cftan wrote:jieheng wrote:insanePaPa wrote:Can someone help my ds on this?

The LCM of 6, 12 & n is 660. Find all possible values of n.

6 = 2 * 3

12=2² * 3

660 = 2² * 3 * 5 * 11

n must include 5 * 11

n = 5 * 11 = 55

n = 2 * 5 * 11 = 110

n = 2² * 5 * 11 = 220

n = 3 * 5 * 11 = 165

n = 2 * 3 * 5 * 11 = 330

n = 2² * 3 * 5 * 11 = 660

Can someone help to explain. Thanks.

NOTE that the way to pick LCM Of several numbers is to take the Highest Number of each prime factor among the three, and then multiply them.

LCM, which is 660, contains two number of 2s, so all three numbers must contain <= two number of 2s, with at least one number having two number of 2s. Since 12 already got two number of 2s,

so n can have zero or one or two number of 2s

Same for 3, LCM contains one 3, so n can have zero or one 3.

LCM contains one 5, but the other two numbers does not have 5, so 5 must come from n. n must contain exactly one 5.

LCM contains one 11, the other two numbers does not have 11, so n must contain one 11.

Combining all these, n must be 5*11 or 5*11*2, 5*11*2*2, 5*11*3, 5*11*2*3, 5*11*2*2*3

This type of questions are quite common in RI and other IP school worksheets. Some are even tougher than this, where they can combine LCM and HCF

Like

"A,B,C are three different integers. A=18,B=60. The lowest common multiple of A,B,C is 540. The highest common factor of A,B,C is 3. Find the possible values of C."

Students need to have a good understanding of how to pick LCM and HCF from prime factorization.

Jerry Guo

Math Program Director -- MaxiMath

We specialize in teaching IP school Math, including Euler Program (RI)

http://www.maximath.sg