If $$N=3^{10}+3^{11}+3^{12}+3^{13},$$​then how many positive factors does N have ?

Correct option is D

Given:
​$$N = 3^{10} + 3^{11} + 3^{12} + 3^{13}$$​​
Formula Used:
Number of factors of p^a × q^b = (a+1)(b+1)
Solution:
First, factor out the lowest power of 3 from the given expression.
​$$N = 3^{10}(1 + 3^1 + 3^2 + 3^3)$$​​
​$$= 3^{10}(1 + 3 + 9 + 27)$$​​
​$$= 3^{10}(40)$$​​
Now, find the prime factorization of 40.
​$$40 = 2^3 × 5^1$$​​
Substitute this back into the expression for N.
​$$N = 2^3 × 3^{10} × 5^1$$​​
The number of positive factors is calculated by adding 1 to each exponent and multiplying them.
Total factors = $$(3 + 1)(10 + 1)(1 + 1)$$​​
= 4 × 11 × 2 = 88
Final Answer
So the correct answer is (d)