Let the number of columns be x.
x is the largest number, which should divide both 104 and 96
104 = 96 x 1 + 8 1
96 = 8 x 12 + 0
∴  HCF of 104 and 96 is 8
Hence, 8 columns are required.

The required number is the LCM of 1, 2, 3, 4, 5, 6, 7, 8,9,10
∴  LCM = 2 x 2 x 3 x 2 x 3 x 5 x 7
 =2520

Let         x = 571
\(\Rightarrow \quad \sqrt { x } =\sqrt { 571 } \)
Now 571 lies between the perfect squares of (23)² and (24)²
Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19 ,23.
Since 571 is not divisible by any of the above numbers. So, 571 is a prime number.

Denominator = 500
= 2² x 5³
Decimal expansion, \(\frac { 257 }{ 500 } =\frac { 257\times 2 }{ 2\times { 2 }^{ 2 }\times { 5 }^{ 3 } } =\frac { 514 }{ 10^{ 3 } } \)
= 0.514

Let  \(5\sqrt { 6 } \)  be a rational number, which can be put in a/b, where b\(\neq \)0; a and b are co-prime
\(5\sqrt { 6 } =\frac { a }{ b } \)
\(\sqrt { 6 } =\frac { a }{ 5b } \)
⇒  \(\sqrt { 6 } \) =rational
But, we know that  \(\sqrt { 6 } \)   is an irrational number.
Thus, our assumption is wrong.
Hence, \(\sqrt { 6 } \) is an irrational number

Length = 8 m 50 cm = 850 cm
breadth = 6 m 25 cm = 625 cm
height = 4 m 75 cm = 475 cm
length of the longest rod is equal to

HCF (625, 850) = 25
∵ 25 divides 475
∴ HCF(62, 850, 475)=25

(15)ⁿ can end with the digit 0 only if (15)" is divisible by 2 and 5.
But prime factors of (15)ⁿ are 3ⁿ x 5ⁿ 
By Fundamental theorem of Arithmetic, there is no natural number II for which (15)ⁿ ends with the digit zero.

\({ 7 }^{ n }=(1\times 7)^{ n }={ 1 }^{ n }\times { 7 }^{ n }\)
So the only prime in the factorization of 7ⁿ  is 7, 1 not 2 or 5.
∴ 7ⁿ cannot end with the digit zero.

No.
15 does not divide 175.
LCM of two numbers should be exactly divisible by their HCF.
∴ Two numbers cannot have their HCF as 15 and LCM as 175

Let a be any positive integer Yz
By division algorithm
a=6q + r, where \(0\le r<6\)
∴ a = 6q,6q + 1,6q + 2,6q + 3,6q + 4,6q + 5
But a is an even integer
∴  a = 6q,6q + 2, 6q+ 4

\(1200=4\times 3\times (2\times 5)^{ 2 }\)
 \(={ 2 }^{ 4 }\times 3\times { 5 }^{ 2 }\)
The smallest natural number is 3.

Hence HCF (1,656,4,025)= 23

(7 x 13 x 11) + 11 = 11 x (7 x 13 + 1)
= 11 x (91 + 1)
=l1 x 92 = l1 x 2 x 2 x 23
and (7 x 6 x 5 x 4 x 3 x 2 x 1) + 3
= 3 (7 x 6 x 5 x 4 x 2 x 1 + 1)
= 3 x (1681)= 3 x 41 x 41
which is a composite number (more than one prime factor)

\(3\times 12\times 101+4=4(3\times 3\times 101+1)\)
 = 4(909+1)
 = 4(910)
= a composite number
[∵  Product of more than two factors]

∴ Composite number, x = 6762

\(a=\frac { 9,009 }{ 3,003 } =3\)
\(b=\frac { 1,001 }{ 143 } =7\)
Since 143 = 11 x 13, so c=11 or 13
and  d=13 or 11.

x =11,130

y = 5 x 13 = 65
and x = 3 x 195 = 585

Given HCF (306, 1,314) =18
LCM (306, 1,314) = ?
Let  a = 306
b = 1,314
We know that
a x b = LCM (a, b) x HCF (a, b)
⇒ 306 x 1,314 = LCM (a, b) x 18
\(\Rightarrow \quad LCM(a,b)=\frac { 306\times 1,314 }{ 18 } \)
∴ LCM (306, 1,314) = 22,338

240 = 228 x 1 + 12
and 228 = 12 x 19 + 0
⇒  HCF of 240 and 228 = 12

90 = 2 x 3² x 5
and 144 = 2⁴ x 3²
HCF = 2 x 3² = 18
LCM = 2⁴ x 3² x 5 = 720

HCF of K
k.2
k.3
k.2²
k.5 is k

13; x = - 2, y =1

Required number=LCM (1, 2, 3, 4, 5)
60

(i) \(\frac{11}{30}\)   (ii) \(\frac{100}{3}\)

Rational

No

19000

36

138

Prime factors of 66 = 2 x 3 x 11 = 2¹ x 3¹ x 11¹
Prime factors of 88 = 2 x 2 x 2 x 11 = 2³ x 11¹
and prime factors of 110 = 2 x 5 x 11 = 2¹ x 5¹ x 11¹
Then, HCF of 66, 88 and 110 = 2 x 11 = 22
\(\therefore \)  Required number of rows = \(\frac { 66 }{ 22 } +\frac { 88 }{ 22 } +\frac { 110 }{ 22 } \)
 = 3 + 4 + 5 = 12

Let a = n² - 1, where n = 1, 3, 5, ...
At n = 1, then a = (1)² - 1 = 1 - 1 = 0
At n = 3, then a = (3)² - 1 = 9 - 1 = 8
At n = 5, then a = (5)² - 1 = 25 - 1 = 24
which is divisible by 8.
Hence, n is an odd integer.

Here, 327.7081 is terminating. So, it represents a rational number.
Thus, \(327.7081=\frac { 3277081 }{ 10000 } =\frac { p }{ q } \)
Here, q = 10⁴ = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
= 2⁴ x 5⁴ = (2 x 5)⁴
So, the prime factors of q are 2 and 5.

\(\because \) smallest odd composite number = 9
and smallest odd prime number = 3.
\(\therefore \) HCF of 9 and 3 = 3
and LCM of 9 and 3 = 9
Now, if an odd number p divides q², then p is one of the factors of q², i.e. q² = pm, for some integer m.    .... (i) 
Now, q³ = q² . q \(\Rightarrow \) q³ = pm . q   [from Eq.(i)]
\(\Rightarrow \)  q³ = p (mq)
\(\Rightarrow \)  p is a factor of q³ also \(\Rightarrow \)  p divides q³

By Euclid's division algorithm,
a = bq+r
Take b=4
∴ Since \(0\le r<4\) ,  r=0, 1, 2, 3
So, a = 4q,4q + 1, 4q + 2,4q + 3
Clearly, a = 4q,4q + 2 are even, as they are divisible by 2.
Therefore A cannot be 4q, 4q + 2 as a is odd. But 4q + 1, 4q + 3 are odd, as they are not divisible by 2.