches = 24 inches

Number of tiles =

=

= 20 tiles

NOTE: Always find the HCF of the given values to find their maximum.

To find the number of biscuit packets and pastries, we should find HCF of 15 and 12

Prime factors of 15 = 3 × 5

Prime factors of 12 = 2 × 2 × 3

Therefore HCF of 15 and 12 is: 3

Number of pastries = = 5

Number of biscuit packets = = 4

To find the largest number of animals, we should find HCF of 105, 140 and 175

Prime factors of 105 = 3 × 5 × 7

Prime factors of 140 = 2 × 2 × 5 × 7

Prime factors of 175 = 5 × 5 × 7

Therefore HCF of 105, 140 and 175 is: 5 × 7 = 35

Hence, 35 animals went in each trip

Given: The length, breadth and height of a room are 8 m and 25 cm, 6 m 75 cm and 4 m 50 cm.

To find: the longest rod which can measure the three dimensions of the room exactly.

Solution:

To find the length of largest rod, we should find HCF of 8m and 25 cm, 6 m 75 cm and 4 m 50 cm

As 1 m = 100 cm

⇒ 8m and 25 cm = 8 × 100 cm + 25 cm
= 800 cm + 25 cm
= 825 cm

⇒ 6 m and 75 cm = 6 × 100 cm + 75 cm
= 600 cm + 75 cm
= 675 cm

Length = 825 cm; Breadth = 675 cm; Height = 450 cm

Prime factors of 825 = 3 × 5 × 5 × 11

Prime factors of 675 = 3 × 3 × 3 × 5 × 5

Prime factors of 450 = 2 × 3 × 3 × 5 × 5

Therefore HCF of 825, 675 and 450 is: 3 × 5 × 5 = 75

Hence, the length of rod is: 75 cm

(i) Prime factor of 420 = 2 × 2 × 3 × 5 × 7

(ii) Prime factor of 468 = 2 × 2 × 3 × 3 × 13

(iii) Prime factor of 945 = 3 × 3 × 5 × 7

(v) Prime factor of 7325 = 5 × 5 × 293

(i) Prime factors of 20570 = 2 × 5 × 11 × 11 × 17

(ii) Prime factors of 58500 = 2 × 2 × 3 x 3 × 5 x 5 × 13

(iii) Prime factors of 45470971 = 7 × 7 × 13 x 13 × 17 × 17 × 19

Numbers are of two types - prime and composite. Prime numbers has only two factors namely 1 and the number itself whereas composite numbers have factors other than 1 and itself.

For example: 7 is a prime number as it can be divided by 1 and 7 only whereas 14 is a composite number as it can be divided by 7, 2 and 1

From the question it can be observed that,

7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) [ taking 13 common ]

= 13 × 78

= 13 × 13 × 6

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 x 1 + 1)

= 5 × (1008 + 1)

= 5 ×1009

In the above we observe that 1009 cannot be factorized further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorization must include primes 2 and 5 both as 10 = 2 ×5

Prime factorization of 6ⁿ = (2 ×3) ⁿ

In the above equation it is observed that 5 is not in the prime factorization of 6ⁿ

By Fundamental Theorem of Arithmetic Prime factorization of a number is unique. So 5 is not a prime factor of 6ⁿ.

Hence, for any value of n, 6ⁿ will not be divisible by 5.

Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

(i) Prime factors of 26 = 2 × 13

Prime factors of 91 = 7 ×13

Hence LCM of 26 and 91 = 2 × 7 × 13 = 182

HCF of 26 and 91 is = 13

LCM × HCF = 182 × 13 = 2366

Product of two numbers = 26 × 91 = 2366

Hence from above result we got LCM × HCF = Product of the numbers

(ii) Prime factors of 510 = 2 × 3 × 5 × 17

Prime factors of 92 = 2 × 2 × 13

Hence LCM of 92 and 510 = 2 × 2 × 3 × 5 × 17 × 23 = 23460

HCF of 92 and 510 is = 2

LCM × HCF = 23460 × 2 = 46920

Product of two numbers = 92×510 = 46920

Hence from above result we got LCM × HCF = Product of the numbers

(iii) Prime factors of 336 = 2 × 2 × 2 × 2 × 3 × 7

Prime factors of 54 = 2 × 3 × 3 × 3

Hence LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 3024

HCF of 54 and 336 is = 2 × 3 = 6

LCM × HCF = 3024 × 6 = 18144

Product of two numbers = 336 × 54 = 18144

Hence from above result we got LCM × HCF = Product of the numbers

(i) Prime factors of 12 = 2 × 2 × 3

Prime factors of 15 = 3 × 5

Prime factors of 21 = 3 ×7

Hence LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 = 420

HCF of 12, 15 and 21 = 3

(ii) Prime factors of 17 = 1 × 17

Prime factors of 23 = 1 ×23

Prime factors of 29 = 1 × 29

Hence LCM of 17, 23 and 29 = 11339

HCF of 17, 23 and 29 = 1

(iii) Prime factors of 8 = 1 × 2 × 2 × 2

Prime factors of 9 = 1 × 3 × 3

Prime factors of 25 = 1 × 5 × 5

Hence LCM of 8, 9 and 25 = 1800

HCF of 8, 9 and 25 = 1

(iv) Prime factors of 40 = 2 × 2 × 2 × 5

Prime factors of 36 = 2 × 2 × 3 × 3

Prime factors of 126 = 2 × 3 × 3 × 7

Hence LCM of 40, 36 and 126 = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520

HCF of 40, 36 and 126 = 2

(v) Prime factors of 12 = 2 × 2 × 3

Prime factors of 84 = 2 × 2 × 3 × 7

Prime factors of 90 = 2 × 3 × 3 × 5

Hence LCM of 12, 84 and 90 = 2 × 2 × 3 × 3 × 5 × 7 = 1260

HCF of 12, 84 and 90 = 2 × 3 = 6

(vi) Prime factors of 15 = 3 × 5

Prime factors of 24 = 2 × 2 × 2 × 3

Prime factors of 36 = 2 × 2 × 3 × 3

Hence LCM of 15, 24 and 36 = 2 × 2 × 2 × 3 × 3 × 5 = 360

HCF of 15, 24 and 36 = 3

We know that LCM × HCF = Product of the numbers

Therefore LCM = = = 22338

NO

LCM should be divisible by HCF since HCF is the common factor of both the numbers. But, in this case 380 is not divisible by 16 therefore, the two numbers are not possible which have 16 as their HCF and 380 as their LCM

We know that LCM × HCF = Product of the numbers

Therefore Other Number = = = 435

We know that LCM × HCF = Product of the numbers

Therefore Other Number = = = 192

Given: The LCM and HCF of two numbers are 180 and 6 respectively. One of the numbers is 30.

To find: The other number

Solution:

We know that:

LCM × HCF = Product of the numbers

LCM × HCF = a × b ...... (1)

Where a and b are numbers

One number i.e a is 30.

LCM is 180.

HCF is 6.

Therefore, the other number is 36.

To find the smallest number we should find LCM of 468 and 520

Prime factors of 468 = 2 × 2 × 3 × 3 × 13

Prime factors of 520 = 2 × 2 × 2 × 5 × 13

Hence LCM of 468 and 520 = 2 × 2 ×2 × 3 × 3 ×5×13 = 4680

Therefore the smallest number which when increased by 17 is exactly divisible by both 520 and 468= LCM – 17 ⇒ 4680 – 17 = 4663

Greatest number of 6 digits is 999999.

Required number must be divisible by the LCM of 24, 15 and 36 i.e., by 360.
Now on dividing six digit greatest number by LCM we get 279 as remainder.

Therefore the greatest number of 6 digits exactly divisible by 24, 15 and 36
= Six digit greatest number – remainder
= 999999 – 279

= 999720

To find : The number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21

Solution : To be exactly divisible by each of 8, 15 and 21,

the required number must be divisible by the LCM of 8, 15 and 21 i.e.by 840.

