Let n be any positive integer.
By Eucild's division lemma, 11= 5q + r, \(0\le r<5\)
n = Sq, Sq + 1, Sq + 2,5q + 3 or 5q + 4, where \(q\epsilon w\) q is a whole number
now    n² = (5q)² = 25q² = 5(5q²) = 5m
n²= (5q + 1)²= 25q² + 10q + 1= 5m + 1
n² = (5q + 2)²= 25q² + 20q + 4 = 5m+ 4
Similarly  n²= (5q + 3)² = 5m + 4
and   n² =(5q + 4)² = 5m + 1
Thus square of any positive integer cannot be of the form 5m + 2 or 5m + 3.

Required number
= HCF of (1657 - 6) and (2037 - 5)
= HCF of (1651 and 2032
By using Euclid's division lemma, we get
2032 = (1651 x 1) + 381
Here, divisor is 1651 and remainder is 381
Again, by using Euclid's division lemma, we get
1651 = (381 x 4) + 127
Here, divisor is 381 and remainder is 127.
Again, by using Euclid's division lemma, we get
381 = (127 x 3) + 0
Here, the remainder is zero.
So, HCF of 1651 and 2032 is 127.
Hence, the required number is 127 which on dividing 1657 and 2037 leaves remainders 6 and 5, respectively.

Let the two number are a and b.
Given, HCF(a, b) = 2
and product (a x b) = 120
We know that, 
HCF (a, b) x LCM (a, b) = Product of a and b
\(\therefore \)   2 x LCM (a, b) = 120
\(\Rightarrow \)  LCM (a, b) = \(\frac{120}{2}\) = 6
Hence, the required LCM is 60.

We have, (3 x 5 x 7) + 7 = 105 + 7 = 112
\(\therefore \)  Prime factors of 112 = 2 x 2 x 2 x 2 x 7 = 2⁴ x 7
So, it is the product of prime factors 2 and 7.
Hence, it is a composite number.

On applying Euclid's division lemma for 960 and 432, we get
960 = (432 x 2) + 96
Here, remainder = 96 \(\neq \) 0,
so take new dividend as 432 and divisor as 96.
Then, we get 432 = (96 x 4) + 48
Here, remainder = 48 \(\neq \) 0,
so take new dividend as 96 and divisor as 48.
Then. we get 96 = (48 x 2) + 0
Here, the remainder is 0 (zero) and last divisor is 48.
Hence, HCF of 960 and 432 is 48.

n³-n = n n²-1)
= n (n+1)(n-1)
=( n-1) n(n+1)
= product of threeconsecutive positive integers
Now, we ave to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer a is of the form 3q, 3q + 1 or 3q + 2 for some integer q.
Let a, a + 1, a + 2 be any three consecutive integers.
Case I: if  a=3q
a(a + l)(a + 2) = 3q(3q + 1)(3q + 2)
 = 3q (2r)
= 6qr, which is divisible by 6.
(∵ Product of two consecutive integers (3q + 1) and (3q + 2) is an even integer, say 2r)
Case II: If   a=3q + 1
∴ a(a + l)(a + 2) = (3q + 1)(3q + 2)(3q + 3)
= (2r) (3)(q + 1)
= 6r(q + 1),
which is divisible by 6
Case III: If    a=3q + 2
ஃ a(a + l)(a + 2) = (3q + 2)(3q + 3)(3q + 4)
= multiple of 6 for ever
= 6r (say),
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6.

(i) 36  
(ii) 42

Hint LCM of 48. 72 and 108 = 2 x 2 x 2 x 2 x 3 × 3 x 3 = 432
Therefore, after 432 s, they will change simultaneously. We know that 60 s = 1min
⇒ 1s = 1/60 min
⇒ 432 s = 432/60 min =7min 12 s
Hence, the lights change simultaneously at 7 : 07 : 12 am.

Given, HCF of 657 and 963 = 657 x + 963 (- 15)   .... (i) 
By using Euclid's division lemma, we get
963 = (657 x 1) + 306
Here, divisor is 657 and remainder is 306.
so, again applying Euclid's division lemma, we get
657 = (306 x 2) + 45
Here, divisor is 306 and remainder is 45.
So, again applying Euclid's division lemma, we get
306 = (45 x 6) + 36
Here, divisor is 45 and remainder is 36.
So, again applying Euclid's division lemma, we get
45 = (36 x 1)  + 9
Here, divisor is 36 and remainder is 9.
So, again applying Euclid's division lemma, we get
36 = (9 x 4) + 0
Here, remainder is zero and last divisor is 9.
\(\therefore \)  HCF of 657 and 963 = 9   ... (ii)
From Eqs.(i) and Eqs. (ii),
9 = 657 x  + 963 (- 15)
\(\Rightarrow \)  657x = 9 + 963 x 15 = 9 + 14445
\(\Rightarrow \)  657x = 14454  \(\Rightarrow \)  \(x=\frac { 14454 }{ 657 } =22\)

