Asked by Logan

How many even natural-number factors does $n = 2^2 \cdot 3^1 \cdot 7^2$ have?
Answered by Writeacher
Logan/Jake/Zach -- pick a name and keep it, please, so the math tutors can keep track of your progress.
Answered by oobleck
how many factors involving only 3 and 7 are there?
Those factors are all odd.
So double that number, since each must be multiplied by 2^1 or 2^2
Answered by Logan
I dont get that
Answered by oobleck
when you multiply two odd numbers, you get another odd number.
Your number is 2^2 * 3^1 * 7^2
All of the odd factors are
3
3*7
3*7*7
7*7
To get all of the even factors, multiply those by 2 and 4
Then the other even factors are just 2 and 4 by themselves.
No, I'm not going to do the counting for you.
Answered by dsakf spjgiokksdsiosji
just guessed it it was 12. thank me later aops students
Answered by Lemon_Bread
greengreengreen oh yellow!
Answered by e
e
Answered by e
12
Answered by Bot
The mathematical constant $e$ is a mathematical constant approximately equal to $2.71828$. It is the base of the natural logarithm, and it appears in many mathematical and scientific applications. Its decimal representation goes on infinitely without repeating, like $\pi$. The number $e$ arises naturally in calculus in the context of exponential functions and rates of change.
Answered by Bot
12 is a positive integer that can be expressed as the product of the primes $2^2\cdot3^1$. It has a total of 6 positive factors: 1, 2, 3, 4, 6, and 12. All of these factors are also divisors of 12.