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Question Number 137269 by mohssinee last updated on 31/Mar/21

Commented by mohssinee last updated on 31/Mar/21

$n \in N^{*}$
	Answered by aleks041103 last updated on 31/Mar/21

	$C h e c k f o r n = 1$ $40^{1} 1! = 40 ∣ 120 = 5! = (5.1)!$ $S u p p o s e i t s t r u e f o r s o m e n > 1 .$ $T h e n :$ $(5 (n + 1))! = (5 n + 5)! =$ $= (5 n + 1) (5 n + 2) (5 n + 3) (5 n + 4) (5 n + 5) (5 n)! =$ $= 5 (5 n + 1) (5 n + 2) (5 n + 3) (5 n + 4) (n + 1) (5 n)!$ $S i n c e 40^{n} n! ∣ (5 n)! \Rightarrow (5 n)! = m . 40^{n} n!$ $\Rightarrow (5 (n + 1))! = 5 (5 n + 1) (5 n + 2) (5 n + 3) (5 n + 4) (n + 1)! 40^{n} m$ ${L e t}^{'} s l o o k a t t w o c a s e s$ $1) n \equiv 1 (m o d 2) a n d n \equiv 1 o r 3 (m o d 4)$ $5 n \equiv 1 (m o d 2) a n d 5 n \equiv 1 o r 3 (m o d 4)$ $\Rightarrow 5 n + 1 \equiv 2 o r 0 (m o d 4)$ $5 n + 3 \equiv 0 o r 2 (m o d 4)$ $T h e r e f o r e b o t h a r e d e v i s i b l e b y 2$ $a n d o n e i s d e v i s i b l e b y 4 .$ $\Rightarrow 5.2.4 = 40 ∣ 5 (5 n + 1) (5 n + 2) (5 n + 3) (5 n + 4)$ $\Rightarrow 5 (5 n + 1) (5 n + 2) (5 n + 3) (5 n + 4) = 40 k$ $2) n \equiv 0 o r 2 (m o d 4)$ $A n a l o g o u s l y$ $5 (5 n + 1) (5 n + 2) (5 n + 3) (5 n + 4) = 40 k$ $T h e n$ $(5 (n + 1))! = 40 . k . 40^{n} . (n + 1)! . m$ $\Rightarrow (m k) 40^{n + 1} (n + 1)! = (5 (n + 1))!$ $\Rightarrow 40^{n + 1} (n + 1)! ∣ (5 (n + 1))!$ $B y i n d u c t i o n, t h e s t a t e m e n t i s t r u e!$
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