Question 166199This question is from textbook
: I just want to double check to make sure I did this right.
a. How many divisors does 5³ have? What are they?
5³ = 125
Divisors would be 1 , 5, 125
b. The prime factorization of a real number x is 3² * 11³. How many divisors does x have? List the divisors? What is the value of x?
3² = 9
Divisors would be 1, 3, 9
11³ = 1,331
Divisors would be 1, 11, 121, 1331
x = 9 * 1331
x = 11,979
This question is from textbook
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! 1) How many divisors (factors) has  ?
The factors (divisors) of 125 are:
1, 5, 25, 125
2) The value of x = 11979. You are correct here.
The factors (divisors) of 11079 are:
1, 3, 9, 11, 33, 99, 121, 363, 1089, 1331, 3993, 11979
Notice that if you multiply the first and last of the factors (1 * 11979 = 11979), you will get the starting number (11979).
If you then multiply the second and next-to-last of the factors (3 * 3993 = 11979), again you get the starting number. Continue like this in pairs.
9*1331 = 11979
11+1089 = 11979
33*363 = 11979
99*121 = 11979
As you can see, there are many more than you thought!
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