AP ICET 2019 Shift 2 Paper

Question : 83 of 200

Marks: +1, -0

The number of divisors excluding 1 and itself of the number 8625 is

Solution:

8625=5^{3} × 3^{1} × 23^{1}

The number of divisors excluding 1 and itself

=[(3+1)(1+1)(1+1)-2]=14

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