Question

if in a set of three natural numbers x, y, z; x is the h.c.f. of all the three numbers then

Answer 1

Keynote 1
If a = bq + r, then HCF(a,b) = HCF(b,r) ; a>b>0 [ EUCLIDEAN Algorithm ]

Keynote 2
HCF(a, b) = HCF(a + b, b) = HCF(a - b, b) ; a>b

Keynote 3
HCF(na, nb) = n * HCF(a,b) and LCM(na, nb) = n * LCM(a,b)

Keynote 4
LCM of first X natural numbers/LCM of first Y natural numbers = Product of primes between X and Y multiplied by additional powers of primes existing in LCM of first Y natural numbers ; X > Y
example LCM of (1,2,3,4,5,6,7,8) / LCM of (1,2,3,4,5) = 7 * 2 = 14
The 2 multiplied is for extra two from 8 that appears in numerator while 2^2 = 4 is already accounted for in denominator.
LCM(1,2,…,16)/LCM(1,2,3,4,5) = 7 * 11 * 13 * (2^2) * 3
2^2 is for two extra from 16 = 2^4 which were not present in highest power of 2 in LCM of 1 to 5
[4 = 2^2 was the highest power ]

Keynote 5
LCM (a, b) * HCF (a, b) = Product of a and b
If HCF of a set of numbers is 1 then their product is equal to their LCM...

Keynote 6
LCM of fractions = LCM of numerators/HCF of denominators AND
HCF of fractions = HCF of numerators/LCM of denominators.

Keynote 7
HCF (a + b, LCM (a, b)) = HCF (a, b)

Keynote 8
LCM >= Numbers >= HCF always
HCF is a factor of LCM always
If HCF = LCM => all numbers are equal

Keynote 9
n! is always divisible by LCM of first n natural numbers.
Furthermore,
LCM of first n natural numbers = n! for n = 1, 2, 3
LCM of first n natural numbers = n!/2 for n = 4,5
and can be extended similarly.............


Hope it helps u.......