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Question Number 158843 by mnjuly1970 last updated on 09/Nov/21

Commented by mr W last updated on 09/Nov/21

$(1)$ $n o t t r u e!$ $e x a m p l e : x = 4, y = 6$ $6 x + 11 y = 24 + 66 = 90 \equiv 0 m o d (30)$ $b u t x + 7 y = 4 + 42 = 46 ≢ 0 m o d (30)$
Commented by mnjuly1970 last updated on 09/Nov/21

$y e s y o u a r e r i g h t . . . t h x m r W$
	Answered by Rasheed.Sindhi last updated on 09/Nov/21

	$Q_{2} 5^{9} + 64 = {(5^{3})}^{3} + {(4)}^{3}$ $= (5^{3} + 4) ({(5^{3})}^{2} - (5^{3}) (4) + {(4)}^{2})$ $= (125 + 4) (15625 - 500 + 16)$ $= 129 \times 15141$ $∴ 5^{9} + 64 i s c o m p o s i t e .$
	Answered by Rasheed.Sindhi last updated on 09/Nov/21

	$Q_{3} N = n^{5} - 5 n^{3} + 4 n$ $= (n - 2) (n - 1) (n) (n + 1) (n + 2)$ $(i) ∙ n m o d u l o 3 :$ $n = 3 k : O b v i o u s l y 3 ∣ n \Rightarrow 3 ∣ N$ $n = 3 k + 1 : n + 2 = 3 k + 3 \Rightarrow 3 ∣ (n + 2) \Rightarrow 3 ∣ N$ $n = 3 k + 2 : n + 1 = 3 k + 3 \Rightarrow 3 ∣ (n + 1) \Rightarrow 3 ∣ N$ $∴ 3 ∣ N . . . . . . . . . . . . . . . . . . . . . . . . . . . A$ $(ii) ∙ n m o d u l o 2$ $n = 2 k : N = (2 k - 1) (2 k - 2) (2 k) (2 k + 1) (2 k + 2)$ $= 8 (4 k^{3} - 5 k^{2} + k) \Rightarrow 8 ∣ N$ $n = 2 k + 1 : N = (2 k) (2 k - 1) (2 k + 1) (2 k + 2) (2 k + 3)$ $= 32 k^{5} + 80 k^{4} + 40 k^{3} - 20 k^{2} - 12 k$ $= 8 (4 k^{5} + 10 k^{4} + 5 k^{3} - \frac{k (5 k + 3)}{2})$ $C a n b e p r o v e d t h a t k (5 k + 3) i s$ $d i v i s i b l e b y 2 i n b o t h c a s e s : (i) k \in E$ $o r (i i) k \in O$ $∴ 8 ∣ N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B$ $(iii) ∙ n m o d u l o 5 :$ $n = 5 k \Rightarrow 5 ∣ n \Rightarrow 5 ∣ N$ $n = 5 k + 1 \Rightarrow n - 1 = 5 k \Rightarrow 5 ∣ n - 1 \Rightarrow 5 ∣ N$ $n = 5 k + 2 \Rightarrow n - 2 = 5 k \Rightarrow 5 ∣ n - 2 \Rightarrow 5 ∣ N$ $n = 5 k + 3 \Rightarrow n + 2 = 5 (k + 1) \Rightarrow 5 ∣ n + 2 \Rightarrow 5 ∣ N$ $n = 5 k + 4 \Rightarrow n + 1 = 5 (k + 1) \Rightarrow 5 ∣ n + 1 \Rightarrow n ∣ N$ $∴ 5 ∣ N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C$ $A & B & C \Rightarrow 120 ∣ N$
	Commented by mnjuly1970 last updated on 09/Nov/21

	$t h a n k s a l o t s i r$
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