Now on dividing 110000 by 840 we get 800 as remainder.

Therefore the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21

is 110000 – Remainder

⇒ 110000 – 800 = 109200

The smallest number would be the LCM of 28 and 32

LCM of 28 and 32 = 224

Therefore the required smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively would be:

Number = LCM – (sum of the remainders)

⇒ 224 – (12 + 8)

⇒ 224 – 20 = 204

Therefore the required smallest number would be the LCM of the numbers

Prime factors of 35 = 5 × 7

Prime factors of 56 = 2 × 2 × 2×7

Prime factors of 91 = 7 × 13

LCM of 35, 56 and 91 ⇒ 2 × 2 × 2 × 5 × 7×13 = 3640

Number = LCM + Remainder

⇒ 3640 + 7 = 3647

The least number would be the LCM of the numbers from 1 to 10

We want LCM(1,2,3,4,5,6,7,8,9,10)

We can ignore 1, since any counting number is divisible by 1.

We prime factor each of the counting numbers from 2 to 10

2 = 2
3 = 3
4 = 2×2
5 = 5
6 = 2×3
7 = 7
8 = 2×2×2
9 = 3×310 = 2×5

The LCM of all those must have as many factors of
each prime that appears in any factorization

2 appears at most 3 times as a factor of 8

3 appears at most 2 times as a factor of 9

5 appears at most 1 time as a factor if 5 and 10

7 appears at most 1 time as a factor of 7

So the LCM has

3 factors of 2, 2 factors of 3, and 1 factor each of 5 and 7

LCM = 2×2×2×3×3×5×7 = 2520

Hence, the least number that is divisible by all the numbers between 1 and 10 (both inclusive) is 2520

Length of the courtyard
= 18 m 72 cm
= [18(100) + 72] cm [As, 1 m = 100 cm]
= 1872 cm

The breadth of the courtyard
= 13 m 20 cm
= [13(100) + 20] cm
= 1320 cm

To find the maximum edge of the tile we need to calculate HCF of length and breadth,

Using Euler’s division lemma: a = pq + r where 0≤r<p

1872 = 1320 × 1 + 552

As 'r' is not equals to 0,

So apply Euler's division on 1320 and 552,

1320 = 552 × 2 + 216

552 = 216 × 2 + 120

216 = 120×1 + 96

120 = 96 ×1 + 24

96 = 24 × 4 + 0

Therefore HCF of 1872 and 1320 is 24

Maximum edge can be 24 cm.

Number of tile = = = 4290 tiles

Given: A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48, 60 and 72 km a day.

To find: When will they meet again

Solution:


Circumference of the circular field is 36 km.

Use the formula:

Time taken by first cyclist = = = 7.5 days

As 1 day = 24 hours


⇒ 7.5×24 = 180 Hrs

Similarly,

Time taken by second cyclist = = 6 days


⇒ 6×24 = 144 Hrs

Time taken by third cyclist = = 5 days


⇒ 5×24 = 120

we need to find the minimum time at which they will meet.

So we need to find LCM of the time they all take.

Now LCM of 180, 144 and 120 is 720 hours,

Therefore the time when all three will meet again = = 30 days

Note: To find the minimum of given values always find their L.C.M.

To calculate minimum distance Minimum distance we should calculate LCM

Prime factors of 80 = 2×2×2 × 2 ×5

Prime factors of 85 = 5 ×17

Prime factors of 90 = 2×3×3×5

LCM of 80, 85 and 90 ⇒ 2 ×2 × 2 ×2 × 3×3 ×5 × 17 = 12240

Minimum distance is 12240 cm,

As 1 m = 100 cm

So 12240 = 120 m 40 cm

Therefore the minimum distance each should walk so that he can cover the distance in complete steps is 122 m 40 cm

(i) Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

√2 =

is a rational number as p and q are integers. Therefore √2 is rational which contradicts the fact that √2 is irrational.

Hence, our assumption is false and we can say that is irrational.

