LCM Question -- solving for possible m and n

Question

Could someone please help me with the following question:

Find positive integers m and n such that 9 divides m, 15 divides n, and LCM(m,n) = 330. If this is not possible, give a brief explanation as to why it is impossible.

I wrote that the LCM would be at most 135 since 9 x 15 is 135. Thus, it is not possible to satisfy all 3 conditions. Is this reasoning correct?

Edit: I see that my reasoning is wrong. How do I go about answering this question then?

The information I have: 9*k = m, 15*t = n, m * q = 330, and n * r = 330.

Playing around with your equations gives $9kq = 330$. Do you notice anything strange about that? — paw88789, Oct 14 '14 at 17:22
Yes! 330 is not divisible by 9! In general, if k divides m, is it true that k must divide every multiple of m? — mylasthope, Oct 14 '14 at 17:43
It is true that if $k$ divides $m$ ($m=kq$), then $k$ divides every multiple of $m$ ($ma=kaq$). — paw88789, Oct 14 '14 at 17:56

score 1 · Answer 1 · answered Oct 14 '14 at 05:18

1

Your argument is wrong. If $m=9$ and $n=60$, then $\mathrm{lcm}(m,n)=180$. So it is not "at most $135$."

answered Oct 14 '14 at 05:18

Thomas Andrews

177,126

score 0 · Answer 2 · answered Oct 14 '14 at 05:18

0

You might have $m > 9$ or $n > 15$; equality is not required.

Note that, since $330$ is even, at least one of $m$ and $n$ must also be even.

This should be enough to get you started.

answered Oct 14 '14 at 05:18

marty cohen

107,799

Ok. I think i have it. Please let me know if this is correct. Since we know 9 divides m and 15 divides n, 9 and 15 should divide the lcm(m,n). However, 9 does not divide 330. Thus, m does not exist. M = 9 * k. 330 = m * q. 330 = 9 * k * q. – mylasthope Oct 14 '14 at 17:38

LCM Question -- solving for possible m and n

2 Answers2