We have, (7 x 11 x 13) + 13 = 1001 + 13 = 1014
\(\Rightarrow \) 1014 = 2 x 3 x 13 x 13
We know that, a number is called composite number, if it has atleast one factor other than 1 and the number itself. Here, 1014 is the product of more than two prime numbers, i.e. 2, 3 and 13.
So, it is a composite number.
Now, (7 x 6 x 5 x 4 x 3 x 2 x 1) + 5 = 5045
\(\Rightarrow \) 5045 = 5 x 1009
Thus, it is the product of prime factors 5 and 1009.
Hence, it is also a composite number.

Consider an arbitrary positive integer y. Then, by taking y as dividend and 3 as divisor, we can write y as
y = 3q + r, where \(0\le r<3\)
[by Euclid's division lemma]
i.e. r = 0, 1, 2
Now, if r = 0, then y = 3q
if r = 1, then y = 3q + 1
and if r = 2, then y = 3q + 2
Thus, the positive integer y is of the form 3q, 3q + 1 or 3q + 2.
Consider, y = 3q and squaring on both sides, we get
y² = (3q)²  \(\Rightarrow \)  y² + 9q² = 3(3q)² = 3m .... (i)
[taking 3q² = m, where m is some integer]
Consider, y = 3q + 1 and squaring on both sides, we get
y² = (3q + 1)²  
\(\Rightarrow \)  y² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1   .... (ii)
[taking 3q² + 2q = m, where m is some integer]
Consider, y = 3q + 2 and squaring on both sides, we get
y² = (3q + 2)²  \(\Rightarrow \)  y² = 9q² + 12q + 4
\(\Rightarrow \)  y² = (9q² + 12q + 3) + 1 = 3(3q² + 4q + 1) + 1
\(\Rightarrow \)  y² = 3m + 1  .... (iii)
From Eqs. (i), (ii) and (iii), we get y² = 3m or 3m + 1
Thus, the square of any positive integer can be either of the form 3m or 3m + 1, for some integer m.

Given, HCF = 113, LCM = 56952 and one number = 904
Let another number = x
\(\therefore \quad HCF=\frac { x\times 904 }{ LCM } \)
\(\Rightarrow \quad 113=\frac { x\times 904 }{ 56952 } \)
\(\Rightarrow \quad x=\frac { 113\times 56952 }{ 904 } =7119\)

 P = ab² = a x b x b and q = a³b  = a x a x a x b
LCM (p, q) = a³ b²

Let a be any positive integer.
We know that, any odd positive integer is of the form 2q + 1, where q is a whole number.
\(\therefore \)  a = 2q + 1
\(\Rightarrow \)  a² = (2q + 1)²           [squaring both sides]
\(\Rightarrow \)   a² = 4q(q + 1) + 1            ...(i)
Note that q(q + 1) is either '0' or even, for any whole number q.
So, let q(q + 1) = 2m where m is a whole number.
From Eq.(i), we get  a² = 4(2m) + 1 = 8m + 1

True, because the product of any two consecutive numbers, say n(n + 1) will always be even as one out of n or (n+1) must be even.

\(\frac { 3 }{ 8 } =\frac { 3 }{ { 2 }^{ 3 } } \)
 \(=\frac { { 3\times 5 }^{ 3 } }{ { 2 }^{ 3 }\times { 5 }^{ 3 } } \)
 \(=\frac { 375 }{ 10^{ 3 } } =\frac { 375 }{ 1000 } \)
 \(=0.375\)

The decimal expansion of a rational number terminates, if the denominator of rational no p/q, when p and q are co-primes and q can be expressed as 2^m5ⁿ  where In and n are non-negative integers.
e.g. \(\frac { 3 }{ 10 } =\frac { 3 }{ 2^{ 1 }\times 5^{ 1 } } =0.3\)

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

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

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

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

HCF of 3³ x 5 and 3² x 5²
= 3² x 5
= 9 x 5
= 45

The smallest prime number is 2 and the smallest composite number is 2² Hence, required HCF (2²,2) = 2.

The given number ends in 5. Hence it is a multiple of 5. Therefore it is a composite number

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

36

\(\because \) 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