# algebra word problem

• Jan 14th 2008, 10:37 PM
sarahh
algebra word problem
A person takes 2 medications throughout the day and night. One medication is to be taken every 6 hours and the other is to be taken every 4 hours. If the person begins taking both medications at 7:00A.M. and takes both medications on schedule, how many hours later will it be when they next take both medications at the same time?
• Jan 14th 2008, 10:46 PM
Jhevon
Quote:

Originally Posted by sarahh
A person takes 2 medications throughout the day and night. One medication is to be taken every 6 hours and the other is to be taken every 4 hours. If the person begins taking both medications at 7:00A.M. and takes both medications on schedule, how many hours later will it be when they next take both medications at the same time?

the number of hours when the person will take both medications together again will be the LCM (the least common multiple) of 4 and 6. what is that? and why is that?
• Jan 14th 2008, 11:04 PM
sarahh
LCM is 12 but I'm not sure why we can say it's the LCM..
• Jan 14th 2008, 11:40 PM
Jhevon
Quote:

Originally Posted by sarahh
LCM is 12 but I'm not sure why we can say it's the LCM..

did you try drawing a chart to see a picture of what's going on?

Code:

 <---------><---------><---------->  <---------------><--------------->  |  |  |  |  |  |  |  |  |  |  |  |
the | represents the hours.

<---> represent the number of hours it can be divided into. there are 12 hours here. in the first line, you see three <---> symbols, that span 4 of the hours at a time. we see that it takes 3 of these symbols to span the 12 hours.

in the second line, there are 2 <--> symbols each spanning 6 hours at a time. we see that the arrow heads line up at the beginning when we have zero hours (when the person just takes the pills) and again at the end of 12 hours (the next time the person takes the two pills together).

so we had to find a number where both 6 and 4 divide into at the same time, because at these numbers, the arrow heads will line up as you see. since we wanted the very next time the arrow heads line up, we wanted the LCM
• Jan 15th 2008, 08:11 AM
topsquark
In general, one way to find the LCM of two numbers a and b, is to find the prime factorization of each number, then construct the LCM by adding each prime number of the greatest power in each factorization for all distinct primes in the two factorizations.

I'd better provide an example.

Find the LCM of 72 and 240.

The prime factorization of 72 is $2^3 \cdot 3^2$.

The prime factorization of 240 is $2^4 \cdot 3 \cdot 5$.

First look at the 2. There are 3 twos in 72 and 4 twos in 240. We pick the greatest number of these in the factorizations, which is 4. So we need a factor of $2^4$ in our LCM.

Now look at the 3. There are 2 threes in the factorization of 72 and 1 in the factorization of 240. So we need a factor of $3^2$ in the LCM.

Finally, look at the 5. There are no fives in the factorization of 72 and 1 in the factorization of 240. So we need a factor of $5^1$ in the LCM.

Thus the LCM of 72 and 240 is $2^4 \cdot 3^2 \cdot 5^1 = 720$.

In the case of the problem at hand we need to find the LCM of 4 and 6, note that $4 = 2^2$ and $6 = 2 \cdot 3$, so the LCM will be $2^2 \cdot 3 = 12$. (I readily admit that for this problem it's easier to see intuitively rather than employing the factorization method.)

-Dan