Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.
Find the number of values of k in \(12^{12}\) the lcm of the positive integers \(6^{6}\), \(8^{8}\) and k.
Lcm
Algebra
Integers
Answer: is 25.
AIME I, 1998, Question 1
Elementary Number Theory by Sierpinsky
here \(k=2^{a}3^{b}\) for integers a and b
\(6^{6}=2^{6}3^{6}\)
\(8^{8}=2^{24}\)
\(12^{12}=2^{24}3^{12}\)
lcm\((6^{6},8^{8})\)=\(2^{24}3^{6}\)
\(12^{12}=2^{24}3^{12}\)=lcm of \((6^{6},8^{6})\) and k
=\((2^{24}3^{6},2^{a}3^{b})\)
=\(2^{max(24,a)}3^{max(6,b)}\)
\(\Rightarrow b=12, 0 \leq a \leq 24\)
\(\Rightarrow\) number of values of k=25.