N2=24×36×78Total number of factors = (4+1)(6+1)(8+1)=5×7×9=315 1...............⏟156N................⏟156N2Hence answer is = 156-58 = 98Let 3α7β be such factor then α+β must be even ≠0N54=2×74 , number of factors = 2×5=10

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Let $N = 2^{2} \times 3^{3} \times 7^{4}$

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Number of factors of N which are of form 4k+1, $(k \in N)$ is 18

Number of factors of N which are of form 4k+1, $(k \in N)$ is 9

Number of factors of $N^{2}$ which are less than N and are not the factors of N is 98

Number of factors of N which are multiple of 54 is 10

answer is A, B, C.

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Detailed Solution

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

Total number of factors $= (4 + 1) (6 + 1) (8 + 1) = 5 \times 7 \times 9 = 315$
$1 \underset{156}{\underset{⏟}{...............}} N \underset{156}{\underset{⏟}{................}} N^{2}$
Hence answer is = 156-58 = 98
$L e t 3^{α} 7^{β}$ be such factor then $α + β$ must be even $\neq 0$
$\frac{N}{54} = 2 \times 7^{4}$ , number of factors = $2 \times 5 = 10$