Find 2digit no that has Hcf 24and lcm is 144
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we are Given lcm.144 and hcf is 24
as we know that
Let the numners that we have the two numbers e and f respectively.
if we can multiply lcm with hcf .it will be equal to product of the two number .
so here we can find lcm,
so here we can say that lcm depends upon the (e,f) .now we have
(e,f) and hcf are 3456 and 24 so own.We can say that (e, f) = (24, 144) ; (48, 72) ; (72,
48) ; (144. 24)
The one of the two digit numbers are 24 and , 48 or 72 whose the lcm. is 144 and hcf
is 24
Answered by
1
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
(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....
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
(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
==================================================================
Hope this will help you....
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