MHB What Is the Correct Method to Find the LCM of 12 and 56?

  • Thread starter Thread starter Casio1
  • Start date Start date
  • Tags Tags
    Multiple
AI Thread Summary
The correct method to find the least common multiple (LCM) of 12 and 56 involves prime factorization. The prime factorization of 12 is \(2^2 \cdot 3\) and for 56, it is \(2^3 \cdot 7\). The LCM is determined by taking the highest power of each prime factor, resulting in \(LCM(12, 56) = 2^3 \cdot 3 \cdot 7 = 168\). It is important to continue factoring until all factors are prime, and there are alternative methods to find the LCM, such as using the greatest common divisor (GCD). Understanding these methods ensures accurate calculations of the LCM.
Casio1
Messages
86
Reaction score
0
I did actually think I had this off to a T sort of speak, but seems I have run into a problem and am now unsure?

Find the LCM of 12 and 56.

This is how I do them but if incorrect I would appreciate the correct notation being pointed out to me;)

12/2 = 6 My prime factors of 12 are therefore 2, 6

56/2 = 28, and 28/2 = 14, and 14/2 = 7

my prime factors are; 2, 6, 7, 14

My common factors being 2

my lowest common multiple being = 1176?

12/4 = 3, therefore 3,4

56/14 = 4 therefore 4,14

common factors are 4

Lowest common multiple is 3 x 4 x 14 = 168
:D
so I am assuming there is a specific notation (method) to finding the LCM's without mistakes?
 
Mathematics news on Phys.org
Casio said:
my prime factors are; 2, 6, 7, 14

6 and 14 cannot be prime factors, for they are not prime!

\(12 = 2\cdot6 = 2\cdot2\cdot3 = 2^2\cdot3\)

\(56 = 2\cdot28 = 2\cdot2\cdot14=2\cdot2\cdot2\cdot7=2^3\cdot7\)

Now we take the highest power of each factor. Our least common multiple is therefore \(\mathrm{LCM}(12,56) = 2^3\cdot3\cdot7 = 168\).
 
Reckoner said:
6 and 14 cannot be prime factors, for they are not prime!

\(12 = 2\cdot6 = 2\cdot2\cdot3 = 2^2\cdot3\)

\(56 = 2\cdot28 = 2\cdot2\cdot14=2\cdot2\cdot2\cdot7=2^3\cdot7\)

Now we take the highest power of each factor. Our least common multiple is therefore \(\mathrm{LCM}(12,56) = 2^3\cdot3\cdot7 = 168\).

Thanks for point that out to me I completely missed that point, i.e. 14 is not a prime.

So should I take it then that the correct way to prime factorise intergers is to divide them always by prime numbers?

90 / 2 = 45, then 45 / 3 = 15, then 15 / 5 = 3
 
Casio said:
So should I take it then that the correct way to prime factorise intergers is to divide them always by prime numbers?

90 / 2 = 45, then 45 / 3 = 15, then 15 / 5 = 3
You can factor the integers however you like. The key is that you don't stop factoring until all the factors are prime. When you get to that point, you will have found the prime factorization of the integer, which is what we want.

So we could factor 90 as \(90 = 9\cdot10\), but we have to keep going, because 9 and 10 are not prime:

\(90 = 9\cdot10 = (3\cdot3)\cdot(2\cdot5) = 2\cdot3^2\cdot5\).

We stop here because 2, 3, and 5 are all prime.Of course, there are other methods for calculating the least common multiple that do not require factorization. If you only have two numbers, one way to find their LCM is to divide the product of the numbers by their greatest common divisor:

\[\mathrm{LCM}(a, b) = \frac{ab}{\mathrm{GCD}(a, b)}.\]

Another quick algorithm that's easy to do mentally is to take successive multiples of the bigger number until you find one that is divisible by the other number(s). For example, to find the LCM of 12 and 56, we would take multiples of 56:

\(56\cdot1 = 56\), which is not divisible by 12
\(56\cdot2 = 112\), which is not divisible by 12
\(56\cdot3 = 168\), which is divisible by 12. 168 is our least common multiple.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top