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3 years ago

Suppose LCM(a,b) = 20160, a=60. Find the smallest possible value of b.

Mathematics

2 answers:

Black_prince [1.1K]3 years ago

7 0

Answer:

Step-by-step explanation:

1260=22*32*5*7

60=22*3*5

Therefore we need the smallest number divisible by seven and nine, which is 7*9=63.

Mars2501 [29]3 years ago

5 0

Answer:

see below

Step-by-step explanation:

given:

LCM(a,b) = 20160, a=60

find:

smallest possible value of b

solution;

60, b = 20,160

b = 20,160 / 60

b = 336

60 x 336 = 20,160

= 2⁶ x 3² x 5 x 7

= (2² x 3 x 5) x (2⁴ x 3 x 7)

= (60) x (336)

LCM(60,336) = 1680

consider the 2² x 3 to get the prime number:

= 336 x 12

= 4032

therefore,

LCM(60,4032) = 20160

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