(ii) Let assume that 7√5 is rational therefore it can be written in the form of

where p and q are integers and q≠0. Moreover, let p and q have no common divisor > 1.

7√5 = for some integers p and q

Therefore √5 =

is rational as p and q are integers, therefore √5 should be rational.

This contradicts the fact that √5 is irrational.

Therefore our assumption that 7√5 is rational is false. Hence 7√5 is irrational.

(iii) Let assume that 6 + √2 is rational therefore it can be written in the form of

where p and q are integers and q≠0. Moreover, let p and q have no common divisor > 1.

6 + √2 = for some integers p and q

Therefore √2 = - 6

Since p and q are integers therefore - 6 is rational, hence √2 should be rational. This contradicts the fact that √2 is irrational. Therefore our assumption is false. Hence, 6 + √2 is irrational.

(iv) Let us assume that 3 - is rational

Therefore 3 - can be written in the form of where p and q are integers and q≠0

3 - =

⇒ -3 =

⇒ =

Since p, q and 3 are integers therefore is rational number

But we know is irrational number, Therefore it is a contradiction.

Hence 3 - is irrational

(i) Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

√7=

is a rational number as p and q are integers. This contradicts the fact that √7 is irrational, so our assumption is incorrect. Thereforeis irrational

(ii) Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

√5=

is a rational number as p and q are integers. This contradicts the fact that √5 is irrational, so our assumption is incorrect. Thereforeis irrational

(iii) Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

√2= -4

-4 is a rational number as p and q are integers. This contradicts the fact that √2 is irrational, so our assumption is incorrect. Therefore is irrational.

(iv) Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

√2=

is a rational number as p and q are integers. This contradicts the fact that √2 is irrational, so our assumption is incorrect. Therefore is irrational.

Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

√3= 2 -

is a rational number as p and q are integers. This contradicts the fact that √3 is irrational, so our assumption is incorrect. Therefore is irrational.

Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

√2= - 3

is a rational number as p and q are integers. This contradicts the fact that √3 is irrational, so our assumption is incorrect. Therefore is irrational.

Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

5√2= - 4

is a rational number as p and q are integers. This contradicts the fact that √2 is irrational, so our assumption is incorrect. Therefore is irrational.

To prove : is an irrational number.

Solution:

Let assume that is rational.

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

2√3= 5 -


⇒ √3 =

is a rational number as p and q are integers. This contradicts the fact that √3 is irrational,
so our assumption is incorrect. Therefore is irrational.

Note: Sometimes when something needs to be proved, prove it by contradiction.
Where you are asked to prove that a number is irrational prove it by assuming that it is rational number
and then contradict it.

Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write =

2√3= + 1

is a rational number as p and q are integers. This contradicts the fact that √3 is irrational, so our assumption is incorrect. Therefore is irrational.

Given:

To prove: is irrational.

Proof:

Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write

= √5

is a rational number as p and q are integers. This contradicts the fact that √5 is irrational, so our assumption is incorrect. Therefore is irrational.

Note: Sometimes when something needs to be proved, prove it by contradiction.
Where you are asked to prove that a number is irrational prove it by assuming that it is rational number
and then contradict it.

Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write √5 = - √3

(√5)² = ( - √3)²

5 = - +3

5 - 3 = -

- 2 =

= √3

is a rational number as p and q are integers. This contradicts the fact that √3 is irrational, so our assumption is incorrect. Therefore is irrational.

Given:

To prove: is irrational.

Proof:

Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0



Therefore we can write,



Squaring both sides we get,



Apply the formula ( a - b )² = a² + b² - 2ab in ,

is a rational number as p and q are integers.

This contradicts the fact that √3 is irrational, so our assumption is incorrect.

Therefore is irrational.

Note: Sometimes when something needs to be proved, prove it by contradiction.
Where you are asked to prove that a number is irrational prove it by assuming that it is rational number
and then contradict it.

Let assume that √p is rational

Therefore it can be expressed in the form of , where a and b are integers and b≠0

Therefore we can write √p =

(√p)² = ( )²

P =

a² = pb²

Since a² is divided by b², therefore a is divisible by b.

