counting factors of N

Given that

$N = 2^2 \times 3^3 \times 7^4$

$\Rightarrow N^2 = 2^4 \times 3^6 \times 7^8$

Number of positive factors of $N = (2+1)(3+1)(4+1) = 60$

Number of positive factors of $N^2 = (4+1)(6+1)(8+1) = 315$ which includes $N$ and half each of remaining as $< N$ and $> N$ .

Number of positive factors of $N^2$ which are $\leq N = (315 + 1) / 2 = 158$

Number of positive factors of $N^2$ which are less than $N$ ,are not factors of $N = 158 - 60 = \boxed {98}$

TOTAL FACTORS OF N^2=315 . FOR EVERY VALYE BETWEEN 1 AND N THERE EXIST ANOTHER VALUE BETWEEN N AND N^2 SUCH THAT THERE GEOMETRIC MEAN IS N. THUS IF WE REMOVE 1,N,N^2 FROM THE 315 WE WILL HAVE 1............N............N^2 ,,,,,,,,, 156 TERMS BETWEEN 1 AND N ,AND, N AND N^2 TOTAL FACTORS OF N=60 . REMOVING 1 AND N FROM FACTORS WE GET FACTORS OF N=58 .NOW THE ANSWER =156-58=98