Find-n-N-n-100-4-2-n-n-3-Note-that-S-denotes-the-cardinality-or-number-of-elements-of-a-set-S-

Question Number 112011 by Aina Samuel Temidayo last updated on 05/Sep/20

Find ∣ {n \in N ∣ n ⩽ 100, 4! ∣ 2^{n} - n^{3}} ∣ .

(Note that ∣ S ∣ denotes the cardinality

or number of elements of a set, S) .

Answered by Rasheed.Sindhi last updated on 07/Sep/20

n m a y b e o d d o r e v e n .

^{▸} n = 2 k + 1

2^{n} - n^{3} = 2^{2 k + 1} - {(2 k + 1)}^{3}

= 2 . 4^{k} - (8 k^{3} + 1 + 3 (2 k) (2 k + 1))

= 2 . 4^{k} - 8 k^{3} - 1 - 12 k^{2} - 6 k

= 2 . 4^{k} - 8 k^{3} - 12 k^{2} - 6 k - 1

= a n o d d n u m b e r (n o t d i v i s i b l e

b y 4)

^{▸} n = 2 k

2^{n} - n^{3} = 2^{2 k} - {(2 k)}^{3}

= 4^{k} - 8 k^{3}

= 4 (4^{k - 1} - 2 k^{3})

^{▸} H e n c e o n l y f o r n \in E

4 ∣ 2^{n} - n^{3}

^{▸} E v e n n u m b e r s u p t o 100 i s 50

T h i s p r o v e s t h a t

∣ {n \in N ∣ n ⩽ 100, 4! ∣ 2^{n} - n^{3}} ∣= 50

R e c a l l t h a t I c o n s i d e r d

N = {1, 2, 3, \dots}

(I f w e c o n s i d e r

N = {0, 1, 2, 3, \dots}

T h e n t h e a n s w e r i s 51)

Commented by Rasheed.Sindhi last updated on 07/Sep/20

S o r r y I {c a n}^{'} t d o i t . H o w e v e r

w i t h t h e h e l p o f c a l c u l a t o r, t h e

n u m b e r s a r e 4, 10, 16, \dots ., 6 n - 2

S o t h e n u m b e r o f t h e s e n u m b e r s

u p t o 100 i s 17

Commented by kaivan.ahmadi last updated on 05/Sep/20

h i d e a r r a s h e e d

t h e n u m b e r i n q u e s t i o n i s 4! = 4 \times 3 \times 2 \times 1

Commented by Rasheed.Sindhi last updated on 05/Sep/20

M i s r e a d i n g s i r m i s r e a d i n g!

I^{'} l l r e v i e w m y a n s w e r .

T h a n X d e a r k a i v a n!

Commented by Aina Samuel Temidayo last updated on 06/Sep/20

Please I^{'} m still waiting for the

correction . Thanks .

Answered by Rasheed.Sindhi last updated on 09/Sep/20

4! ∣ 2^{n} - n^{3}

24 ∣ 2^{n} - n^{3}

2^{n} \equiv n^{3} (m o d 24)

L e t {\begin{cases} 2^{n} \equiv u (m o d 24) \\ n^{3} \equiv v (m o d 24 \end{cases}

[\begin{matrix} n & \underset{(u)}{2^{n} m o d 24} & \underset{(v)}{n^{3} m o d 24} \\ 1 & 2 & 1 \\ 2 & 4 & 8 \\ 3 & 8 & 3 \\ 4 & 16 & 16 & u = v \\ 5 & 8 & 5 \\ 6 & 16 & 0 \\ 7 & 8 & 7 \\ 8 & 16 & 8 \\ 9 & 8 & 9 \\ 10 & 16 & 16 & u = v \\ 11 & 8 & 11 \\ 12 & 16 & 0 \\ 13 & 8 & 13 \\ 14 & 16 & 8 \\ 15 & 8 & 15 \\ 16 & 16 & 16 & u = v \\ 17 & 8 & 17 \\ 18 & 16 & 0 \\ 19 & 8 & 19 \\ 20 & 16 & 8 \\ 21 & 8 & 21 \\ 22 & 16 & 16 & u = v \\ 23 & 8 & 23 \\ 24 & 16 & 0 \end{matrix}]

V a l u e s o f n f o r w h i c h u = v o r

2^{n} \equiv n^{3} (m o d 24) o r i n o t h e r w o r d s

4! ∣ 2^{n} - n^{3} a r e :

4, 10, 16, \dots (6 n - 2) . .

T h i s i s a n A P u p t o 100 o f w h i c h

a r e 17 t e r m s .

∴ 4! ∣ 2^{n} - n^{3} h a s 17 s o l u t i o n s

f o r n \in N \land n ⩽ 100 .

Commented by Aina Samuel Temidayo last updated on 07/Sep/20

Can you please complete it ?

Commented by Rasheed.Sindhi last updated on 09/Sep/20

T h e A n s w e r i s n o w c o m p l e t e .

S o r r y f o r l a t e .

Commented by Aina Samuel Temidayo last updated on 09/Sep/20

Thanks .