Let N = $2^{2} \times 3^{3} \times 7^{4}$
How many factors of $N^{2}$ which are less than N are not the factors of N?

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Solution

The correct option is C 98
Given $N = 2^{2} \times 3^{3} \times 7^{4}$
The number of factors of $N$ are $3 \times 4 \times 5 = 60$
$N^{2} = 2^{4} \times 3^{6} \times 7^{8}$
The number of factors of $N^{2}$ are $5 \times 7 \times 9 = 315$
Among $315 - 1 = 314$ factors , half are less than $N$ , which implies $157$ factors of $N^{2}$ are less than $N$
Number of Factors of $N$ which are less than $N$ are $60 - 1 = 59$
Therefore number of factors of $N^{2}$ which are less than $N$ and not the factors of $N$ are $157 - 59 = 98$

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