Find-maximum-n-such-that-12-n-divides-100-

Question Number 57960 by mr W last updated on 15/Apr/19

F i n d m a x i m u m n s u c h t h a t 12^{n} d i v i d e s

100! .

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19

12 = 3 \times 2^{2}

h i g h e s t p o w e r o f 3 c o n t a i n e d i n 100! i s

I [\frac{100}{3}] + I [\frac{100}{3^{2}}] + I [\frac{100}{3^{3}}] + I [\frac{100}{3^{4}}]

= 33 + 11 + 3 + 1

= 48

h i g e s t p o w e r o f 2 c o n t a i n e d i n 100!

I [\frac{100}{2}] + I [\frac{100}{2^{2}}] + I [\frac{100}{2^{3}}] + [\frac{100}{2^{4}}] + I [\frac{100}{2^{5}}] + I [\frac{100}{2^{6}}]

= 50 + 25 + 12 + 6 + 3 + 1

= 97

s o h i g h e s t p o w e r o f 12 \to 12^{n}

t h e v a l u e o f n = 48 [c o m m o n m i n i m u m v a l u e

b e t w e e n h i g h e s t p o w e r o f f a c t o r 3 a n d 2]

s i r p l s c h e c k \dots

Commented by mr W last updated on 15/Apr/19

c o r r e c t, t h a n k y o u v e r y m u c h s i r!

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19

m o s t w e l c o m e s i r

Commented by mr W last updated on 15/Apr/19

t h e a n s w e r i s a n o t h e r i f t h e q u e s t i o n

i s n o t 100! b u t 99! . a m i r i g h t ?

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19

I [\frac{99}{3}] + I [\frac{99}{3^{2}}] + I [\frac{99}{3^{3}}] + I [\frac{99}{3^{4}}]

= 33 + 11 + 3 + 1 = 48

I [\frac{99}{2}] + I [\frac{99}{2^{2}}] + I [\frac{99}{2^{3}}] + I [\frac{99}{2^{4}}] + I [\frac{99}{2^{5}}] + I [\frac{99}{2^{6}}]

= 49 + 24 + 12 + 6 + 3 + 1

= 95

s o a n s i s 48 \dots s i r p l s c h d c k

Commented by mr W last updated on 15/Apr/19

12^{n} = 3^{n} \times 2^{2 n}

f o r p o w e r o f 3 : n ⩽ 48

f o r p o w e r o f 2 : 2 n ⩽ 95 \Rightarrow n ⩽ 47

\Rightarrow n_{m a x} = 47

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19

y e s s i r t h a n k y o u f o r y o u r t h o u g h t p r o c e s s

q u i t e d e e p \dots t h a n k s a g a i n \dots