Let a = kc

(kc)² = pb²

K²c² = pb²

Here also b is divided by c, therefore b² is divisible by c². This contradicts that a and b are co-primes. Hence is an irrational number.

Since it is given that p and q are prime positive integer.

Let us assume that is a rational number of the form ,





Squaring both sides we get,

Since p, q are integers therefore is rational number

But we know is irrational number, Therefore it is a contradiction.

Hence is irrational

(i) 8 = 2³ since the denominator is of the form 2^m therefore the expression is terminating.

(ii) 441 = 3² × 7² since the denominator is not in the form of 2^m and 5ⁿ Therefore the given expression is Non-terminating repeating

(iii) =

10 = 2×5 therefore the denominator is of the form 2^m × 5ⁿ

Hence, the given expression is Terminating

(iv) = =

30 = 2×3×5

Since the denominator is not of the form 2^m × 5ⁿ

Hence, the given expression is Non-terminating repeating Terminating

(v)

Since the denominator is not in the form 2^m × 5ⁿ

Hence, the given expression is Non-terminating repeating Terminating

(i) = = the denominator is in the form of

⇒ = 0.375

(ii) = = the denominator is in the form of

⇒ = 0.104

(iii) = = the denominator is in the form of

⇒ = 0.0875

(iv) = = the denominator is in the form of

⇒ = 23.3408

(v) = = the denominator is in the form of

⇒ = 0.0004128

(i) 43.123456789 = =

Since Prime factorization of the denominator is in the form, where m, n are non-negative integers.

(ii) Since has non-terminating repeating decimal expression, therefore Prime factorization of the denominator contains factors other than 2 or 5.

(iii) Since has non-terminating repeating decimal expression, therefore Prime factorization of the denominator contains factors other than 2 or 5.

(iv) Since has non-terminating repeating decimal expression, therefore Prime factorization of the denominator contains factors other than 2 or 5.

Euclid’s Division Lemma: Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b. (It is a technique to compute the Highest Common Factor (HCF) of two given positive integers.)

The prime factorization of 144 is as follows:

144 = 2 × 2 × 2 × 2 × 3 × 3

⇒ 144 = 2⁴ × 3²

We know that the exponent of a number a^m is m.

∴ The exponent of 2 in the prime factorization of 144 is 4.

Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

We know that LCM of two or more numbers is always divisible by their HCF.

1200 is divisible by 600, 200 and 400 but not by 500.

98 = 49 × 2 = 7² × 2

98 is factorised as a product of primes i.e. 7² × 2.

If any number ends with the digit 0, it should be divisible by 10,

i.e. it will be divisible by 2 and 5.

Prime factorization of n is given as 2³ × 3⁴ × 5⁴ × 7.

It can be observed that there is (2 × 5) × (2 × 5) × (2 × 5)

⇒ 10 × 10 × 10 = 1000

Thus, there are 3 zeros in n.

The prime factorization of 144 is as follows:

144 = 2 × 2 × 2 × 2 × 3 × 3

⇒ 144 = 2⁴ × 3²

We know that the exponent of a number a^m is m.

The exponent of 2 in the prime factorization of 144 is 4.

The prime factorization of 196 is as follows:

196 = 2 × 2 × 7 × 7

⇒ 98 = 2² ×7²

We know that the exponent of a number a^m is m.

∴The sum of powers = 2 + 2 = 4

The prime factorization of 98 is as follows:

98 = 2 × 7 × 7

⇒ 98 = 2¹ ×7²

We know that the exponent of a number a^m is m.

∴The sum of powers = 1 + 2 = 3

The sum of the exponents of prime factors in the prime factorization of 98 is 3.

Let x = p/q be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

The maximum power of 2 or 5 in the given rational number is 2.

So, it will terminate after 2 places of decimals.

If any number ends with the digit 0, it should be divisible by 10,

i.e. it will be divisible by 2 and 5.

Prime factorization of n is given as 2³ × 3² × 5² × 7.

