Asked by picklepeter
Suppose LCM(a,b) = 20160, a=60. Find the smallest possible value of b.
Answers
Answered by
oobleck
20160 = 60*336 = 2^6 * 3^2 * 5 * 7 = (2^2 * 3 * 5) * (2^4 * 3 * 7)
Now, LCM(60,336) = 1680
You need to include the other 2^2 * 3 to get all the necessary prime powers.
336*12 = 4032
LCM(60,4032) = 20160
Now, LCM(60,336) = 1680
You need to include the other 2^2 * 3 to get all the necessary prime powers.
336*12 = 4032
LCM(60,4032) = 20160
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