Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including 1 and itself. Find (n/(75))


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Question Number 11075 by Joel576 last updated on 10/Mar/17

$Let n be the smallest positive integer that is$ $a multiple of 75 and has exactly 75 positive$ $integral divisors, including 1 and itself .$ $Find \frac{n}{75}$
Commented by FilupS last updated on 11/Mar/17

$n = 75 k$ $n has 75 divisors$ $n = 1 \times 75 \times k$ $75 = 3 \times 5^{2}$ $∴ n = 1 \times 3 \times 5 \times 5 \times k$ $∴ k h a s 75 - 4 d i v i s o r s e x c l u d i n g 1 a n d k$ $let k = 2^{71}$ $n = 75 \times 2^{71}$ $n = 3 \times 5^{2} \times 2^{71}$ $∴ \frac{n}{75} = k = 2^{71}$
Commented by mrW1 last updated on 14/Mar/17

$T h e a n s w e r t o t h i s q u e s t i o n d e p e n t s$ $v e r y m u c h o n h o w t o u n d e r s t a n d t h e$ $c o n d i t i o n ‘ ‘ i t h a s e x a c t l y 75 {d i v i s o r s}^{″} .$ ${L e t}^{'} s l o o k a t a n e a s i e r c a s e t h a t i t h a s$ $e x a c t l y 5 d i v i s o r s . A c c o r d i n g t o y o u r$ $c o n s i d e r a t i o n a b o v e t h e a n s w e r s h o u l d$ $b e$ $n = 1 \times 3 \times 5 \times 5 \times 2^{1} = 150$ $B u t t h i s n u m b e r h a s n o t e x a c t l y o n l y$ $5 d i v i s o r s, b u t m u c h m o r e . A l l f o l l o w i n g$ $n u m b e r s a r e d i v i s o r s o f i t :$ $1 / 3 / 5 / 2$ $15 / 6$ $25 / 10$ $75 / 30 / 50$ $150$ $- - - - -$ $t h a t m e a n s t h e n u m b e r n = 150 h a s$ $i n f a c t n o t e x a c t l y 5 d i v i s o r s, b u t$ $e x a c t l y 12 d i f f e r e n t d i v i s o r s .$ $A n o r t h e r p r o b l e m i s i f i d e n t i c a l d i v i s o r s$ $a r e a l l o w e d o r n o t . S i n c e 1 i s d e f i n i t e l y$ $a l l o w e d, a c c o r d i n g t o y o u r c o n s i d e r a t i o n$ $t h e a n s w e r s h o u l d b e$ $n = 1 \times 3 \times 5 \times 5 \times 1^{71} = 75$
Commented by FilupS last updated on 11/Mar/17

$a h h . I was thinking of factors!$
Commented by mrW1 last updated on 11/Mar/17

$I m a d e f o l l o w i n g c o n s i d e r a t i o n :$ $n = p_{1} \times p_{2} \times p_{3} . . . . \times p_{m} = \underset{k = 1}{\prod^{m}} p_{k}$ $p_{k} = t h e k - t h p r i m e n u m b e r .$ $n h a s f o l l o w i n g k i n d s o f d i v i s o r s :$ $- e a c h o f t h e m p r i m e n u m b e r s : C_{1}^{m} = \frac{m!}{1! (m - 1)!} = m$ $- p r o d u c t o f e v e r y 2 o f t h e m p r i m e n u m b e r s : C_{2}^{m} = \frac{m!}{2! (m - 2)!}$ $- p r o d u c t o f e v e r y 3 o f t h e m p r i m e n u m b e r s : C_{3}^{m} = \frac{m!}{3! (m - 3)!}$ $. . . . . .$ $- p r o d u c t o f a l l o f t h e m p r i m e n u m b e r s : C_{m}^{m} = \frac{m!}{m! (m - m)!} = 1$ $t h a t m e a n s t h e t o t a l n u m b e r o f d i v i s o r s i s$ $\underset{k = 1}{\sum^{m}} C_{k}^{m} = \underset{k = 1}{\sum^{m}} \frac{m!}{k! (m - k)!} = 2^{m} - 1$
Commented by FilupS last updated on 12/Mar/17

