Use the factor tree to write the prime factorization of the number. 1872 HELP!!!?

Question

Use the factor tree to write the prime factorization of the number.
1872

Thanks

Answer 1

1872
2 x 936
2 x 2 x 468
2 x 2 x 2 x 234
2 x 2 x 2 x 2 x 117
2 x 2 x 2 x 2 x 3 x 39
2 x 2 x 2 x 2 x 3 x 3 x 13

Answer 2

First , you have to divide 1872 by 2. So your top row on your notebook paper has 1872. The second row of you note book paper has 2 and 936. ( because these are the two factors of 1872) the 2 is a prime factor, but 936 is not prime so you now have to divide 936, into ( there are lots of factors ,but lets do the 2's first) 2 and 468. So the third row is 2 and 468. Keep going down and factor out all the 2's , then start factoring out with a 3. Most people put a carat ^ under each number they divide

1872
^
2 936
^
2 468
^
2 234
^
A prettier way:

1872
^
8 234
^ ^
2 4 26 9
^ ^ ^
2 2 2 13 3 3
So there are four 2's two 3's and one 13
This way uses less paper, but the one on top is simpler

Answer 3

560 56 x 10 7 x 8 2 x 5 2 x 4 2 x 2 So the prima factorization is 7 x 2 x 2 x 2 x 2 x 5 or 7 x 2(to the fourth) x 5

Answer 4

1872
2 x 936
2 x 2 x 468
2 x 2 x 2 x 234
2 x 2 x 2 x 2 x 117
2 x 2 x 2 x 2 x 3 x 39
2 x 2 x 2 x 2 x 3 x 3 x 13

Answer 5

We can describe a recursive algorithm to perform such factorizations: given a number n

if n is prime, this is the factorization, so stop here.
if n is composite, divide n by the first prime p1. If it divides cleanly, recurse with the value n/p1. Add p1 to the list of factors obtained for n/p1 to get a factorization for n. If it does not divide cleanly, divide n by the next prime p2, and so on.
Note that we need to test only primes pi such that .

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Example
Suppose we wish to factorize the number 9438.
9438/2 = 4719 with a remainder of 0, so 2 is a factor of 9438. We repeat the algorithm with 4719.
4719/2 = 2359 with a remainder of 1, so 2 is NOT a factor of 4719. We try the next prime, 3.
4719/3 = 1573 with a remainder of 0, so 3 is a factor of 4719. We repeat the algorithm with 1573.
1573/3 = 524 with a remainder of 1, so 3 is NOT a factor of 1573. We try the next prime, 5.
1573/5 = 314 with a remainder of 3, so 5 is NOT a factor of 1573. We try the next prime, 7.
1573/7 = 224 with a remainder of 5, so 7 is NOT a factor of 1573. We try the next prime, 11.
1573/11 = 143 with a remainder of 0, so 11 is a factor of 1573. We repeat the algorithm with 143.
143/11 = 13 with a remainder of 0, so 11 is a factor of 143. We repeat the algorithm with 13.
13/11 = 1 with a remainder of 2, so 11 is NOT a factor of 13. We try the next prime, 13.
13/13 = 1 with a remainder of 0, so 13 is a factor of 13. We stop when we reached 1.
Thus working from top to bottom, we have 9438 = 2 × 3 × 11 × 11 × 13.

[edit]
Code
Here is some code in Python for finding the factors of numbers less than 2147483647: