Question 166199
1) How many divisors (factors) has {{{5^3}}}?
{{{5^3 = 125}}}
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!