Question 1095883
.
<pre>
The integer number 130 has this decomposition into the product of these prime numbers:

130 = {{{2}}}.{{{5}}}.{{{13}}} = {{{2^1}}}.{{{5^1}}}.{{{13^1}}}.


Note that the indexes are 1 for 2, 1 for 5 and 1 for 13.


Now add 1 to each of the three indexes:

1 + 1 = 2  (for 2);

1 + 1 = 2  (for 5);

1 + 1 = 2  (for 13).


Now the number of all divisors for 130 (including 1 and including 130 itself) is the product 2*2*2 = 8.


They are

1   = {{{2^0*5^0*13^0}}},

2   = {{{2^1*5^0*13^0}}},

5   = {{{2^0*5^1*13^0}}},

10  = {{{2^1*5^1*13^0}}},

13  = {{{2^0*5^0*13^1}}},

26  = {{{2^1*5^0*13^1}}},

65  = {{{2^0*5^1*13^1}}},

130 = {{{2^1*5^1*13^1}}}.
</pre>

The answer to your problem is the option a):  there are 8 divisors.