It can be observed that there is (2 × 5) × (2 × 5)

⇒ 10 × 10 = 100

Thus, there are 2 zeros in n.

The number of consecutive zeros in n is 2.

Let us take p₁ = 5 and p₂ = 3

Then p₁² – p₂² = 25 – 9 = 16

16 is an even number.

We know that LCM = Product of the greatest power of each prime factor, involved in the numbers.

So, LCM(a, b) = p³q²

We know that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.

Here a × b = 1080 and HCF = 30

∴ LCM = (a × b)/ HCF

⇒ LCM = 1080/30

⇒ LCM = 36

The LCM of given product of two numbers is 36.

We know that HCF = Product of the smallest power of each common prime factor in the numbers.

So, HCF(a, b) = pq

Let x = p/q be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

We know that HCF = Product of the smallest power of each common prime factor in the numbers.

So, HCF(a, b) = pq²

Let x = p/q be a rational number, such that the prime factorization of q is not of the form 2ⁿ 5^m, where n, m are non-negative integers. Then x has a decimal expansion which is non-terminating repeating (recurring).

We know that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.

Here LCM = 36, HCF = 2 and b = 18

Then, 2 × 36 = a × 18

a = (2 × 36) / 18

a = 4

Let us take the last factors of the tree 3 and 7.

Then 3 × 7 = 21

Now we get the factors in the next level as 21 and 2.

So, 21 × 2 = 42

The missing entries are 42, 21.

We know that HCF = Product of the smallest power of each common prime factor in the numbers.

The prime factorization of

95 = 5 × 19

And 152 = 2³ × 19

∴ HCF = 19

Let be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non-negative integers. Then 'x' has a decimal expansion which terminates after n or m places, whichever is maximum

The maximum power of 2 or 5 in the given rational number is 4.

So, it will terminate after 4 places of decimals.

The decimal expansion of the given rational number will terminate after 4 places of decimals.

We know that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.

Here HCF = 13, a = 26 and b = 169

Then,

13 × LCM = 26 × 169

LCM = (26 × 169) / 13

LCM = 338

Let be a rational number, then x is terminating if and only if, the denominator q has the form 2ⁿ 5^m

The denominator is not of the form 2ⁿ 5^m as it has 7² as its factor.

Therefore, Given rational number has a non-terminating decimal representation.

Given: number
To find: Whether the given number is rational or irrational
Solution:

Factorize 45 and 20.

⇒

⇒

⇒

⇒

⇒ 3 + 3 = 6

We know that a rational number is defined as the number which can be written in the form of p/q.
As 6 can be written as 6/1.

The given number after simplification gives a rational number.

We know that LCM = Product of the greatest power of each prime factor, involved in the numbers.

Since the power of 3 in LCM is 2,

C = 3²× 5

Let x = p/q be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

The denominator 1250 can also be written as 2 × 5⁴.

The maximum power of 2 or 5 in 1250 is 4.

So, it will terminate after 4 places of decimals.

An algorithm is a series of well-defined steps which gives a procedure for solving a type of problem. The word algorithm comes from the name of the 9th-century Persian mathematician al-Khwarizmi.

Given: If p and q are co-prime numbers.

To find: p² and q² are

Solution:

Two numbers are co-prime if their HCF is 1 i.e they have no number common other than 1.

Let us take p = 4 and q = 5.

As 4 and 5 has no common factor other than 1,

p and q are co-prime.

Now p² = 16 and q² = 25,

So p² and q² are also co-prime.

A lemma is a proven statement used for proving another statement.

We know that HCF = Product of the smallest power of each common prime factor in the numbers.

For any two prime numbers, one of the common prime factors will be 1.

Let x = p/q be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

(i) Here q = 225

225 can be written as 3²×5²

Since it is in the form of 5^m, it is a terminating decimal.

(ii) Here q = 18

18 can be written as 2 × 3²

Since 3 is also there and it is not in the form of 2ⁿ5^m, it is not a terminating decimal.

(iii) Here q = 21

21 can be written as 3 × 7

Since it is not in the form of 2ⁿ5^m, it is not a terminating decimal.

