an integer n between 1 and 98 , inclusive is to be chosen at random. what is the probability that n(n+1) will be divisible by 3


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Question Number 103606 by bemath last updated on 16/Jul/20

$a n i n t e g e r n b e t w e e n 1 a n d 98,$ $i n c l u s i v e i s t o b e c h o s e n a t$ $r a n d o m . w h a t i s t h e p r o b a b i l i t y$ $t h a t n (n + 1) w i l l b e d i v i s i b l e b y 3$
Commented by bemath last updated on 16/Jul/20

$t h a n k y o u a l l$
	Answered by OlafThorendsen last updated on 16/Jul/20

	$1 st case :$ $n = 3 p, p = 1 . . 32$ $32 possibilities$ $n (n + 1) can be divisible by 3$ $2 nd case :$ $n = 3 p + 1 then n + 1 = 3 p + 2$ $n (n + 1) = (3 p + 1) (3 p + 2)$ $n (n + 1) {can}^{'} t be divisible by 3$ $3 rd case :$ $n = 3 p + 2, p = 0 . . 32$ $33 possibilities$ $then n + 1 = 3 p + 3 = 3 (p + 1)$ $n (n + 1) = 3 (3 p + 2) (p + 1)$ $n (n + 1) can be divisible by 3$ $Total result :$ $32 + 0 + 33 = 65$ $probability = \frac{65}{98} \approx 66, 3 %$
	Answered by floor(10²Eta[1]) last updated on 16/Jul/20
	$for n(n+1) be divisible by 3, we have 2 cases: 1)n is a multiple of 3 2)n+1 is a multiple of 3 [1, 98]→98 terms 1)AP: (3, 6, 9, ..., 96)→{a_n =3n} 3n=96⇒n=32(the AP has 32 terms) so the probability that n is a multiple of 3 is: P_1 =((32)/(98))=((16)/(49)) 2)AP: (2, 5, 8, ..., 95)→32 terms probability of n+1 multiple of 3: P_2 =((32)/(98))=((16)/(49)) ⇒probability that n(n+1) is divisible by 3: P_1 +P_2 =((32)/(49))≈65%$
	$for n (n + 1) be divisible by 3, we have 2$ $cases :$ $1) n is a multiple of 3$ $2) n + 1 is a multiple of 3$ $[1, 98] \to 98 terms$ $1) AP : (3, 6, 9, . . ., 96) \to {a_{n} = 3 n}$ $3 n = 96 \Rightarrow n = 32 (the AP has 32 terms)$ $so the probability that n is a multiple of$ $3 is : P_{1} = \frac{32}{98} = \frac{16}{49}$ $2) AP : (2, 5, 8, . . ., 95) \to 32 terms$ $probability of n + 1 multiple of 3 :$ $P_{2} = \frac{32}{98} = \frac{16}{49}$ $\Rightarrow probability that n (n + 1) is divisible$ $by 3 :$ $P_{1} + P_{2} = \frac{32}{49} \approx 65 %$
	Commented by OlafThorendsen last updated on 16/Jul/20

	$Why n = 98 cannot be chosen in$ $2) ?$
	Commented by floor(10²Eta[1]) last updated on 16/Jul/20

	$yeah {you}^{'} re right i miss that : /$
	Answered by bobhans last updated on 16/Jul/20

	$n o t e t h a t n (n + 1) i s n o t d i v i s i b l e b y 3$ $p r e c i s e l y w h e n n \equiv 1 (m o d 3); t h a t i s n$ $h a s r e m a i n d e r 1 u p o n d i v i s i o n b y 3 . f r o m$ $1 t o 98 i n c l u s i v e t h e r e a r e 33 s u c h v a l u e s$ $o f n . h e n c e t h e d e s i r e d p r o b a b i l i t y e q u a l s$ $t o 1 - \frac{33}{98} = \frac{65}{98} .$ $[p (A) = 1 - p (A^{c})] ◻$
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