Question 541373
<pre>
30, 72, and N

There must exist positive integers A, B, and C such that 

30·72 = NA
  30N = 72B
  72N = 30C

Simplifying:

2160 =  NA
  5N = 12B
 12N =  5C  

Solving each for N:

N = {{{2160/A}}}
N = {{{12B/5}}}
N = {{{5C/12}}}

Set the last two right sides equal, since both are equal to N

{{{12B/5}}} = {{{5C/12}}}

144B = 25C

The smallest B and C can be are B = 25 and C = 144

Substituting in

N = {{{12B/5}}} = {{{12*25/5}}} = 60 
N = {{{5C/12}}} = {{{5*144/12}}} = 60

That works because

30, 72, and 60

30*72 = 2160 = 60*36
30*60 = 1800 = 73*25
60*72 = 4320 = 30*144

So the answer is N = 60

Edwin</pre>