1. ## lowest common multiple

The lowest common multiple of 2, x and 21 is 126. Find the smallest possible value of x

2. $126 = 2\cdot3\cdot21$.

3. that was what i did!! hi-5

but wait the answer is not 3 but 9.18 or 63

$126 = 2\cdot3\cdot21$.
The lowest common multiple of 2, 3 and 21 is 42.

You just need to find the smallest number that is a factor of 126, but which 2 OR 21 are not factors, and which that number is not a factor of 2 OR 21 . The smallest number that fits those criteria is 63.

5. I didn't quite understand that... perhaps a working would be nice

and with reference to "and which that number is not a factor of 2 OR 21" , isnt 21 a factor of 63?

6. Originally Posted by Punch
and with reference to "and which that number is not a factor of 2 OR 21" , isnt 21 a factor of 63?
Yes, but 2 isn't. Hence the OR.

I don't know if I could show any working, I really just worked it out by common sense and logic. Put it this way; it can't be 3 because the lowest common multiple of 2, 3 and 21 is 42, not 126. Maybe the easiest way to try and explain it is to divide the LCM by 2, which gives you 63. Since 2 doesn't go into 63, the next multiple of 63 which is divisible by both 2 and 21 is 126, so x = 63.

You could divide 126 by 3, which gives 42, but the LCM of 2, 21 and 42 is 42. Similarly you could keep dividing 126 by integers until you reach another number which isn't divisible by both 2 and 21, and that would be your x, but the only number which satisfies the criteria I gave in the previous post is 63 (well, so does 126 itself, but you are asked to find the smallest possible number for x).

I hope this makes it a little clearer. I'll hunt around for a formula or I might just have to make one up.

EDIT:

Here's a way to check that the number you've found is correct.

$lcm(a,b)=\frac{|a\cdot b|}{gcd(a,b)}$ where $gcd$ is the greatest common divisor of $a$ and $b$. Here, choose one of your given factors (2 or 21), and the factor you'd like to check (in this case, 63). It's best to test with both given factors, as we'll see.
So we get:
$lcm(2,63)=\frac{2\cdot63}{1}=2\cdot63=126$

However if you used 21 instead of 2, you would get:
$lcm(21,63)=\frac{21\cdot63}{21}=63$

So you see why it's important to test both factors.

This formula is originally designed to find the LCM, not to work the other way around, but it does provide a good way to test your answer.

Hope this helps.

7. well , nevertheless the smallest value of x is 9. LCM of 2,9,21 is 126

8. Originally Posted by luhter
well , nevertheless the smallest value of x is 9. LCM of 2,9,21 is 126
I agree, the correct answer is 9.

$126 = 2^1\cdot3^2\cdot7^1$

The relevant primes are 2, 3, and 7.

$2=2^1\cdot3^0\cdot7^0$

$21=2^0\cdot3^1\cdot7^1$

So we can write

$x=2^a\cdot3^b\cdot7^c$

and

$\text{lcm}(2,x,21)=2^{\max(1,0,a)}\cdot3^{\max(0,1 ,b)}\cdot7^{\max(0,1,c)}$

We need x = 9 in order to get max(0,1,b) = 2.