$You wrote :$ $‘ ‘ n = 1 \times 3 \times 5 \times 5 \times 1^{71} = 75^{″}$ $If we assume he meant 1 only once$ $(due to the fact that x = x {(1)}^{\infty}),$ $would the answer be :$ $n = 1 \times 3 \times 5 \times 5 \times 2^{71} = 75 \times 2^{71} ? ? ? ?$
Commented by mrW1 last updated on 12/Mar/17

$T h e d i f f i c u l t y l i e s o n h o w t o m a k e$ $t h e n u m b e r t o h a v e e x a c t l y 75 d i v i s o r s :$ $- t o g e t m o r e d i v i s o r s w e s h o u l d t a k e$ $m o r e d i f f e r e n t p r i m e n u m b e r s$ $- t o k e e p t h e n u m b e r s m a l l w e s h o u l d$ $u s e m o r e 2^{'} s .$ $T h e g e n e r a l f o r m u l a i s$ $n = 2^{n_{1}} \times 3^{n_{2}} \times 5^{n_{3}} \times 7^{n_{4}} \times 11^{n_{5}} . . .$ $w i t h n_{1}, n_{4}, n_{5}, . . . ⩾ 0, n_{2} ⩾ 1, n_{3} ⩾ 2$ $T h e q u e s t i o n i s t o d e t e r m i n e$ $n_{1}, n_{2}, n_{3}, . . .$ $t o k e e p n a s s m a l l a s p o s s i b l e a n d$ $t o h a v e e x a c t l y 75 d i v i s o r s .$ $F o r e x a m p l e :$ $n = 2^{11} \times 3 \times 5^{2} = 153600 h a s 72 d i v i s o r s$ $n = 2^{12} \times 3 \times 5^{2} = 307200 h a s 78 d i v i s o r s$ $n = 2^{3} \times 3^{2} \times 5^{2} \times 7 = 12600 h a s 72 d i v i s o r s$ $n = 2^{4} \times 3^{2} \times 5^{2} \times 7 = 25200 h a s 90 d i v i s o r s$ $n = 2^{2} \times 3 \times 5^{2} \times 7 \times 11 = 23100 h a s 72 d i v i s o r s$ $n = 2^{3} \times 3 \times 5^{2} \times 7 \times 11 = 46200 h a s 96 d i v i s o r s$ $W e s e e {i t}^{'} s n o t t h e r i g h t w a y t o u s e$ $o n l y f a c t o r s 2 . E . g . 2^{11} \times 3 \times 5^{2} i s m u c h$ $b i g g e r t h a n 2^{3} \times 3^{2} \times 5^{2} \times 7, b u t b o t h$ $h a v e 72 d i v i s o r s .$
	Answered by mrW1 last updated on 16/Mar/17

