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How can I evaluate L.C.M for the three numbers $\;\;2,\;2.5,\;3\;\;?$

EDIT: this is Ti89 calculator; and the answer is $15$ , why could that be possible? enter image description here

user373141
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  • @Mr. Xcoder what do you think is the right answer $15$ or $30$ ? – user373141 Aug 25 '17 at 20:25
  • @amWhy Can you please delete your comment now that I've deleted the answer (or at least remove the ping)? – Mr. Xcoder Aug 25 '17 at 20:39
  • @amWhy can you have another look sir ? – user373141 Aug 25 '17 at 20:39
  • @Mr. Xcoder can you have another look sir? – user373141 Aug 25 '17 at 20:41
  • @user373141 In my *humble* opinion the answer is $30$. Probably that program doesn't work for rationals, or I am just dumb. – Mr. Xcoder Aug 25 '17 at 20:41
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    No need to delete your answer @Mr.Xcoder. It is correct. $2$ does not divide 15, so 15 is not correct. Rather, $30$ is the least number divisible by all three numbers. – amWhy Aug 25 '17 at 20:45
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    The least common multiple of $2, 2.5$ is not 5; clearly 5 is not a multiple of 2, but rather, 10 is the lcm because both $2\mid 10, ;2.5 \mid 10.$ now what is $\text{lcm }(10, 3)?$ – amWhy Aug 25 '17 at 20:51
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    The calculator is incorrect. I imagine it figures that if d = gcd(a,b,c) then lcm(a,b,c) = (a/d x b/d x c/d )x d. So gcd (2,2.5, 3) = 1 and so lcm(2,2.5,3) = 2x2.5x3 = 15. This is incorrect. lcm(a,b,c) = (a/d x b/d x c/d)x d does not work if the a,b,c, are not integers. – fleablood Aug 25 '17 at 21:04
  • @amWhy Reposted. – Mr. Xcoder Aug 26 '17 at 09:54

2 Answers2

2

You can rewrite them as fractions:

  • $2=\frac{2}{1}$
  • $2.5=\frac{5}{2}$
  • $3=\frac{3}{1}$

For rational numbers, their LCM is the LCM of numerators when they're written as a fraction divided by the GCD of their denominators.

In this case, because $\gcd(1,2,1) = 1$:

$$\text{lcm}(2,2.5,3)=\text{lcm}(2,5,3)$$

Since they are all prime, the lcm is their product: $$\text{lcm}(2,5,3)=2\cdot 5\cdot 3 = 30$$

Hence, $\text{lcm}(2,2.5,3) = 30$.

Mr. Xcoder
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0

There is one potentially confusing aspect of least common multiples is that the terms don't have to be integers. This is actually okay, because the least common multiple only has to be multiplied by an integer. It doesn't have to be an integer itself.

Ex. The least common multiple of $\frac 12$ and $\frac 14$ is $\frac 12$ because $\frac 12 = 1*\frac 12 = 2*\frac 14$ and so $\frac 12$ is a common multiple and it is the least (positive) common multiple.

So we need to find $k = m*2 = n*\frac 52 = p*3$ where $m,n,p$ are positive integers.

So $\frac k3 = \frac {m*2}{3}=\frac {n*5}{2*3} = p$ must be an integer so $3$ must divide $m$ and $6$ must divide $n$ and $3$ must divide $k$.

Let $m = 3m'$ and $n = 6n'$ and $k = 3k'$ and we have

$k' = 2m' = 5n' = p$

So $\frac {k'}2 = m' =\frac {5n'}2= \frac p2$ is an integer so $2$ must divide $k'$,$n'$ and $p$.

Let $k' = 2\overline k$ ,$n' = 2\overline n$ and $p = 2p'$ and we have

$\overline k = m' = 5\overline n = p'$

$\frac {\overline k}5 = \frac {m'}5 = \overline n = \frac {p'}5$ is an integer.

So $5$ divides $\overline k$, $m'$ and $p'$.

Let $\overline k = 5v; m' = 5\overline m; p'=5\overline p$ and we have

$v = \overline m = \overline p$ is an integer.

The smallest such positive integer is $v = 1$.

So $k = 3*k' = 3*2\overline k = 3*2*5 v = 3*2*5*1 = 30$.

So the least common multiple is $30$ and indeed:

$30 = 15*2 = 12*2.5 = 10*3$.

Also note $\gcd(15,12, 10) = 1$. That is not a coincidence.

fleablood
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