Question

Polpak, can you tell me how you got that answer?

Answer 1

Polpak, did u just delete my answer...

Answer 2

no.

Answer 3

Your answer to what?

Answer 4

The weird one about the ellipse center..

Answer 5

can you delete someone else's answer?

Answer 6

Nope. I was fiddling with wolframalpha

Answer 7

oh you found the lcm of 14, 40 and 42=840

Answer 8

I can yes.

Answer 9

mods can .... if need be

Answer 10

Strange...

Answer 11

ooh how did you arrange for that pray tell?

Answer 12

arrange to be a moderator? They asked. I said ok.

Answer 13

oh i see. got it. you have to be chosen. sigh...

Answer 14

i typed an answer the other day and posted it; but when i went back to see it in all its glory; it was gone ...

i cried for like 3 hours

Answer 15

Sorry devon. I'll explain using the 'chart' method.

Answer 16

Polpak did you get my reply

Answer 17

Apologies, Polpak, he repeated the question quickly...

Answer 18

ok thanks

Answer 19

@devon do you know how to find lcm of the following three numbers
\[2^2\times 3\times 5\]
\[2^3\times 5^5,\]
\[3\times 7\]

Answer 20

kind of

Answer 21

that is do you know how to find the lcm once the numbers are factored?

Answer 22

you need each factor you see to the highest power you see in any ONE factor

Answer 23

so in the example i sent , you will need a 2, a 3, a 5 and a 7

Answer 24

ok

Answer 25

2 must be raised to the power of 3
3 must be raised to the power of 1
5 must be raised to the power of 5
7 must be raised to the power of 1

Answer 26

\[\begin{array}{|c|cccc}\# & 2 & 3& 5 & 7\\14 & 1 & 0 & 0 & 1 \\ 40 & 3 & 0 & 1 & 0 \\ 42 & 1 & 1 & 0 & 1 \end{array}\]

Now we take the MAX from each of the factor columns and find the product of that many of each factor:
\[2^3 \cdot 3^1 \cdot 5^1 \cdot 7^1 = 840\]

Answer 27

so for the three numbers i sent the lcm would be
\[2^3\times 3\times 5^5\times 7\]

Answer 28

5^1, not 5^5

Answer 29

wow i have never seen that

Answer 30

@poplak i meant for the problem i made up

Answer 31

Oh I see. Sorry

Answer 32

Myn, yeah that's right for 2 numbers a and b, I'm not sure if it generalizes for more than 2.

Answer 33

u getting all this, Devon:-)

Answer 34

Have to think about that.

Answer 35

it doesn't polpak i just checked

Answer 36

Yeah I thought not.

Answer 37

Good to know though

Answer 38

\[lcm(a,b)=\frac{a*b}{\gcd(a,b)}\]
i wonder if we can find a formula for a case of three

Answer 39

it will be cool to see see a formula for n a case of n numbers

Answer 40

We want to find \[lcm(a,b,c)\]

So we can find the lcm of a,b
doing \[lcm(a,b)=\frac{a*b}{\gcd(a,b)}\]
and we can find the lcm of a,c
doing \[lcm(a,c)=\frac{a*c}{\gcd(a,c)}\]
finally we can find the lcm of b,c
doing \[lcm(b,c)=\frac{b*c}{\gcd(b,c)}\]

Answer 41

\begin{array}{|c|cccc} \# & 2 & 3& 5 & 7 \\ \hline 14 & 1 & 0 & 0 & 1 \\ 40 & 3 & 0 & 1 & 0 \\ 42 & 1 & 1 & 0 & 1 \\\hline \text{Max multiplicity} &3 & 1 & 1 & 1 \end{array}
\[\implies LCM(14,40,42) = 2^3 \times 3^1 \times 5^1 \times 7^1 = 840\]

Answer 42

Just having fun with tables ;p

Answer 43

\[lcm(14,40)=\frac{14*40}{\gcd(14,40)}, lcm(14,42)=\frac{14*42}{\gcd(14,42)}, lcm(40,42)=\frac{40*42}{\gcd(40,42)}\]

Answer 44

\[lcm(14,40)=\frac{560}{2}=280\]
\[lcm(14,42)=\frac{588}{14}=42\]
\[lcm(40,42)=\frac{1680}{4}=420\]

Answer 45

nope don't see how to make a formula for three