(iv) Here q = 250

250 can be written as 2 × 5³

Since it is in the form of 2ⁿ5^m, it is a terminating decimal.

We know that LCM = Product of the greatest power of each prime factor, involved in the numbers.

If p and q are two prime numbers, then p × q is their LCM.

The prime factors of a + b

= 3 + 7 = 10 = 2 × 5

So the least prime factor is 2.

Since, the given number is non-terminating recurring,

It is a rational number.

We know that the factors of a prime number are 1 and the number itself.

The total number of factors of a prime number is 2.

Every composite number can be expressed (factorized) as a product of primes.

If we multiply we will get a perfect square 9 which is a rational number.

Smallest composite number = 4

= 2 × 2

Smallest prime number = 2

We know that HCF = Product of the smallest power of each common prime factor in the numbers.

HCF (4, 2) = 2

The HCF of the smallest composite number and the smallest prime number is 2.

Let x = p/q be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

So 1/3 should be multiplied by 3/10 so that it is in the form of 2ⁿ5^m.

9 2ⁿ – 4 2ⁿ is of the form a 2ⁿ — b 2ⁿ.

It is divisible by both a - b and a + b.

So, 9 2ⁿ – 4 2ⁿ is divisible by both 9 - 4 = 5 and 9 + 4 = 13.

True.

LCM of two or more numbers is always divisible by their HCF.

Example: Let us take two numbers 6 and 20.

The prime factorization of

6 = 2¹ × 3¹ and 20 = 2² × 5¹

HCF = 2

LCM = 2² × 3 × 5 = 60

∴ 2 is a factor of 60.

For any n ∈ N, e and 5ⁿ end with 6 and 5 respectively.

Therefore, 6ⁿ – 5ⁿ always ends with 6 - 5 = 1.

True

When numbers are equal, LCM and HCF of two rational numbers are equal.

False

Prime numbers are always odd numbers and the sum of odd numbers is even.

Example: Let two prime numbers be 3 and 5.

Sum of 3 and 5 = 3 + 5 = 8 which is not a prime number.

Given:The sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF.
To find: The product of two numbers.
Solution: Let LCM be x and HCF be y.

Then, by given condition,Sum of LCM and HCF
x + y = 1260 (1)

Substituting x = 900 + y in equation (1)

900 + y + y = 1260

900 + 2y = 1260

2y = 360

y = 180

Then, x = 900 + 180 = 108

Thus, LCM = 1080 and HCF = 180

We know that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.

1080 × 180 = 194400

True

Since n(n + 1)(n + 2) is divisible by 2 and 3, then it is divisible by 6.

Example: Let us take three consecutive numbers 10, 11, 12.

Product = 10 × 11 × 12 = 1320

∴ 1320 is divisible by 6.

Any prime number greater than 3 is of the form 6k 1, where k is a natural number

and (6k 1)2 = 36k² 12k + 1 = 6k (6k 2) + 1

Thus, the remainder is 1.

True

Let a = bq + r. b = 2, q = m

r = 0.

a = 2m

True

Let a = bq + r: b = 2, q = m

-2<r<0 i.e., r = -1

a = 2m – 1 for odd integer.

False

The product of two irrational is sometimes an irrational number.

Example: If we multiply an irrational number with 0 we will get the product as 0 which is a rational number.

False

The sum of two irrational numbers is sometimes an irrational number.

Example: If the two irrational numbers have a sum zero, then the sum is a rational number.

Let us take n = 1.

2¹ × 5¹ = 10

So, for no value of n, 2ⁿ × 5ⁿ ends in 5.

No value of n

We know that HCF = Product of the smallest power of each common prime factor in the numbers.

For any two relatively prime numbers, one of the common prime factors will be 1.

We know that LCM = Product of the greatest power of each prime factor, involved in the numbers.

If a and b are two prime numbers, then a × b is their LCM.

False

HCF of two numbers is always a factor of their LCM.

But 12 is not a factor of 350.