	$A f t e r s o m e a t t e m p t s (s e e c o m m e n t s a b o v e)$ $I t h i n k I g o t t h e s o l u t i o n t o t h i s q u e s t i o n .$ $G e n e r a l l y e v e r y p o s i t i v e i n t e g e r c a n$ $b e e x p r e s s e d a s t h e p r o d u c t o f a s e r i a l$ $o f p r i m e n u m b e r s :$ $n = p_{1}^{n_{1}} \times p_{2}^{n_{2}} \times p_{3}^{n_{3}} \times p_{4}^{n_{4}} \times p_{5}^{n_{5}} \times . . . = \underset{k = 1}{\prod^{m}} p_{k}^{n_{k}}$ $w h e r e p_{k} i s t h e k - t h p r i m e n u m b e r$ $a n d m i s t h e n u m b e r o f p r i m e n u m b e r s$ $t o b e n e e d e d, n_{k} ⩾ 0 .$ $T h e n u m b e r o f d i v i s o r s w h i c h n h a s i s$ $D = (n_{1} + 1) (n_{2} + 1) (n_{3} + 1) (n_{4} + 1) . . . = \underset{k = 1}{\prod^{m}} (n_{k} + 1)$ $S i n c e i n o u r c a s e n i s a m u l t i p l e o f 75$ $a n d 75 = 3 \times 5 \times 5, w e h a v e$ $n = 2^{n_{1}} \times 3^{n_{2}} \times 5^{n_{3}} \times 7^{n_{4}} \times 11^{n_{5}} . . .$ $w i t h n_{1}, n_{4}, n_{5}, . . . ⩾ 0, n_{2} ⩾ 1, n_{3} ⩾ 2$ $T h e t a s k i s n o w t o d e t e r m i n e n_{1}, n_{2}, n_{3}, . . .$ $s o t h a t D = 75 a n d n i s a s s m a l l a s p o s s i b l e .$ ${L e t}^{'} s c o n s i d e r a t f i r s t t h e n u m b e r o f$ $d i v i s o r s w h i c h s h o u l d b e e x a c t l y 75, i . e .$ $D = 75 = (n_{1} + 1) (n_{2} + 1) (n_{3} + 1) (n_{4} + 1) . . .$ $T h i s c a n o n l y b e o n e o f f o l l o w i n g 4 c a s e s :$ $C a s e (1) :$ $D = 75 = 3 \times 5 \times 5 = (2 + 1) (4 + 1) (4 + 1)$ $\Rightarrow w e n e e d 3 p r i m e n u m b e r s, e a c h 2, 4, 4 t i m e s$ $C a s e (2) :$ $D = 75 = 5 \times 15 = (4 + 1) (14 + 1)$ $\Rightarrow w e n e e d 2 p r i m e n u m b e r s, e a c h 4, 14 t i m e s$ $C a s e (3) :$ $D = 75 = 3 \times 25 = (2 + 1) (24 + 1)$ $\Rightarrow w e n e e d 2 p r i m e n u m b e r s, e a c h 2, 24 t i m e s$ $C a s e (4) :$ $D = 75 = (74 + 1)$ $\Rightarrow w e n e e d 1 p r i m e n u m b e r 74 t i m e s$ $C a s e (4) i s n o t s u i t a b l e, s i n c e w e n e e d a t l e a s t$ $2 p r i m e n u m b e r s : 3 \times 5^{2} .$ $T o k e e p n s m a l l w e s h o u l d u s e a s m a n y$ $s m a l l e r p r i m e n u m b e r s a s p o s s i b l e .$ $C a s e (1) :$ $3 p r i m e n u m b e r s (e a c h 2, 4, 4 t i m e s)$ $\Rightarrow n = 2^{4} \times 3^{4} \times 5^{2} = 32 400$ $C a s e (2) :$ $2 p r i m e n u m b e r s (e a c h 4, 14 t i m e s)$ $\Rightarrow n = 3^{14} \times 5^{4} = 2 989 355 625$ $C a s e (3) :$ $2 p r i m e n u m b e r s (e a c h 2, 24 t i m e s)$ $\Rightarrow n = 3^{24} \times 5^{2} = 7 060 738 412 025$ $C a s e (1) g i v e s t h e s m a l l e s t n w h i c h i s$ $n = 32400$ $= E N D =$
	Commented by Joel576 last updated on 13/Mar/17

	$H i, mrW 1$ $T h a n k y o u v e r y m u c h f o r y o u r g r e a t s o l u t i o n$ $E v e n m y t e a c h e r, h e {h a v e n}^{'} t f o u n d a n y s o l u t i o n s y e t .$ $G l a d y o u w a n t e d t o s p e n d y o u r t i m e t o f i n d t h e s o l u t i o n .$
	Commented by FilupS last updated on 13/Mar/17

	$I think mrW 1 is the smartest person on$ $here in the widest range of maths . Do you$ $goto collage / university ? I am completely$ $self study . Math is my hobby .$
	Commented by FilupS last updated on 13/Mar/17

	$My strongest topic is in sequences and in$ $number theory . I want to learn linear$ $algebra$
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