Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.

## Lcm and Integer – AIME I, 1998

Find the number of values of k in \(12^{12}\) the lcm of the positive integers \(6^{6}\), \(8^{8}\) and k.

- is 107
- is 25
- is 840
- cannot be determined from the given information

**Key Concepts**

Lcm

Algebra

Integers

## Check the Answer

But try the problem first…

Answer: is 25.

AIME I, 1998, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

here \(k=2^{a}3^{b}\) for integers a and b

\(6^{6}=2^{6}3^{6}\)

\(8^{8}=2^{24}\)

\(12^{12}=2^{24}3^{12}\)

Second Hint

lcm\((6^{6},8^{8})\)=\(2^{24}3^{6}\)

\(12^{12}=2^{24}3^{12}\)=lcm of \((6^{6},8^{6})\) and k

=\((2^{24}3^{6},2^{a}3^{b})\)

=\(2^{max(24,a)}3^{max(6,b)}\)

Final Step

\(\Rightarrow b=12, 0 \leq a \leq 24\)

\(\Rightarrow\) number of values of k=25.

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s