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Question Number 201848 by cortano12 last updated on 14/Dec/23

	Answered by mr W last updated on 14/Dec/23

	$1820 = 2^{2} \times 5 \times 7 \times 13$ $n = (2 + 1) (1 + 1) (1 + 1) (1 + 1) = 24$
	Commented by cortano12 last updated on 14/Dec/23

	$how get 2 + 1 ?$ $if 2^{3} \times 3 \times 7 \times 17$ $then n = (3 + 1) (1 + 1) (1 + 1) (1 + 1)$ $n = 4 \times 2 \times 2 \times 2 = 32 sir ?$
	Commented by mr W last updated on 14/Dec/23

	${d o n}^{'} t y o u a g r e e t h a t$ $1820 = 2^{2} \times 5 \times 7 \times 13 ?$
	Commented by cortano12 last updated on 14/Dec/23

	$yes . but if 2856 = 2^{3} \times 3 \times 7 \times 17$ $then all positive factor is 32 ?$
	Commented by cortano12 last updated on 14/Dec/23

	$what this formula ?$
	Commented by mr W last updated on 14/Dec/23

	$i f N = p_{1}^{n_{1}} p_{2}^{n_{2}} p_{3}^{n_{3}} . . . p_{k}^{n_{k}}$ $w i t h p_{i} = p r i m e n u m b e r, t h e n t h e$ $n u m b e r o f d i v i s o r s o f N i s$ $(n_{1} + 1) (n_{2} + 1) (n_{3} + 1) . . . (n_{k} + 1) .$ $t h i s i s c l e a r . a d i v i s o r (f a c t o r) o f$ $N c a n b e e x p r e s s e d a s$ $p_{1}^{m_{1}} p_{2}^{m_{2}} p_{3}^{m_{3}} . . . p_{k}^{m_{k}}$ $w i t h 0 ⩽ m_{i} ⩽ n_{i} .$ $s o t h e r e a r e t o t a l l y$ $(n_{1} + 1) (n_{2} + 1) (n_{3} + 1) . . . (n_{k} + 1)$ $p o s s i b i l i t i e s .$
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