MHB -c5.LCM and Prime Factorization of A,B,C

[karush]

Build the least common multiple of A, B, and C
Then write the prime factorization of the least common multiple of A, B, and C.
$A = 2 \cdot 3^2 \cdot 7 \cdot 13 \cdot 23^8$
$B = 2 3^5 \cdot 5^9 \cdot 13$
$C = 2 \cdot 5 \cdot 11^8 \cdot 13^3$
$\boxed{?}$

ok this only has a single answer

definition: Prime factorization is a way of expressing a number as a product of its prime factors.
A prime number is a number that has exactly two factors, 1 and the number itself.

so is our first step $A\cdot B \cdot C$

 

[HOI]

No, of course not. ABC would be a "common multiple" of A, B, and C but the "LEAST common multiple" only if all three numbers are relatively prime which is not the case here. I see that A and C have a factor of 2 so the LCM will have a factor of 2. A has a factor of [math]3^2[/math] so the LCM will have [math]3^2[/math] as factor. B has a factor of [math]5^9[/math] and C has a factor of 5 so the LCM will have [math]5^9[/math] as a factor. B has 7 as a factor so the LCM has 7 as a factor. C has a factor of 11 so the LCM will have 11 as a factor. A and B have 13 as a factor and C has [math]13^3[/math] as a factor so the LCM will have [math]13^3[/math] as a factor. A has [math]23^8[/math] as a factor and [math]23^5[/math] so the LCM has [math]23^8[/math] as a factor.

The least common multiple of A, B, and C is [math]2(3^2)(5^9)(7)(11)(13^3)(23^8)[/math].

 

[karush]

sorry I just noticed that $B=2\cdot 3^5\cdot 5^9 \cdot 13$
probably will change every thing
I should just OP the screenshots

Spoiler

 

[HOI]

You still use the same principal:
for the greatest common factor of a set of numbers, take the product of all primes to the smallest power in the given set and for the least common multiple take the product of all primes to the highest power.