Common Core 6th grade math problem Here is the math problem:
The violin part repeats every 3 beats, the cello part repeats every 12 beats, the bass part repeats every 4 beats and the viola part repeats every 9 beats.  What is the shortest number of beats that the movement can last if all the instruments must begin and end in unison?
Thank you
 A: Since $4=2^2$, $12=2^2\cdot 3$ and $9=3^2$, they all divide $36=2^2\cdot 3^2$ which is also known as $36=\operatorname{lcm}(3,4,12,9)$. So $36$ beats is the minimal number of beats before they have all finished their respective sequences.
This is how it is done if you are used to easily find the prime factorizations involved.

A different approach would be to write out multiplication tables for $3,4,12$ and $9$ respectively and find that least common multiple manually that way. You will recognize $36$ as the lowest number occurring in all tables.
A: I’m going to suggest how one might attack such a problem cold, without starting from any special insight.
It can be solved by systematic experimentation: start counting beats (on paper) and mark the beats on which each instrument can end. You’re looking for the first point at which all four instruments can end. The result might look like this:
$$\begin{array}{c|cccc}
\text{Beat}&\text{Violin}&\text{Bass}&\text{Viola}&\text{Cello}\\ \hline
1\\
2\\
3&\bullet\\
4&&\bullet\\
5\\
6&\bullet\\
7\\
8&&\bullet\\
9&\bullet&&\bullet\\
10\\
11\\
12&\bullet&\bullet&&\bullet\\
13\\
14\\
15&\bullet\\
16&&\bullet\\
17\\
18&\bullet&&\bullet\\
19\\
20&&\bullet\\
21&\bullet\\
22\\
23\\
24&\bullet&\bullet&&\bullet\\
25\\
26\\
27&\bullet&&\bullet\\
28&&\bullet\\
29\\
30&\bullet\\
31\\
32&&\bullet\\
33&\bullet\\
34\\
35\\
36&\bullet&\bullet&\bullet&\bullet
\end{array}$$
In the course of making such a table one is likely to notice that whenever the viola can end, so can the violin; this is of course because $9$ is a multiple of $3$. This means that we don’t really need to keep track of the violin: if the viola can end, so can the violin. At that point one might realize that since $12$ is a multiple of $4$, the bass can always end whenever the cello can, so there’s no real reason to keep track of the bass, either. And at that point the penny might drop: we need the number of beats to be a multiple of $9$, so that the viola can end, and a multiple of $12$, so that the cello can end, so we’re looking for the least common multiple of $9$ and $12$ (which will automatically also be a multiple of $3$ and $4$).
A: Since we want all instruments to begin and end in unison AND each instrument must must complete at least one part, start with the instrument that has the most beats in their part, or the cello part with 12 beats.  Divide the number of beats for each instrument to solve for the number of repeats.  Multiply the cello beats by 2, 3, etc. The answer is the number of beats the cello performs when all instrument parts are evenly divisible into the cello part.
cello    violin     bass        viola 

 12      12/3 = 4   12/4 = 3   12/9 = 1 1/3  (incorrect)
 24      24/3 = 8   24/4 = 6   24/9 = 2 2/3  (incorrect)
 36      36/3 = 12  36/4 = 9   36/9 = 4      (Correct!)

The answer is 36.
