Let d(n) denote the number of divisors of n including 1 and itself. Then d(225), d(1125) and d(640) are

Question

Let d(n) denote the number of divisors of n including 1 and itself. Then d(225), d(1125) and d(640) are (A) in AP (B) in HP (C) in GP (D) consecutive integers

Tardigrade Admin · Accepted Answer

Here, 225=32 × 52 ⇒ d(225)=3 × 3=9 1125=32 × 53 ⇒ d(1125)=3 × 4=12 and 640=27 × 5 ⇒ d(640)=8 × 2=16 Hence, 9,12 and 16 are in GP.

Q. Let $d (n)$ denote the number of divisors of $n$ including $1$ and itself. Then $d (225), d (1125)$ and $d (640)$ are

in AP

in HP

in GP

consecutive integers

Here, $225 = 3^{2} \times 5^{2}$
$\Rightarrow d (225) = 3 \times 3 = 9$
$1125 = 3^{2} \times 5^{3}$
$\Rightarrow d (1125) = 3 \times 4 = 12$
and $640 = 2^{7} \times 5$
$\Rightarrow d (640) = 8 \times 2 = 16$
Hence, 9,12 and 16 are in GP.

Q. Let d(n) denote the number of divisors of n including 1 and itself. Then d(225),d(1125) and d(640) are