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-

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} \dots . \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)!}

\dots \dots

- 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}} \dots

w i t h n_{1}, n_{4}, n_{5}, \dots ⩾ 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}, \dots

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 \dots = \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) \dots = \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}} \dots

w i t h n_{1}, n_{4}, n_{5}, \dots ⩾ 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}, \dots

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) \dots

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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