Question

FIND 2 DIGIT NO WHOSE HCF IS 24 AND LCM IS 144

Answer 1

From the question we have given LCM which is 144

Also the HCF which is  24

First of all let assume these two numbers are the a and b here.
So in this case L.C.M.*H.C.F. will e equal to product of the two numbers
We can write which also as L.C.M.= (a*b)/H.C.F
So by putting Values of  LCM And HCF we will have 144 = ab/24
By further solving ab = 144*24
And SO, ab =  2⁷ * 3³ = 3456
Also  H.C.F(ab) = 2³ * 3 = 24 
So we know that L.C.M. depends on the  ab and also on H.C.F. and we know that H.C.F and ab. are 24 and 3456 respectively 
So we can write it as (a, b) = (24, 144);(48, 72);(72, 48);(144, 24)
SO in this case one of  two digit numbers may be these 24 or, 48 or 72 which have  L.C.M.= 144 and H.C.F= 24

Answer 2

Given :

L.C.M= 144

H.C.F. = 24

Let the two numbers be p & q

L.C.M × H.C.F = Product of the two numbers

L.C.M = Product of the two numbers/ H.C.F

L.C.M. = (p×q)/H.C.F.


144 = pq /24

pq= 144× 24

pq= (2×2×2×2×3×3)× (2×2×2×3)

pq= (2⁴×3²)×(2³×3¹)

pq= (2⁴× 2³)× (3²×3¹)

pq= 2⁷ × 3³

pq = 3456

H.C.F. (pq) = 24 = 2³ × 3¹

L.C.M. depends on pq & H.C.F.

L.C.M = pq & H.C.F= 3456 & 24

We can say that (p,q) = (24, 144) ; (48, 72) ; (72, 48) ; (144. 24)

Hence, the two digit numbers are= 24 or, 48 or 72 whose L.C.M = 144 & H.C.F = 24

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Hope this will help you....