Putting arbitrary values of a and b.

IIIustration 1:
Let a = 9 and b = 8
HCF (8 + 9, 9 - 8)
HCF (17, 1) = 1

IIIustration 2:
Let a = 23 and b =17
HCF (17 + 23, 23 -17)
HCF ( 40, 6) = 2

Hence, HCF (a + b, a - b) can either be 1 or 2.

LCM of 2, 4, 6, 8, 10, 12
=(2 x 2 x 3 x 2 x 5) = 120
? After every 2 min, they toll together.
? Number of times they toll together in one hour = (60/2) + 1 times
= 31 times

Let the numbers be 21a and 21b, where a and b are co-primes.
Then, 21a x 21b = (21 x 4641)
? ab = 221
Two co-primes with product 221 are 13 and 17.
? Required number = (21 x 13, 21 x 17)
= (273, 357)

Given that, (x - 6) is the HCF of x² - 2x - 24 and x² - kx - 6 i.e., (x - 6) is a factor of both expressions.
Let f (x₁) = x₁² - 2x₁ - 24
and f (x₂) = x₂² - kx₂ - 6
Now f(x₁) = f (x₂) at (x₁ = x₂ = 6)
? (6)² - 2(6) - 24 = (6)² - k(6) - 6 [by condition]
? 0 = 30 - 6k ? 6 k = 30
? k = 5

HCF (a /HCF (a, b) , b/HCF (a,b) it value is always equal to 1;
IIIustration 1 Let the two positive integers be a = 24 and b = 36.
? HCF (24/HCF (24,36) , 36/HCF (24,36)
? HCF (24/12 , 36/12)
? HCF (2, 3) = 1
IIIustration 2
Let the two positive integers be a = 13 and b = 17
? HCF (13 / HCF (13,17) , 17/HCF (13,17)
? HCF (13/1 , 17/1) = 1

Let p(N) = 8 (N⁵ - N³ + N)
= 4 x 2 x N (N⁴ - N² + 1)
and q(N) = 28 (N⁶ + 1)
= 7 x 4 [( N²)³ + (1)³]
= 4 x 7 (N² + 1) ( N⁴ - N² + 1)
? HCF of p(N) and q (N) = 4 (N⁴ - N² + 1)

Let f (x) = (x⁴ - y⁴)
= (x² - y²) (x² + y²)
= (x - y) (x + y) (x² + y²)
and g(x) = (x⁶ - y⁶)
= (x³)² - (y³)²
= (x³ + y³) (x³ - y³)
= (x + y) (x² - xy + y²) (x - y)(x² + xy + y²)
= (x - y) (x + y) (x² - xy + y²)
(x² + xy + y²)
? HCF of [f(x), g(x)] = (x - y) (x + y)
= x² - y²

Let f(x) = x³ - y² - 2x = x( x² - x- 2)
= x( x² - 2x + x - 2)
= x {x (x - 2) + 1 (x -2)}
= x (x +1) (x - 2)
and g(x) = x³ + x²
= x² (x +1) = x. x (x +1)
? LCM of [ f(x), g(x)] = x (x +1) . x . (x - 2)
= x²(x + 1)(x - 2)
= x²(x² - x - 2)
= x⁴ - x³ - 2x²

a² b⁴ + 2 a² b² = a² b² ( b² +2).......................(i)
and (ab)⁷ - 4 a² b⁹ = a⁷ b⁷ - 4 a² b⁹ = a² b²(a⁵ b⁵ - 4b⁷)................(ii)

from Eqs. (i) and (ii)
HCF = a² b²

Given that
x = 1305, y = 4665, z = 6905

Then ,
|x - y| = |1305 - 4665| = 3360
|y - z] = |4665 - 6905| = 2240
|z - x| =|6905 -1305| = 5600

? Required number = HCF of 3360, 2240 and 5600 = 1120

On dividing 1305 by 1120, remainder is 185.
On dividing 4665 by 1120, remainder is 185.
On dividing 6905 by 1120, remainder is 185.
? Common remainder = 185